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1 | 2.5 mL of 5 2 â M weak monoacidic base ( K b â = 1 Ã 1 0 â 12 at 2 5 â C ) is titrated with 15 2 â M HCl in water at 2 5 â C . The concentration of H + at equivalence point is ( K w â = 1 Ã 1 0 â 14 at 2 5 â C ) | 2-5-ml-of-5-2-m-weak-monoacidic-base-k-b-1-1-0-12-at-2-5-c-is-titrated-with-59646 | <div class="question">$2.5 \mathrm{~mL}$ of $\frac{2}{5} \mathrm{M}$ weak monoacidic base $\left(K_b=1 \times 10^{-12}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$ is titrated with $\frac{2}{15}$ $\mathrm{M} \mathrm{HCl}$ in water at $25^{\circ} \mathrm{C}$. The concentration of $\mathrm{H}^{+}$at equivalence point is $\left(K_w=1 \times 10^{-14}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$</div> | ['Chemistry', 'Ionic Equilibrium', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$3.7 \times 10^{-13} \mathrm{M}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$3.2 \times 10^{-7} \mathrm{M}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$3.2 \times 10^{-2} \mathrm{M}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$2.7 \times 10^{-2} \mathrm{M}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$2.7 \times 10^{-2} \mathrm{M}$</span> </div> | <div class="solution">Weak monoacidic base, e.g. $\mathrm{BOH}$ is neutralised.<br/>$$<br/>\mathrm{BOH}+\mathrm{HCl} \longrightarrow \mathrm{BCl}+\mathrm{H}_2 \mathrm{O}<br/>$$<br/>At equivalence point all $\mathrm{BOH}$ gets converted into salt and remember! the concentration of $\mathrm{H}^{+}$(or $\mathrm{pH}$ of solution) is due to hydrolysis of resultant salt ( $\mathrm{BCl}$, cationic hydrolysis here)<br/>$$<br/>\underset{\mathrm{C}(1-h)}{\mathrm{B}^{+}}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\mathrm{Ch}}{\rightleftharpoons} \mathrm{OH}+\underset{\mathrm{Ch}}{\mathrm{H}^{+}}<br/>$$<br/>Volume of $\mathrm{HCl}$ used up,<br/>$$<br/>V_a=\frac{N_b V_b}{N_a}=\frac{2.5 \times 2 \times 15}{2 \times 5}=7.5 \mathrm{~mL}<br/>$$<br/>Concentration of salt,<br/>$$<br/>\begin{aligned}<br/>{[\mathrm{BCl}] } & =\frac{\text { Concentration of base }}{\text { Total volume }}=\frac{2 \times 25}{5(7.5+2.5)}=\frac{1}{10}=0.1 \\<br/>K_h & =\frac{C h^2}{1-h}=\frac{K_w}{K_b}<br/>\end{aligned}<br/>$$<br/>( $h$ should be estimated whether that can be neglected or not) on calculating $h=0.27$ (significant, not negligible)<br/>$$<br/>\left[\mathrm{H}^{+}\right]=C h=0.1 \times 0.27=2.7 \times 10^{-2} \mathrm{M}<br/>$$<br/>When $\left[\mathrm{H}^{+}\right]$is asked to calculate in connection with neutralisation, it<br/>Should be calculated:<br/>Before neutralisation Using Oslwald's dilution law<br/>During neutralisation $\quad$ Considering buffer solution<br/>Half neutralisation $\mathrm{pH}=\mathrm{p} K_a$ and $\mathrm{p} K_b$<br/>At the end of neutralisation $\quad$ Considering hydrolysis of salt</div> | MarksBatch0.db |
2 | 5.6 L of helium gas at STP is adiabatically compressed to 0.7 L . Taking the initial temperature to be T 1 â , the work done in the process is | 5-6-l-of-helium-gas-at-stp-is-adiabatically-compressed-to-0-7-l-taking-the-initial-temperature-to-45466 | <div class="question">$5.6 \mathrm{~L}$ of helium gas at STP is adiabatically compressed to $0.7 \mathrm{~L}$. Taking the initial temperature to be $T_1$, the work done in the process is</div> | ['Physics', 'Thermodynamics', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{9}{8} R T_1$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{3}{2} R T_1$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{15}{8} R T_1$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{9}{2} R T_1$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{9}{8} R T_1$<br/></span> </div> | <div class="solution">At STP, 22.4 L of any gas is 1 mole.<br/>$$<br/>\therefore \quad 5.6 \mathrm{~L}=\frac{5.6}{22.4}=\frac{1}{4} \text { moles }=n<br/>$$<br/><br/>In adiabatic process,<br/>$$<br/>\begin{array}{rlrl} <br/>& T V^{\gamma-1} & =\text { constant } \\<br/>& \therefore & T_2 V_2^{\gamma-1} & =T_1 V_1^{\gamma-1} \\<br/>& \text { or } & T_2 & =T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}<br/>\end{array}<br/>$$<br/><br/>$$<br/>\begin{aligned}<br/>& \gamma=\frac{C_p}{C_V}=\frac{5}{3} \text { for monoatomic He gas. } \\<br/>& \therefore \quad T_2=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=4 T_1 \\<br/>& \because \text { Further in adiabatic process, } \\<br/>& Q=0 \\<br/>& \therefore \quad W+\Delta U=0 \\<br/>& \text { or } \quad W=-\Delta U \\<br/>& =-n C_V \Delta T \\<br/>& =-n\left(\frac{R}{\gamma-1}\right)\left(T_2-T_1\right) \\<br/>& =-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)\left(4 T_1-T_1\right) \\<br/>& =-\frac{9}{8} R T_1 \\<br/>&<br/>\end{aligned}<br/>$$<br/><br/>$\therefore$ Correct option is (a).<br/>Analysis of Question<br/>(i) From calculation point of view question is moderately tough.<br/>(ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative.</div> | MarksBatch0.db |
3 | 29.2% ( w / w ) HCl stock solution has a density of 1.25 gmL â 1 . The molecular weight of HCl is 36.5 g mol â 1 . The volume ( mL ) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is : | 29-2-w-w-hcl-stock-solution-has-a-density-of-1-25-gml-1-the-molecular-weight-of-hcl-is-67767 | <div class="question">$29.2 \%(\mathrm{w} / \mathrm{w}) \mathrm{HCl}$ stock solution has a density of $1.25 \mathrm{gmL}^{-1}$. The molecular weight of $\mathrm{HCl}$ is $36.5 \mathrm{~g} \mathrm{~mol}^{-1}$. The volume $(\mathrm{mL})$ of stock solution required to prepare a 200 $\mathrm{mL}$ solution of $0.4 \mathrm{M} \mathrm{HCl}$ is :</div> | ['Chemistry', 'Some Basic Concepts of Chemistry', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">8</span> </div> | <div class="solution">Molarity of stock solution of $\mathrm{HCl}$
<br/> <br/>$=\frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}$
<br/> <br/>Let the volume of stock solution required $=V \mathrm{~mL}$
<br/> <br/>Thus, $V \times \frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}=200 \times 0.4$ $\Rightarrow V=8 \mathrm{~mL}$</div> | MarksBatch0.db |
4 | 75.2 g of C 6 â H 5 â OH (phenol) is dissolved in a solvent of K f â = 14 . If the depression in freezing point is 7 K , then find the % of phenol that dimerises. | 75-2-g-of-c-6-h-5-oh-phenol-is-dissolved-in-a-solvent-of-k-f-14-if-the-depression-64934 | <div class="question">$75.2 \mathrm{~g}$ of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$ (phenol) is dissolved in a solvent of $K_f=14$. If the depression in freezing point is $7 \mathrm{~K}$, then find the $\%$ of phenol that dimerises.</div> | ['Chemistry', 'Solutions', 'JEE Advanced', 'JEE Advanced 2006'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">35</span> </div> | <div class="solution">Phenol is dimerised as follows<br/>At $t=0$<br/>$$<br/>2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right)_2<br/>$$<br/>At equilibrium<br/>$$<br/>(1-\alpha) \quad \alpha / 2<br/>$$<br/>Total moles after dimerisation $=1-\alpha+\alpha / 2=1-\alpha / 2$<br/>$$<br/>\begin{aligned}<br/>\text { Van't Hoff factor (i) } & =\frac{\text { Normal molecular wt. of phenol }}{\text { Abnormal molecular wt. of phenol }}=\frac{94}{m_{a b}} \\<br/>m_{a b} & =\frac{1000 \times K_f \times w}{\Delta T_f \times W}<br/>\end{aligned}<br/>$$<br/>Given that<br/>$$<br/>\begin{aligned}<br/>& K_f=14 \\<br/>& w=72.5 \mathrm{gm} \\<br/>& W=1000 \mathrm{gm} \\<br/>& \therefore \quad m_{a b}=\frac{1000 \times 14 \times 72.5}{7 \times 1000}=145 \\<br/>&<br/>\end{aligned}<br/>$$<br/>$$<br/>\therefore \quad i=\frac{94}{145}<br/>$$<br/><br/>$$<br/>\text { (i) }=\frac{\text { No. of particles after dimerisation }}{\text { No. of particles before dimerisation }}=\frac{1-\alpha / 2}{1}<br/>$$<br/>From eq. (i) and (ii)<br/>$$<br/>\begin{aligned}<br/>1-\alpha / 2 & =\frac{94}{145}=0.65 \\<br/>\frac{\alpha}{2} & =1-0.65=0.35<br/>\end{aligned}<br/>$$<br/>Hence, $35 \%$ phenol is present in dimeric form.</div> | MarksBatch0.db |
5 | A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9 à 1 0 â 7 m 2 . If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s â 1 . If the Young's modulus of the material of the wire is n à 1 0 9 Nm â 2 , the value of n is | a-0-1-kg-mass-is-suspended-from-a-wire-of-negligible-mass-the-length-of-the-wire-is-1-m-and-its-cro-90065 | <div class="question">A $0.1 \mathrm{~kg}$ mass is suspended from a wire of negligible mass. The length of the wire is $1 \mathrm{~m}$ and its cross-sectional area is $4.9 \times 10^{-7} \mathrm{~m}^2$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140 \mathrm{rad} \mathrm{s}^{-1}$. If the Young's modulus of the material of the wire is $n \times 10^9 \mathrm{Nm}^{-2}$, the value of $n$ is</div> | ['Physics', 'Mechanical Properties of Solids', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">4</span> </div> | <div class="solution">$\omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{Y A}{l m}}=\sqrt{\frac{\left(n \times 10^9\right)\left(4.9 \times 10^{-7}\right)}{1 \times 0.1}}$<br/>Putting, $\omega=140 \operatorname{rad} \mathrm{s}^{-1}$ in above equation we get,<br/>$$<br/>n=4<br/>$$<br/>$\therefore$ Answer is 4 .</div> | MarksBatch0.db |
6 | A 2 μ F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 , is | a-2-f-capacitor-is-charged-as-shown-in-the-figure-the-percentage-of-its-stored-energy-dissipated-51052 | <div class="question">A $2 \mu \mathrm{F}$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position 2 , is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/-pmB3YN7s_jnLXaMuH5hC5HdHc5c4gtUiS-FBt_xaDw.original.fullsize.png"/><br/></div> | ['Physics', 'Capacitance', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$0 \%$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$20 \%$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$75 \%$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$80 \%$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$80 \%$</span> </div> | <div class="solution">$q_i=C_i V=2 V=q$ (say)<br/>This charge will remain constant after switch is shifted from position 1 to position 2.<br/>$$<br/>\begin{aligned}<br/>U_i & =\frac{1}{2} \frac{q^2}{C_i}=\frac{q^2}{2 \times 2}=\frac{q^2}{4} \\<br/>U_f & =\frac{1}{2} \frac{q^2}{C_f}=\frac{q^2}{2 \times 10}=\frac{q^2}{20}<br/>\end{aligned}<br/>$$<br/>$\therefore$ Energy dissipated $=U_i-U_f=\frac{q^2}{5}$<br/>This energy dissipated $\left(=\frac{q^2}{5}\right)$ is $80 \%$ of the initial stored energy $\left(=\frac{q^2}{4}\right)$.<br/>Analysis of Question<br/>(i) This question is moderately tough.<br/>(ii) In a capacitor circuit, redistribution of charge takes place under following three conditions.<br/>(a) A switch is closed.<br/>(b) A closed switch is opened.<br/>(c) A switch is shifted from one position to another position.<br/>In the redistribution of charge, energy is dissipated.</div> | MarksBatch0.db |
7 | A 20 cm long string, having a mass of 1.0 g , is fixed at both the ends. The tension in the string is 0.5 N . The string is set into vibration using an external vibrator of frequency 100 Hz . Find the separation (in cm ) between the successive nodes on the string. | a-20-cm-long-string-having-a-mass-of-1-0-g-is-fixed-at-both-the-ends-the-tension-in-the-string-i-98746 | <div class="question">A $20 \mathrm{~cm}$ long string, having a mass of $1.0 \mathrm{~g}$, is fixed at both the ends. The tension in the string is $0.5 \mathrm{~N}$. The string is set into vibration using an external vibrator of frequency 100 $\mathrm{Hz}$. Find the separation (in $\mathrm{cm}$ ) between the successive nodes on the string.</div> | ['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">5</span> </div> | <div class="solution">Distance between the successive nodes,<br/>$$<br/>\begin{aligned}<br/>d & =\frac{\lambda}{2}=\frac{v}{2 f} \\<br/>& =\frac{\sqrt{T / \mu}}{2 f}<br/>\end{aligned}<br/>$$<br/>Substituting the values we get<br/>$$<br/>d=5 \mathrm{~cm}<br/>$$</div> | MarksBatch0.db |
8 | A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 3 4 â . A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as | a-ball-is-dropped-from-a-height-of-20-m-above-the-surface-of-water-in-a-lake-the-refractive-index-o-53117 | <div class="question">A ball is dropped from a height of $20 \mathrm{~m}$ above the surface of water in a lake. The refractive index of water is $\frac{4}{3}$. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is $12.8 \mathrm{~m}$ above the water surface, the fish sees the speed of ball as</div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$9 \mathrm{~ms}^{-1}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$12 \mathrm{~ms}^{-1}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$16 \mathrm{~ms}^{-1}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$21.33 \mathrm{~ms}^{-1}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$16 \mathrm{~ms}^{-1}$<br/></span> </div> | <div class="solution">$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 7}=12 \mathrm{~ms}^{-1}$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/9vv6DsUIJzmaAOO7gLXW75DQeYmehCNseQnPs5bdTx4.original.fullsize.png"/><br/><br/>In this case when eye is inside water, $x_{\text {app. }}=\mu x \Rightarrow \frac{d x_{\text {app. }}}{d t}=\mu \cdot \frac{d x}{d t}$ or $\quad v_{\text {app. }}=\mu v=\frac{4}{3} \times 12=16 \mathrm{~ms}^{-1}$</div> | MarksBatch0.db |
9 | A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. If surface BC is frictionless and K A â , K B â and K C â are kinetic energies of the ball at A , B and C respectively, then | a-ball-moves-over-a-fixed-track-as-shown-in-the-figure-from-a-to-b-the-ball-rolls-without-slipping-54634 | <div class="question">A ball moves over a fixed track as shown in the figure. From $A$ to $B$ the ball rolls without slipping. If surface $B C$ is frictionless and $K_A, K_B$ and $K_C$ are kinetic energies of the ball at $A, B$ and $C$ respectively, then<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/c-0kUyAsFC7W3pHpRwq5dJfc6gkT9wgddik055PcPlg.original.fullsize.png"/><br/></div> | ['Physics', 'Work Power Energy', 'JEE Main'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$h_A>h_C ; K_B>K_c$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$h_A>h_c ; K_c>K_A$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$h_A=h_c ; K_B=K_c$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$h_A < h_c ; K_B>K_c$</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$h_A>h_C ; K_B>K_c$<br/></span> </div> | <div class="solution">In smooth part $B C$, due to zero torque, angular velocity and hence the rotational kinetic energy remains constant. While moving from $B$ to $C$ translational kinetic energy converts into gravitational energy.</div> | MarksBatch0.db |
10 | A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. If surface BC is frictionless and K A â , K B â and K C â are kinetic energies of the ball at A , B and C respectively, then | a-ball-moves-over-a-fixed-track-as-shown-in-the-figure-from-a-to-b-the-ball-rolls-without-slipping-18134 | <div class="question">A ball moves over a fixed track as shown in the figure. From $A$ to $B$ the ball rolls without slipping. If surface $B C$ is frictionless and $K_A, K_B$ and $K_C$ are kinetic energies of the ball at $A, B$ and $C$ respectively, then<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/c-0kUyAsFC7W3pHpRwq5dJfc6gkT9wgddik055PcPlg.original.fullsize.png"/><br/></div> | ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$h_A>h_C ; K_B>K_c$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$h_A>h_c ; K_c>K_A$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$h_A=h_c ; K_B=K_c$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$h_A < h_c ; K_B>K_c$</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$h_A>h_C ; K_B>K_c$<br/></span> </div> | <div class="solution">In smooth part $B C$, due to zero torque, angular velocity and hence the rotational kinetic energy remains constant. While moving from $B$ to $C$ translational kinetic energy converts into gravitational energy.</div> | MarksBatch0.db |
11 | A ball of mass 0.2 kg rests on a vertical post of height 5 m . A bullet of mass 0.01 kg , travelling with a velocity v m / s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity v of the bullet is | a-ball-of-mass-0-2-kg-rests-on-a-vertical-post-of-height-5-m-a-bullet-of-mass-0-01-kg-travelling-33601 | <div class="question">A ball of mass $0.2 \mathrm{~kg}$ rests on a vertical post of height $5 \mathrm{~m}$. A bullet of mass $0.01 \mathrm{~kg}$, travelling with a velocity $v \mathrm{~m} / \mathrm{s}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20 \mathrm{~m}$ and the bullet at a distance of $100 \mathrm{~m}$ from the foot of the post. The initial velocity $v$ of the bullet is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/V6qjAqEBQ2prIXraDE2K8miAKuo6vC9S27QOT26tOQo.original.fullsize.png"/><br/></div> | ['Physics', 'Center of Mass Momentum and Collision', 'JEE Main'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$250 \mathrm{~m} / \mathrm{s}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$250 \sqrt{2} \mathrm{~m} / \mathrm{s}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$400 \mathrm{~m} / \mathrm{s}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$500 \mathrm{~m} / \mathrm{s}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$500 \mathrm{~m} / \mathrm{s}$</span> </div> | <div class="solution">Time taken by the bullet and ball to strike the ground is<br/>$$<br/>t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{~s}<br/>$$<br/>Let $v_1$ and $v_2$ are the velocities of ball and bullet after collision.<br/>Then applying<br/>We have, $20=v_1 \times 1$<br/>or $v_1=20 \mathrm{~ms}^{-1}$<br/>$$<br/>100=v_2 \times 1 \text { or } v_2=100 \mathrm{~m} / \mathrm{s}^{-1}<br/>$$<br/>Now, from conservation of linear momentum before and after collision we have,<br/>$0.01 v=(0.2 \times 20)+(0.01 \times 100)$<br/>On solving, we get<br/>$$<br/>v=500 \mathrm{~ms}^{-1}<br/>$$<br/>$\therefore$ Correct answer is (d).<br/>Analysis of Question<br/>Question is moderately lengthy from calculation point of view, otherwise it is simple.</div> | MarksBatch0.db |
12 | A ball of mass 0.2 kg rests on a vertical post of height 5 m . A bullet of mass 0.01 kg , travelling with a velocity v m / s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity v of the bullet is | a-ball-of-mass-0-2-kg-rests-on-a-vertical-post-of-height-5-m-a-bullet-of-mass-0-01-kg-travelling-27996 | <div class="question">A ball of mass $0.2 \mathrm{~kg}$ rests on a vertical post of height $5 \mathrm{~m}$. A bullet of mass $0.01 \mathrm{~kg}$, travelling with a velocity $v \mathrm{~m} / \mathrm{s}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20 \mathrm{~m}$ and the bullet at a distance of $100 \mathrm{~m}$ from the foot of the post. The initial velocity $v$ of the bullet is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/V6qjAqEBQ2prIXraDE2K8miAKuo6vC9S27QOT26tOQo.original.fullsize.png"/><br/></div> | ['Physics', 'Center of Mass Momentum and Collision', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$250 \mathrm{~m} / \mathrm{s}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$250 \sqrt{2} \mathrm{~m} / \mathrm{s}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$400 \mathrm{~m} / \mathrm{s}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$500 \mathrm{~m} / \mathrm{s}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$500 \mathrm{~m} / \mathrm{s}$</span> </div> | <div class="solution">Time taken by the bullet and ball to strike the ground is<br/>$$<br/>t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{~s}<br/>$$<br/>Let $v_1$ and $v_2$ are the velocities of ball and bullet after collision.<br/>Then applying<br/>We have, $20=v_1 \times 1$<br/>or $v_1=20 \mathrm{~ms}^{-1}$<br/>$$<br/>100=v_2 \times 1 \text { or } v_2=100 \mathrm{~m} / \mathrm{s}^{-1}<br/>$$<br/>Now, from conservation of linear momentum before and after collision we have,<br/>$0.01 v=(0.2 \times 20)+(0.01 \times 100)$<br/>On solving, we get<br/>$$<br/>v=500 \mathrm{~ms}^{-1}<br/>$$<br/>$\therefore$ Correct answer is (d).<br/>Analysis of Question<br/>Question is moderately lengthy from calculation point of view, otherwise it is simple.</div> | MarksBatch0.db |
13 | A ball of mass ( m ) 0.5 kg is attached to the end of a string having length ( L ) 0.5 m . The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N . The maximum possible value of angular velocity of ball (in rad/s) is | a-ball-of-mass-m-0-5-kg-is-attached-to-the-end-of-a-string-having-length-l-0-5-m-the-ball-81767 | <div class="question">A ball of mass $(m) 0.5 \mathrm{~kg}$ is attached to the end of a string having length $(L) 0.5 \mathrm{~m}$. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is $324 \mathrm{~N}$. The maximum possible value of angular velocity of ball (in rad/s) is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/0TyV2n8gTki5yMAZr7KgdSjyaPFeAUXVThk_t2iUdQ8.original.fullsize.png"/><br/></div> | ['Physics', 'Motion In Two Dimensions', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>9<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>18<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>27<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>36</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>36</span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/hQ-PiqB_orrFBgFTj13vssd112-QatgeCTIXhpgU2G8.original.fullsize.png"/><br/><br/><br/>$T \cos \theta$ component will cancel $\mathrm{mg}$.<br/>$T \sin \theta$ component will provide necessary centripetal force to the ball towards centre $C$.<br/>$$<br/>\begin{array}{rlrl} <br/>& \therefore & T \sin \theta & =m r \omega^2=m(l \sin \theta) \omega^2 \\<br/>& \text { or } & T & =m l \omega^2 \\<br/>& \therefore & \omega & =\sqrt{\frac{T}{m l}} \\<br/>& \text { or } & \omega_{\max } & =\sqrt{\frac{T_{\max }}{m l}}=\sqrt{\frac{324}{0.5 \times 0.5}} \\<br/>& = & 36 \mathrm{rad} / \mathrm{s}<br/>\end{array}<br/>$$<br/>$\therefore$ Correct option is (d).<br/>Analysis of Question<br/>(i) Question is simple.<br/>(ii) This is called the conical pendulum.<br/>(iii) The interesting fact in this problem is that $\omega$ or $T$ is independent of $\theta$.<br/>$\omega \propto \sqrt{T}$<br/>If $\omega$ is increased, $T$ will also increase.</div> | MarksBatch0.db |
14 | A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2 . Both the curved surface are of the same radius of curvature R = 14 cm . For this bi-convex lens, for an object distance of 40 cm , the image distance will be | a-bi-convex-lens-is-formed-with-two-thin-plano-convex-lenses-as-shown-in-the-figure-refractive-inde-62619 | <div class="question">A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surface are of the same radius of curvature $R=14$ $\mathrm{cm}$. For this bi-convex lens, for an object distance of 40 $\mathrm{cm}$, the image distance will be
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/6u4Po-i8xwAGMmOrimymDpzRZhs_OFnHc8q0NXXMenE.original.fullsize.png"/></div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$-280.0 \mathrm{~cm}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$40.0 \mathrm{~cm}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$21.5 \mathrm{~cm}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$13.3 \mathrm{~cm}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">$40.0 \mathrm{~cm}$</span> </div> | <div class="solution">The focal length $\left(f_{1}\right)$ of the plano-cöneex Tens with $n=1.5$ using lens-maker formula
<br/> <br/>$\frac{1}{f_{1}}=\left(n_{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.5-1)\left[\frac{1}{14}-\frac{1}{\infty}\right]=\frac{1}{28}$
<br/> <br/>The focal length $\left(f_{2}\right)$ of the plano-convex lens with $n=1.2$ $\frac{1}{f_{2}}=\left(n_{2}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.2-1)\left[\frac{1}{\infty}-\frac{1}{-14}\right]=\frac{1}{70}$ Focal length F of the combination
<br/> <br/>$\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{20}$
<br/> <br/>Now, applying lens formula for the combination of lens
<br/> <br/>$\begin{array}{ll}
<br/> <br/>& \frac{1}{V}-\frac{1}{U}=\frac{1}{F} \Rightarrow \frac{1}{V}-\frac{1}{-40}=\frac{1}{20} \quad[\text { Given } \mu=40 \mathrm{~cm}] \\
<br/> <br/>\therefore \quad & V=40 \mathrm{~cm}
<br/> <br/>\end{array}$</div> | MarksBatch0.db |
15 | A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm . A small object is kept at a distance of 30 cm from the lens. The final image is | a-biconvex-lens-of-focal-length-15-cm-is-in-front-of-a-plane-mirror-the-distance-between-the-lens-a-21563 | <div class="question">A biconvex lens of focal length $15 \mathrm{~cm}$ is in front of a plane mirror. The distance between the lens and the mirror is 10 $\mathrm{cm}$. A small object is kept at a distance of $30 \mathrm{~cm}$ from the lens. The final image is</div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>virtual and at a distance of $16 \mathrm{~cm}$ from the mirror<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>real and at a distance of $16 \mathrm{~cm}$ from the mirror<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>virtual and at a distance of $20 \mathrm{~cm}$ from the mirror<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>real and at a distance of $20 \mathrm{~cm}$ from the mirror</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>real and at a distance of $16 \mathrm{~cm}$ from the mirror<br/></span> </div> | <div class="solution">Object is placed at distance $2 f$ from the lens. So first image $I_1$ will be formed at distance $2 f$ on other side. This image $I_1$ will behave like a virtual object for mirror. The second image $I_2$ will be formed at distance $20 \mathrm{~cm}$ in front of the mirror, or at distance $10 \mathrm{~cm}$ to the left hand side of the lens.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/HztNTNxV2XFtsoa1adchjrjvAgpK7fnuGYj4wJoGoSM.original.fullsize.png"/><br/><br/>Now applying lens formula<br/>$$<br/>\begin{array}{rlr} <br/>& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\<br/>& \frac{1}{v}-\frac{1}{+10}=\frac{1}{+15} \\<br/>\text { or } & v=6 \mathrm{~cm}<br/>\end{array}<br/>$$<br/>Therefore, the final image is at distance $16 \mathrm{~cm}$ from the mirror. But, this image will be real.<br/>This is because ray of light is travelling from right to left.<br/>$\therefore$ The correct option is (b).</div> | MarksBatch0.db |
16 | A biconvex lens of focal length f forms a circular image of radius r of sun in focal plane. Then, which option is correct? | a-biconvex-lens-of-focal-length-f-forms-a-circular-image-of-radius-r-of-sun-in-focal-plane-then-wh-79833 | <div class="question">A biconvex lens of focal length $f$ forms a circular image of radius $r$ of sun in focal plane. Then, which option is correct?</div> | ['Physics', 'Ray Optics', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\pi r^2 \propto f$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\pi r^2 \propto f^2$ <br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>If lower half part is covered by black sheet, then area of the image is equal to $\pi r^2 / 2$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>If $f$ is doubled, intensity will increase</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\pi r^2 \propto f^2$ <br/></span> </div> | <div class="solution">$$<br/>r=f \tan \theta<br/>$$<br/>or<br/>$$<br/>\begin{array}{lc}<br/>\text { or } & r \propto f \\<br/>\therefore & \pi r^2 \propto f^2<br/>\end{array}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/yxHNcSn-WGr9bnkl3vlH8HqgkKoxf5WWI77ydI4twbQ.original.fullsize.png"/><br/></div> | MarksBatch0.db |
17 | A binary star consists of two stars A (mass 2.2 M S â ) and B (mass 11 M s â ), where M S â is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is | a-binary-star-consists-of-two-stars-a-mass-2-2-m-s-and-b-mass-11-m-s-where-m-s-is-71459 | <div class="question">A binary star consists of two stars $A$ (mass $2.2 M_S$ ) and $B$ (mass $11 M_s$ ), where $M_S$ is the mass of the sun. They are separated by distance $d$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of $\operatorname{star} B$ about the centre of mass is</div> | ['Physics', 'Gravitation', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">6</span> </div> | <div class="solution">$\frac{L_{\text {Total }}}{L_B}=\frac{\left(I_A+I_B\right) \omega}{I_B \cdot \omega}$ (as $\omega$ will be same in both cases)<br/>$$<br/>\begin{aligned}<br/>& =\frac{I_A}{I_B}+1=\frac{m_A r_A^2}{m_B r_B^2}+1 \\<br/>& \left.=\frac{r_A}{r_B}+1 \quad \quad \text { (as } m_A r_A=m_B r_B\right) \\<br/>& =\frac{11}{2.2}+1 \quad \quad\left(\text { as } r \propto \frac{1}{m}\right) \\<br/>& =6<br/>\end{aligned}<br/>$$<br/>$\therefore$ The correct answer is 6 .</div> | MarksBatch0.db |
18 | A black body of temperature T is inside chamber of temperature T 0 â . Now, the closed chamber is slightly opened to sun such that temperature of black body ( T ) and chamber ( T 0 â ) remains constant. | a-black-body-of-temperature-t-is-inside-chamber-of-temperature-t-0-now-the-closed-chamber-is-63485 | <div class="question">A black body of temperature $T$ is inside chamber of temperature $T_0$. Now, the closed chamber is slightly opened to sun such that temperature of black body $(T)$ and chamber $\left(T_0\right)$ remains constant.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/nmdq3Fzc9vK4geUz_PKd_J9H2Qd2sBve2ZFny1StM6U.original.fullsize.png"/><br/></div> | ['Physics', 'Thermal Properties of Matter', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>Black body will absorb more radiation<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Black body will absorb less radiation<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Black body emit more energy<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>None of these</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>None of these</span> </div> | <div class="solution">Since, the temperature of black body is constant, total heat absorbed $=$ total heat radiated.</div> | MarksBatch0.db |
19 | A block ( B ) is attached to two unstretched S 1 â and S 2 â with spring constants k and 4 k , respectively (see figure I). The other ends are attached to identical supports M 1 â and M 2 â not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (figure II) and released. The block returns and moves a maximum distance y towards wall 2 . Displacements x and y are measured with respect to the equilibrium position of the block B . The ratio â 1 y â is | a-block-b-is-attached-to-two-unstretched-s-1-and-s-2-with-spring-constants-k-and-4-k-r-56544 | <div class="question">A block $(B)$ is attached to two unstretched $S_1$ and $S_2$ with spring constants $k$ and $4 k$, respectively (see figure I). The other ends are attached to identical supports $M_1$ and $M_2$ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block $B$ is displaced towards wall 1 by a small distance $x$ (figure II) and released. The block returns and moves a maximum distance $y$ towards wall 2 . Displacements $x$ and $y$ are measured with respect to the equilibrium position of the block $B$. The ratio $\frac{y}{-1}$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/pyRS7t-PQUH-2VBz541B_XC14w4GSMqF35DFyqE66fo.original.fullsize.png"/><br/></div> | ['Physics', 'Oscillations', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>4<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>2<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{1}{2}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{4}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{1}{2}$<br/></span> </div> | <div class="solution">From energy conservation,<br/>$$<br/>\begin{aligned}<br/>\frac{1}{2} k x^2 & =\frac{1}{2}(4 k) y^2 \\<br/>\frac{y}{x} & =\frac{1}{2}<br/>\end{aligned}<br/>$$<br/>$\therefore$ correct option is (c).</div> | MarksBatch0.db |
20 | A block is moving on an inclined plane making an angle 4 5 â with the horizontal and the coefficient of friction is μ . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 μ , then N is | a-block-is-moving-on-an-inclined-plane-making-an-angle-4-5-with-the-horizontal-and-the-coefficie-97964 | <div class="question">A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define $N=10 \mu$, then $N$ is</div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">5</span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/eySCplq6WaN-Uw5HFXL6SyXyNMZS2BoAPDicdUpjwso.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/A_tLk_dZ6J3lFdcrTO-XBum_JPxphlYFyCaGnc_1-iE.original.fullsize.png"/><br/><br/>$$<br/>\begin{aligned}<br/>& F_1=m g \sin \theta+\mu m g \cos \theta \\<br/>& F_2=m g \sin \theta-\mu m g \cos \theta<br/>\end{aligned}<br/>$$<br/>Given that, $F_1=3 F_2$<br/>or $\left(\sin 45^{\circ}+\mu \cos 45^{\circ}\right)$<br/>$$<br/>=3\left(\sin 45^{\circ}-\mu \cos 45^{\circ}\right)<br/>$$<br/>On solving, we get $\mu=0.5$<br/>$$<br/>\therefore \quad N=10 \mu=5<br/>$$<br/>$\therefore$ Answer is 5 .<br/>Analysis of Question<br/>Question is simple. Only one has to take care of direction and magnitude of friction.</div> | MarksBatch0.db |
21 | A block of base 10 cm à 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3 â . The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0 â . Then, | a-block-of-base-10-cm-10-cm-and-height-15-cm-is-kept-on-an-inclined-plane-the-coefficient-of-fri-50173 | <div class="question">A block of base $10 \mathrm{~cm} \times 10 \mathrm{~cm}$ and height $15 \mathrm{~cm}$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then,</div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>at $\theta=30^{\circ}$, the block will start sliding down the plane<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>the block will remain at rest on the plane up to certain $\theta$ and then it will topple<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>at $\theta=60^{\circ}$, the block will start sliding down the plane and continue to do so at higher angles<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>at $\theta=60^{\circ}$, the block will start sliding down the plane and on further increasing $\theta$, it will topple at certain $\theta$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>the block will remain at rest on the plane up to certain $\theta$ and then it will topple<br/></span> </div> | <div class="solution">Condition of sliding is $m g \sin \theta>\mu m g \cos \theta$<br/>or $\tan \theta>\mu$<br/>or $\quad \tan \theta>\sqrt{3}$<br/>Condition of toppling is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/cM4kLzRjOM5W0qj3dToc8zuVthq00TqewrB6onmam7I.original.fullsize.png"/><br/><br/>Torque of $m g \sin \theta$ about $0>$ torque of $m g \cos \theta$ about<br/>$\therefore \quad(m g \sin \theta)\left(\frac{15}{2}\right)>(m g \cos \theta)\left(\frac{10}{2}\right)$<br/>or $\quad \tan \theta>\frac{2}{3}$<br/>With increase in value of $\theta$, condition of sliding is satisfied first.</div> | MarksBatch0.db |
22 | A block of mass 2 kg is free to move along the x -axis. It is at rest and from t = 0 onwards it is subjected to a time-dependent force F ( t ) in the x direction. The force F ( t ) varies with t as shown in the figure. The kinetic energy of the block after 4.5 s is | a-block-of-mass-2-kg-is-free-to-move-along-the-x-axis-it-is-at-rest-and-from-t-0-onwards-it-is-s-41352 | <div class="question">A block of mass $2 \mathrm{~kg}$ is free to move along the $x$-axis. It is at rest and from $t$ $=0$ onwards it is subjected to a time-dependent force $F(t)$ in the $x$ direction. The force $F(t)$ varies with $t$ as shown in the figure. The kinetic energy of the block after $4.5 \mathrm{~s}$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/nitJht7rqjvVnbZhf_fKlJaXMKMAPrfBW9jbItMnePM.original.fullsize.png"/><br/></div> | ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$4.50 \mathrm{~J}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$7.50 \mathrm{~J}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$5.06 \mathrm{~J}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$14.06 \mathrm{~J}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$5.06 \mathrm{~J}$<br/></span> </div> | <div class="solution">Area under $F$ - $t$ graph $=$ momentum<br/>$$<br/>\begin{aligned}<br/>&=P=\sqrt{2 k m} \\<br/>& \therefore \quad k=\frac{A^2}{2 m} \\<br/>&(A=\text { net area of } F-t \text { graph) } \\<br/>&=\frac{\left\{\left(\frac{4 \times 3}{2}\right)-\left(\frac{1.5 \times 2}{2}\right)\right\}^2}{2 \times 2}=5.0625 \mathrm{~J}<br/>\end{aligned}<br/>$$<br/>$\therefore$ The correct option is (c).</div> | MarksBatch0.db |
23 | A block of mass 2 M is attached to a massless spring with spring-constant k . This block is connected to two other blocks of masses M and 2 M using two massless pulleys and strings. The accelerations of the blocks are a 1 , a 2 and a 3 as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x 0 . Which of the following option(s) is/are correct? [ g is the acceleration due to gravity. Neglect friction] | a-block-of-mass-2-m-is-attached-to-a-massless-spring-with-spring-constant-k-this-block-is-connecte-15478 | <div class="question"><p>A block of mass <math><mn>2</mn><mi mathvariant="normal">M</mi></math> is attached to a massless spring with spring-constant <math><mi mathvariant="normal">k</mi><mo>.</mo></math> This block is connected to two other blocks of masses <math><mi mathvariant="normal">M</mi></math> and <math><mn>2</mn><mi mathvariant="normal">M</mi></math> using two massless pulleys and strings. The accelerations of the blocks are <math><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mtext>â</mtext><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msub></math> and <math><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msub></math> as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is <math><msub><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>.</mo></math> Which of the following option(s) is/are correct?<br/>[ <math><mi mathvariant="normal">g</mi></math> is the acceleration due to gravity. Neglect friction]<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/a8f796a7-fd32-4111-9acf-8877e6c69118-image20.png"/></p></div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2019 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math><msub><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>=</mo><mfrac><mrow><mn>4</mn><mi mathvariant="normal">M</mi><mi mathvariant="normal">g</mi></mrow><mrow><mi mathvariant="normal">k</mi></mrow></mfrac></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">When spring achieves an extension of <math><mfrac><mrow><msub><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac></math> for the first time, the speed of the block connected to the spring is <math><mn>3</mn><mi mathvariant="normal">g</mi><msqrt><mfrac><mrow><mi mathvariant="normal">M</mi></mrow><mrow><mn>5</mn><mi mathvariant="normal">k</mi></mrow></mfrac></msqrt></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>-</mo><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msub></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">At an extension of <math><mfrac><mrow><msub><mrow><mi mathvariant="normal">x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>4</mn></mrow></mfrac></math> of the spring, the magnitude of acceleration of the block connected to the spring is <math><mfrac><mrow><mn>3</mn><mi mathvariant="normal">g</mi></mrow><mrow><mn>10</mn></mrow></mfrac></math></span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><math><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>-</mo><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>3</mn></mrow></msub></math></span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/09b699d9-7e83-4187-8ccd-3ca9c9e42d6c-image21.png"/> <br/>Using string constraint <br/><math><msub><mrow><mi>a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><mfrac><mrow><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><msub><mrow><mi>a</mi></mrow><mrow><mn>3</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac></math> <br/><math><mo>â</mo><mn>2</mn><msub><mrow><mi>a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>=</mo><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><msub><mrow><mi>a</mi></mrow><mrow><mn>3</mn></mrow></msub></math> <br/><math><msub><mrow><mi>a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>-</mo><msub><mrow><mi>a</mi></mrow><mrow><mn>3</mn></mrow></msub><mo>=</mo><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>-</mo><msub><mrow><mi>a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>â</mo><mfenced separators="|"><mrow><mi mathvariant="normal">C</mi></mrow></mfenced></math> is correct. <br/>for other options use <math><mi>m</mi></math> equivalent <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/4dba6f90-2038-4380-ba6c-ceb2c2695efd-image22.png"/> <br/>Equivalent mass <br/>Or Reduced mass <br/><math><msub><mrow><mi>m</mi></mrow><mrow><mi>e</mi><mi>q</mi></mrow></msub><mo>=</mo><mfrac><mrow><mn>2</mn><msub><mrow><mi>m</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>Ã</mo><msub><mrow><mi>m</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow><mrow><msub><mrow><mi>m</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>+</mo><msub><mrow><mi>m</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></mfrac></math> (for a two-mass system) <br/><math><mo>=</mo><mfrac><mrow><mn>2</mn><mfenced separators="|"><mrow><mn>2</mn><mi>m</mi></mrow></mfenced><mo>Ã</mo><mi>m</mi></mrow><mrow><mn>2</mn><mi>m</mi><mo>+</mo><mi>m</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>4</mn><mi>m</mi></mrow><mrow><mn>3</mn></mrow></mfrac></math> <br/><math><mo>â</mo><mi>T</mi><mo>=</mo><mfenced separators="|"><mrow><mfrac><mrow><mn>4</mn><mi>m</mi></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></mfenced><mo>.</mo><mi>g</mi></math> <br/><math><mo>â´</mo><mn>2</mn><mi>T</mi><mo>=</mo><mn>2</mn><mo>.</mo><mfrac><mrow><mn>4</mn><mi>m</mi><mi>g</mi></mrow><mrow><mn>3</mn></mrow></mfrac><mo>=</mo><mfenced separators="|"><mrow><mfrac><mrow><mn>8</mn><mi>m</mi></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></mfenced><mi>g</mi></math> <br/><math><mo>â</mo><msub><mrow><mi>m</mi></mrow><mrow><mi>e</mi><mi>q</mi></mrow></msub><mo>=</mo><mfrac><mrow><mn>8</mn><mi>m</mi></mrow><mrow><mn>3</mn></mrow></mfrac></math> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/6ad10ded-dcf6-48f2-bf60-a13a548aa9e5-image23.png"/> <br/><em>Using mechanical energy conservation</em> <br/><math><mo>â</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mi>k</mi><msubsup><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow><mrow><mn>2</mn></mrow></msubsup><mo>=</mo><mfrac><mrow><mn>8</mn><mi>m</mi><mi>g</mi></mrow><mrow><mn>3</mn></mrow></mfrac><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></math> <br/><math><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>=</mo><mfrac><mrow><mn>16</mn><mi>m</mi><mi>g</mi></mrow><mrow><mn>3</mn><mi>k</mi></mrow></mfrac><mo>â</mo><mfenced separators="|"><mrow><mi mathvariant="normal">A</mi></mrow></mfenced></math> <em>is incorrect</em> <br/><em>Option</em> <math><mfenced separators="|"><mrow><mi mathvariant="normal">B</mi></mrow></mfenced><mo>â</mo><msub><mrow><mi>V</mi></mrow><mrow><mfrac><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></msub><mo>=</mo><msub><mrow><mi>V</mi></mrow><mrow><mi mathvariant="normal">m</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">x</mi></mrow></msub><mo>=</mo><mfrac><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><mi>Ï</mi><mo>=</mo><mfrac><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msqrt><mfrac><mrow><mi>k</mi></mrow><mrow><mn>2</mn><mi>m</mi><mo>+</mo><mfrac><mrow><mn>8</mn><mi>m</mi></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></mfrac></msqrt><mo>=</mo><mfrac><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac><msqrt><mfrac><mrow><mn>3</mn><mi>k</mi></mrow><mrow><mn>14</mn><mi>m</mi></mrow></mfrac></msqrt><mo>=</mo><mi>g</mi><msqrt><mfrac><mrow><mn>32</mn></mrow><mrow><mn>21</mn><mi>k</mi></mrow></mfrac></msqrt></math> <br/>Option <math><mfenced separators="|"><mrow><mi mathvariant="normal">D</mi></mrow></mfenced><mo>â</mo><msub><mrow><mi>a</mi></mrow><mrow><mfrac><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></msub><mo>=</mo><mfrac><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>4</mn></mrow></mfrac><msup><mrow><mi>Ï</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mfrac><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>4</mn></mrow></mfrac><mfrac><mrow><mn>3</mn><mi>k</mi></mrow><mrow><mn>14</mn><mi>m</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>3</mn><mi>k</mi><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>42</mn><mi>m</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>8</mn><mi>g</mi></mrow><mrow><mn>21</mn></mrow></mfrac></math> <br/>For a simple harmonic motion <math><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>â</mo></math> maximum displacement from extreme to extreme <br/><math><mo>â</mo><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub><mo>=</mo><mn>2</mn><mi>A</mi><mo>â</mo><mi>A</mi><mo>=</mo><mfrac><mrow><msub><mrow><mi>x</mi></mrow><mrow><mn>0</mn></mrow></msub></mrow><mrow><mn>2</mn></mrow></mfrac></math> <br/>and <math><mi>Ï</mi><mo>=</mo><msqrt><mfrac><mrow><mi>k</mi></mrow><mrow><mi>m</mi></mrow></mfrac></msqrt><mo>=</mo><msqrt><mfrac><mrow><mn>3</mn><mi>k</mi></mrow><mrow><mn>14</mn><mi>m</mi></mrow></mfrac></msqrt></math> <br/><math><mo>â´</mo><msub><mrow><mi>v</mi></mrow><mrow><mi mathvariant="normal">m</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">x</mi></mrow></msub><mo>=</mo><mi>A</mi><mo>.</mo><mi>Ï</mi></math> <br/>and <math><mi>a</mi><mo>=</mo><msup><mrow><mi>Ï</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>.</mo><mi>x</mi></math></div> | MarksBatch0.db |
24 | A block of mass 0.18 kg is attached to a spring of force constant 2 N / m . The coefficient of friction between the block and the floor is 0.1. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m / s is v = 10 N â . Then, N is | a-block-of-mass-0-18-kg-is-attached-to-a-spring-of-force-constant-2-n-m-the-coefficient-of-frict-39134 | <div class="question">A block of mass $0.18 \mathrm{~kg}$ is attached to a spring of force constant $2 \mathrm{~N} / \mathrm{m}$. The coefficient of friction between the block and the floor is 0.1. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of $0.06 \mathrm{~m}$ and comes to rest for the first time. The initial velocity of the block in $\mathrm{m} / \mathrm{s}$ is $v=\frac{N}{10}$. Then, $N$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/6Z7Mom2YXU3qPy7T4tuDo3vBvksgUzsCc5KY2gv2Pis.original.fullsize.png"/><br/></div> | ['Physics', 'Work Power Energy', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">4</span> </div> | <div class="solution">Decrease in mechanical energy<br/>$=$ Work done against friction<br/>$\therefore \quad \frac{1}{2} m v^2-\frac{1}{2} k x^2=\mu m g x$<br/>or $v=\sqrt{\frac{2 \mu m g x+k x^2}{m}}$<br/>Substituting the values, we get<br/>$$<br/>v=0.4 \mathrm{~ms}^{-1}=\left(\frac{4}{10}\right) \mathrm{ms}^{-1}<br/>$$<br/>$\therefore$ Answer is 4 .<br/>Analysis of Question<br/>(i) Question is simple.<br/>(ii) If $\mu_s$ and $\mu_k$ two values of coefficient of friction are given, then we will take $\mu_k$.</div> | MarksBatch0.db |
25 | A block of mass m is on an inclined plane of angle θ . The coefficient of friction between the block and the plane is μ and tan θ > μ . The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P 1 â = m g ( sin θ â μ cos θ ) to P 2 â = m g ( sin θ + μ cos θ ) , the frictional force f versus P graph will look like | a-block-of-mass-m-is-on-an-inclined-plane-of-angle-the-coefficient-of-friction-between-the-bloc-27808 | <div class="question">A block of mass $m$ is on an inclined plane of angle $\theta$. The coefficient of friction between the block and the plane is $\mu$ and $\tan \theta>\mu$. The block is held stationary by applying a force $P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_1=m g(\sin \theta-\mu \cos \theta) \quad$ to $P_2=m g(\sin \theta+\mu \cos \theta)$, the frictional force $f$ versus $P$ graph will look like<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/A1ojR6zLwSTbNLaw_xmpcC01mA0ISZdeJy8a7wgIg1I.original.fullsize.png"/><br/></div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3J8uASe6iZWz0DOcTxIPzedKr7ohmEX9FABVqk8QJ8o.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/2_rTr0Fl2IAtefXZcHvyiFZZaGPSfcwbFS78QgVmZcQ.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5QaJVuKln-jR-hVq-AZYJMVL9aaTSqhWrWc_CSxYnWA.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/q8dV5585ipXNhs0oLdXFOr58RjnItmRWxkg1FY04Ups.original.fullsize.png"/><br/></span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3J8uASe6iZWz0DOcTxIPzedKr7ohmEX9FABVqk8QJ8o.original.fullsize.png"/><br/></span> </div> | <div class="solution">When<br/>$$<br/>\begin{aligned}<br/>P & =m g(\sin \theta-\mu \cos \theta) \\<br/>f & =\mu m g \cos \theta \quad \text { (upwards) } \\<br/>\text { when } \quad P & =m g \sin \theta \\<br/>f & =0<br/>\end{aligned}<br/>$$<br/>and when $P=m g(\sin \theta+\mu \cos \theta)$<br/>$$<br/>f=\mu m g \cos \theta \text { (downwards) }<br/>$$<br/>Hence friction is first positive, then zero and then negative.<br/>$\therefore$ correct option is (a).</div> | MarksBatch0.db |
26 | A bob of mass M is suspended by a massless string of length L . The horizontal velocity v at position A is just sufficient to make it reach the point B . The angle θ at which the speed of the bob is half of that at A , satisfies | a-bob-of-mass-m-is-suspended-by-a-massless-string-of-length-l-the-horizontal-velocity-v-at-positio-22404 | <div class="question">A bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $v$ at position $A$ is just sufficient to make it reach the point $B$. The angle $\theta$ at which the speed of the bob is half of that at $A$, satisfies<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/yB_mlfJxewK2ISXeI6nPyfMNBOljptoFdvbYpqzvEPk.original.fullsize.png"/><br/></div> | ['Physics', 'Motion In Two Dimensions', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\theta=\frac{\pi}{4}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{\pi}{4} < \theta < \frac{\pi}{2}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{\pi}{2} < \theta < \frac{3 \pi}{4}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{3 \pi}{4} < \theta < \pi$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{3 \pi}{4} < \theta < \pi$</span> </div> | <div class="solution">$v=\sqrt{5 g L}$<br/>Solving Eqs. (i), (ii) and (iii) we get,<br/>$$<br/>\cos \theta=-\frac{7}{8} \quad \text { or } \theta=\cos ^{-1}\left(-\frac{7}{8}\right)=151^{\circ}<br/>$$<br/>$\therefore$ correct option is (d).</div> | MarksBatch0.db |
27 | A bob of mass m , suspended by a string of length l 1 , is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m , suspended by a string of length l 2 , which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio l 1 l 2 is | a-bob-of-mass-m-suspended-by-a-string-of-length-l-1-is-given-a-minimum-velocity-required-to-comp-50092 | <div class="question">A bob of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math>, suspended by a string of length <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>l</mi><mn>1</mn></msub></math>, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math>, suspended by a string of length <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>l</mi><mn>2</mn></msub></math>, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msub><mi>l</mi><mn>1</mn></msub><msub><mi>l</mi><mn>2</mn></msub></mfrac></math> is</div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2013 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">5</span> </div> | <div class="solution"><p style="text-align: justify;"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/EM0012552_AnsExp1_1..jpg"/><br/>As shown in the figure shown above, the minimum speed required by the bob of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> attached to the string of length <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>l</mi><mn>1</mn></msub></math> in order to complete the full circle is <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>5</mn><mi>g</mi><msub><mi>l</mi><mn>1</mn></msub></msqrt></math>.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/EM0012552_AnsExp1_2..jpg"/><br/>When the bob goes to the highest point, its speed becomes <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mi>g</mi><msub><mi>l</mi><mn>1</mn></msub></msqrt></math>. Here, it strikes with the identical mass, which is tied to the string of length <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>l</mi><mn>2</mn></msub></math>. The speed delivered to the object is <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mi>g</mi><msub><mi>l</mi><mn>1</mn></msub></msqrt></math>. This should be equal to <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>5</mn><mi>g</mi><msub><mi>l</mi><mn>2</mn></msub></msqrt></math> so that this bob could complete a full circle.   <br/><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mi>g</mi><msub><mi>l</mi><mn>1</mn></msub></msqrt><mo>=</mo><msqrt><mn>5</mn><mi>g</mi><msub><mi>l</mi><mn>2</mn></msub></msqrt></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>â</mo><mi>g</mi><msub><mi>l</mi><mn>1</mn></msub><mo>=</mo><mn>5</mn><mi>g</mi><msub><mi>l</mi><mn>2</mn></msub></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>â</mo><mfrac><msub><mi>l</mi><mn>1</mn></msub><msub><mi>l</mi><mn>2</mn></msub></mfrac><mo>=</mo><mn>5</mn></math></p></div> | MarksBatch0.db |
28 | A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m / s 2 . The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is 10 P â . The value of P is | a-boy-is-pushing-a-ring-of-mass-2-kg-and-radius-0-5-m-with-a-stick-as-shown-in-the-figure-the-stick-22732 | <div class="question">A boy is pushing a ring of mass $2 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$ with a stick as shown in the figure. The stick applies a force of $2 \mathrm{~N}$ on the ring and rolls it without slipping with an acceleration of $0.3 \mathrm{~m} / \mathrm{s}^2$. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is $\frac{P}{10}$. The value of $P$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/pT01g3-QjNyJlpULpdmqYgGlziOBCvSjJ_Q1CqlXk3E.original.fullsize.png"/><br/></div> | ['Physics', 'Rotational Motion', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">4</span> </div> | <div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/bZnr6gPHUpcK9KztXUjb3pyuleRikKc7WEgAbAi8fIU.original.fullsize.png"/><br/><br/>There is no slipping between ring and ground. Hence, $f_2$ is not maximum. But there is slipping between ring and stick. Therefore, $f_1$ is maximum. Now, let us write the equations.<br/>$$<br/>\begin{aligned}<br/>I & =m R^2=(2)(0.5)^2 \\<br/>& =\frac{1}{2} \mathrm{kgm}^{-2} \\<br/>N_1-F_2 & =m a \\<br/>\text { or } N_1-F_2 & =(2)(0.3)=0.6 \mathrm{~N} \\<br/>a & =R \alpha=\frac{R \tau}{I} \\<br/>& =\frac{R\left(f_2-f_1\right) R}{I}=\frac{R^2\left(f_2-f_1\right)}{I} \\<br/>\therefore \quad 0.3 & =\frac{(0.5)^2\left(f_2-f_1\right)}{(1 / 2)} \\<br/>\text { or } \quad f_2-f_1 & =0.6 \mathrm{~N} \\<br/>N_1^2+f_1^2 & =(2)^2=4<br/>\end{aligned}<br/>$$<br/>Further $\quad F_1=\mu N_1=\left(\frac{P}{10}\right) N_1$<br/>Solving above four equation, we get $P \simeq 3.6$<br/>Therefore, the correct answer should be 4 .<br/>Analysis of Question<br/>(i) Question is moderately tough from concept point of view. But calculations are lengthy.<br/>(ii) One has to think about the two components of the force applied by the stick.<br/>(iii) Answer comes an integer when you consider only $N_1$.</div> | MarksBatch0.db |
29 | A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is | a-circuit-is-connected-as-shown-in-the-figure-with-the-switch-s-open-when-the-switch-is-closed-the-75616 | <div class="question">A circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed, the total amount of charge that flows from $Y$ to $X$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/gVNmoMqCTlpc9kjg3AHBkN3h3FpoDB_xQ_LN2wyvCFU.original.fullsize.png"/><br/></div> | ['Physics', 'Capacitance', 'JEE Main'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>zero<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$54 \mu \mathrm{C}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$27 \mu \mathrm{C}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$81 \mu \mathrm{C}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$27 \mu \mathrm{C}$<br/></span> </div> | <div class="solution">From $Y$ to $X$ charge flows to plates $a$ and $b$.<br/>$$<br/>\left(q_0+q_b\right)_i=0,\left(q_0+q_b\right)_f=27 \mu \mathrm{C}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/U9y-IpNMTnta5hFfTFwh-2QOoznAart4eldMPFAg-5M.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/qLd1-LHBQQ3Yec5ukMhmKqut2Kcs75i6baaqnE6p4O8.original.fullsize.png"/><br/><br/>$\therefore 27 \mu \mathrm{C}$ charge flows from $Y$ to $X$.<br/>$\therefore$ Correct option is (c)</div> | MarksBatch0.db |
30 | A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is | a-circuit-is-connected-as-shown-in-the-figure-with-the-switch-s-open-when-the-switch-is-closed-the-76341 | <div class="question">A circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed, the total amount of charge that flows from $Y$ to $X$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/gVNmoMqCTlpc9kjg3AHBkN3h3FpoDB_xQ_LN2wyvCFU.original.fullsize.png"/><br/></div> | ['Physics', 'Capacitance', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>zero<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$54 \mu \mathrm{C}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$27 \mu \mathrm{C}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$81 \mu \mathrm{C}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$27 \mu \mathrm{C}$<br/></span> </div> | <div class="solution">From $Y$ to $X$ charge flows to plates $a$ and $b$.<br/>$$<br/>\left(q_0+q_b\right)_i=0,\left(q_0+q_b\right)_f=27 \mu \mathrm{C}<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/U9y-IpNMTnta5hFfTFwh-2QOoznAart4eldMPFAg-5M.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/qLd1-LHBQQ3Yec5ukMhmKqut2Kcs75i6baaqnE6p4O8.original.fullsize.png"/><br/><br/>$\therefore 27 \mu \mathrm{C}$ charge flows from $Y$ to $X$.<br/>$\therefore$ Correct option is (c)</div> | MarksBatch0.db |
31 | A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg is placed as shown. The coefficient of friction between the block and all surfaces of groove in contact is μ = 2/5 . The disc has an acceleration of 25 m / s 2 . Find the acceleration of the block with respect to disc. | a-circular-disc-with-a-groove-along-its-diameter-is-placed-horizontally-a-block-of-mass-1-kg-is-pla-72553 | <div class="question">A circular disc with a groove along its diameter is placed horizontally. A block of mass $1 \mathrm{~kg}$ is placed as shown. The coefficient of friction between the block and all surfaces of groove in contact is $\mu=2 / 5$. The disc has an acceleration of $25 \mathrm{~m} / \mathrm{s}^2$. Find the acceleration of the block with respect to disc.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/HKqQm79q-3Vu0FuEB78jPiwvRnJXqOICCd975832LI8.original.fullsize.png"/><br/></div> | ['Physics', 'Laws of Motion', 'JEE Advanced', 'JEE Advanced 2006'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">10</span> </div> | <div class="solution">Normal reaction in vertical direction $N_1=m g$<br/>Normal reaction from side to the groove $N_2=m a \sin 37^{\circ}$<br/>Therefore, acceleration of block with respect to discs<br/>$$<br/>a_r=\frac{m a \cos 37^{\circ}-\mu N_1-\mu N_2}{m}<br/>$$<br/>Substituting the values we get, $\quad a_r=10 \mathrm{~m} / \mathrm{s}^2$</div> | MarksBatch0.db |
32 | A circular wire loop of radius R is placed in the x â y plane centered at the origin O . A square loop of side a ( a R ) having two turns is placed with its centre at z = 3 â R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 4 5 â with respect to the z -axis. If the mutual inductance between the loops is given by 2 p /2 R μ 0 â a 2 â , then the value of p is | a-circular-wire-loop-of-radius-r-is-placed-in-the-x-y-plane-centered-at-the-origin-o-a-square-38180 | <div class="question">A circular wire loop of radius $R$ is placed in the $x-y$ plane centered at the origin $O$. A square loop of side $a(a< < R)$ having two turns is placed with its centre at $z=\sqrt{3} R$ along the axis of the circular wire loop, as shown in figure.<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/JqNcOFDBbBi_8pytzODy6YCrM91SCU7qIL1ioD2YVLw.original.fullsize.png"/><br/> <br/> <br/>The plane of the square loop makes an angle of $45^{\circ}$ with respect to the $z$-axis. If the mutual inductance between the loops is given by $\frac{\mu_{0} a^{2}}{2^{p / 2} R}$, then the value of $p$ is</div> | ['Physics', 'Electromagnetic Induction', 'JEE Main'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7</span> </div> | <div class="solution">The magnetic field due to current carrying wire at the location of square loop is
<br/> <br/>$B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{\left(R^{2}+3 R^{2}\right)^{3 / 2}}=\frac{\mu_{0} i}{16 R}$
<br/> <br/>The mutual induction
<br/> <br/>$\begin{array}{l}
<br/> <br/>M=\frac{N \phi}{i}=\frac{2}{i}\left[\frac{\mu_{0} i}{16 R} \times a^{2} \cos 45^{\circ}\right] \\
<br/> <br/>\therefore \quad M=\frac{\mu_{0} a^{2}}{2^{\frac{7}{2}} R}
<br/> <br/>\end{array}$</div> | MarksBatch0.db |
33 | A circular wire loop of radius R is placed in the x â y plane centered at the origin O . A square loop of side a ( a R ) having two turns is placed with its centre at z = 3 â R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 4 5 â with respect to the z -axis. If the mutual inductance between the loops is given by 2 p /2 R μ 0 â a 2 â , then the value of p is | a-circular-wire-loop-of-radius-r-is-placed-in-the-x-y-plane-centered-at-the-origin-o-a-square-80632 | <div class="question">A circular wire loop of radius $R$ is placed in the $x-y$ plane centered at the origin $O$. A square loop of side $a(a< < R)$ having two turns is placed with its centre at $z=\sqrt{3} R$ along the axis of the circular wire loop, as shown in figure.<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/JqNcOFDBbBi_8pytzODy6YCrM91SCU7qIL1ioD2YVLw.original.fullsize.png"/><br/> <br/> <br/>The plane of the square loop makes an angle of $45^{\circ}$ with respect to the $z$-axis. If the mutual inductance between the loops is given by $\frac{\mu_{0} a^{2}}{2^{p / 2} R}$, then the value of $p$ is</div> | ['Physics', 'Electromagnetic Induction', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7</span> </div> | <div class="solution">The magnetic field due to current carrying wire at the location of square loop is
<br/> <br/>$B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{\left(R^{2}+3 R^{2}\right)^{3 / 2}}=\frac{\mu_{0} i}{16 R}$
<br/> <br/>The mutual induction
<br/> <br/>$\begin{array}{l}
<br/> <br/>M=\frac{N \phi}{i}=\frac{2}{i}\left[\frac{\mu_{0} i}{16 R} \times a^{2} \cos 45^{\circ}\right] \\
<br/> <br/>\therefore \quad M=\frac{\mu_{0} a^{2}}{2^{\frac{7}{2}} R}
<br/> <br/>\end{array}$</div> | MarksBatch0.db |
34 | A composite block is made of slabs A , B , C , D and E of different thermal conductivities (given in terms of a constant, K ) and sizes (given in terms of length, L ) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then, in steady state | a-composite-block-is-made-of-slabs-a-b-c-d-and-e-of-different-thermal-conductivities-given-in-25555 | <div class="question">A composite block is made of slabs $A, B, C, D$ and $E$ of different thermal conductivities (given in terms of a constant, $K$ ) and sizes (given in terms of length, $L$ ) as shown in the figure. All slabs are of same width. Heat $Q$ flows only from left to right through the blocks. Then, in steady state<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/E0zfPXPVD6anEcAXH-PQnBX7KAEIvnco_WbswydRJsM.original.fullsize.png"/><br/></div> | ['Physics', 'Thermal Properties of Matter', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>heat flow through $A$ and $E$ slabs are same<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>heat flow through slab $E$ is maximum<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>temperature difference across slab $E$ is smallest<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>heat flow through $C=$ heat flow through $B$ + heat flow through $D$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>heat flow through $A$ and $E$ slabs are same<br/>, <br/>temperature difference across slab $E$ is smallest<br/>, <br/>heat flow through $C=$ heat flow through $B$ + heat flow through $D$</span> </div> | <div class="solution">Thermal resistance<br/>$$<br/>\begin{aligned}<br/>& R=\frac{l}{K A} \\<br/>& \therefore \quad R_A=\frac{L}{(2 K)(4 L w)}=\frac{1}{8 K w} \\<br/>& \text { (Here, } w=\text { width) } \\<br/>& R_B=\frac{4 L}{3 K(L w)}=\frac{4}{3 K w} \\<br/>& R_C=\frac{4 L}{(4 K)(2 L w)}=\frac{1}{2 K w} \\<br/>& R_D=\frac{4 L}{(5 K)(L w)}=\frac{4}{5 K w} \\<br/>& R_E=\frac{L}{(6 K)(L w)}=\frac{1}{6 K w} \\<br/>& R_A: R_B: R_C: R_D: R_E \\<br/>&=15: 160: 60: 96: 12<br/>\end{aligned}<br/>$$<br/>So, let us write, $R_A=15 R, R_B=160 R$ etc and draw a simple electrical circuit as shown in figure.<br/><br/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/AGHLRh_Fhiti6oMHUfQ1FIukEOkNLjb8FmuxSaIUu0A.original.fullsize.png"/><br/><br/><br/><br/>$H=$ Heat current $=$ Rate of heat flow.<br/>$$<br/>H_A=H_E=H<br/>$$<br/>[let]<br/>$\therefore$ Option (a) is correct.<br/>In parallel, current distributes in inverse ratio of resistance.<br/>$$<br/>\begin{aligned}<br/>& \therefore H_B: H_C: H_D=\frac{1}{R_B}: \frac{1}{R_C}: \frac{1}{R_D} \\<br/>& =\frac{1}{160}: \frac{1}{60}: \frac{1}{96} \\<br/>& =9: 24: 15 \\<br/>& \therefore \quad H_B=\left(\frac{9}{9+24+15}\right) H=\frac{3}{16} H \\<br/>& H_C=\left(\frac{24}{9+24+15}\right) H=\frac{1}{2} H \\<br/>& \text { and } \quad H_D=\left(\frac{15}{9+24+15}\right) H=\frac{5}{16} H \\<br/>&<br/>\end{aligned}<br/>$$<br/><br/>$$<br/>H_C=H_B+H_D<br/>$$<br/>$\therefore$ Option (d) is correct.<br/>Temperature difference (let us call it $T$ )<br/>$=($ Heat current $) \times($ Thermal resistance $)$<br/>$T_A=H_A R_A=(H)(15 R)=15 \mathrm{HR}$<br/>$T_B=H_B R_B=\left(\frac{3}{16} H\right)(160 R)=30 H R$<br/>$T_C=H_C R_C=\left(\frac{1}{2} H\right)(60 R)=30 H R$<br/>$T_D=H_D R_D=\left(\frac{5}{16} H\right)(96 R)=30 H R$<br/>$T_E=H_E R_E=(H)(12 R)=12 \mathrm{HR}$<br/>Here, $T_E$ is minimum. Therefore, option<br/>(c) is also correct.<br/>$\therefore$ Correct options are (a), (c) and (d).<br/>Analysis of Question<br/>(i) From calculation point of view, question is difficult otherwise question is simple.<br/>(ii) In heat transfer, questions are mainly asked from conduction and radiation topic.</div> | MarksBatch0.db |
35 | A compound M p â X q â has cubic close packing (ccp) arrangement of X . Its unit cell structure is shown below. The empirical formula of the compound is | a-compound-m-p-x-q-has-cubic-close-packing-ccp-arrangement-of-x-its-unit-cell-structure-48724 | <div class="question">A compound $\mathrm{M}_{\mathrm{p}} \mathrm{X}_{\mathrm{q}}$ has cubic close packing (ccp) arrangement of $\mathrm{X}$. Its unit cell structure is shown below. The empirical formula of the compound is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/0OaasNwsPUCQ9ofdG8mPkRg9E7U5S4FL1zoVZBkEghQ.original.fullsize.png"/></div> | ['Chemistry', 'Solid State', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\mathrm{MX}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\mathrm{MX}_{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{M}_{2} \mathrm{X}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{M}_{5} \mathrm{X}_{14}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{MX}_{2}$</span> </div> | <div class="solution">No. of $M$ atoms $=\frac{1}{4} \times 4+1=1+1=2$
<br/> <br/>No. of $X$ atoms $=\frac{1}{2} \times 6+\frac{1}{8} \times 8=3+1=4$
<br/> <br/>So, formula $=M_{2} X_{4}=M X_{2}$</div> | MarksBatch0.db |
36 | A cubical region of side a has its centre at the origin. It encloses three fixed point charges, â q at ( 0 , â a /4 , 0 ) , + 3 q at ( 0 , 0 , 0 ) and â q at ( 0 , + a /4 , 0 ) . Choose the correct options(s) | a-cubical-region-of-side-a-has-its-centre-at-the-origin-it-encloses-three-fixed-point-charges-26810 | <div class="question">A cubical region of side $a$ has its centre at the origin. It encloses three fixed point charges, $-q$ at $(0,-a / 4,0),+3 q$ at $(0,0,0)$ and $-q$ at $(0,+a / 4,0)$. Choose the correct options(s)
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/u23N_x2xO_WSWDxMB8NBywx72Fb4WfQkP4LhZc8YNx8.original.fullsize.png"/><br/></div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">The net electric flux crossing the plane $x=+a / 2$ is equal to the net electric flux crossing the plane $x=-$ $a / 2$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">The net electric flux crossing the plane $y=+a / 2$ is more than the net electric flux crossing the plane $y=-a / 2$.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">The net electric flux crossing the entire region is $\frac{q}{\varepsilon_{0}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">The net electric flux crossing the plane $z=+a / 2$ is equal to the net electric flux crossing the plane $x=+a / 2$.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">The net electric flux crossing the plane $x=+a / 2$ is equal to the net electric flux crossing the plane $x=-$ $a / 2$, The net electric flux crossing the entire region is $\frac{q}{\varepsilon_{0}}$, The net electric flux crossing the plane $z=+a / 2$ is equal to the net electric flux crossing the plane $x=+a / 2$.</span> </div> | <div class="solution">Due to symmetry, the net electric flux passing through $x=+\frac{a}{2}$, $x=-\frac{a}{2}, z=+\frac{a}{2}$ is same
<br/> <br/>According to Gauss's theorem, net electric flux net electric flux crossing through any closed surface $\phi=\frac{q_{\text {in }}}{\varepsilon_{0}}=$ $\frac{-q+3 q-q}{\varepsilon_{0}}=\frac{q}{\varepsilon_{0}}$.</div> | MarksBatch0.db |
37 | A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it, the correct statement(s) is(are) | a-current-carrying-infinitely-long-wire-is-kept-along-the-diameter-of-a-circular-wire-loop-without-19618 | <div class="question">A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it, the correct statement(s) is(are)</div> | ['Physics', 'Electromagnetic Induction', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">The emf induced in the loop is zero if the current is constant.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">The emf induced in the loop is finite if the current is constant.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">The emf induced in the loop is zero if the current decreases at a steady rate.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">The emf induced in the loop is infinite if the current decreases at a steady rate.</span> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">The emf induced in the loop is zero if the current is constant., The emf induced in the loop is zero if the current decreases at a steady rate.</span> </div> | <div class="solution">If the current is constant, the emf induced in the loop zero. Emf will be induced in the circular wire loop when flux through it changes with time.
<br/> <br/>$\mathrm{e}=-\frac{\Delta \phi}{\Delta \mathrm{t}}$
<br/> <br/>when the current is constant, the flux changing through it will be zero.
<br/> <br/>Also, if the current decreases at steady rate, the emf. induced in the loop is zero.
<br/> <br/>When the current is decreasing at a steady rate then the change in the flux (decreasing inwards) on the right half of the wire is equal to the change in flux (decreasing outwards) on the left half of the wire such that $\Delta \phi$ through the circular loop is zero.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/pMb-i-hx_6TMOAXEfVyhqx49W_QA3YpqMVkpEO5sdg4.original.fullsize.png"/></div> | MarksBatch0.db |
38 | A cylindrical cavity of diameter a exists inside a cylinder of diameter 2 a as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by 12 N â μ 0 â a J , then the value of N is | a-cylindrical-cavity-of-diameter-a-exists-inside-a-cylinder-of-diameter-2-a-as-shown-in-the-figure-39143 | <div class="question">A cylindrical cavity of diameter a exists inside a cylinder of diameter $2 a$ as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density $J$ flows along the length. If the magnitude of the magnetic field at the point $P$ is given by $\frac{N}{12} \mu_{0} a J$, then the value of $N$ is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/YfqH0WzwUDjeUhuXA7otppktNta3TUFOyi7I8kErGug.original.fullsize.png"/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">5</span> </div> | <div class="solution">Current density $J=\frac{\text { current }}{\text { area }}=\frac{I}{A} \Rightarrow I=\mathrm{JA}$ Magnetic field $\mathrm{B}_{\mathrm{R}}$ after removing cavity $(\mathrm{C})$ $\mathrm{B}_{\mathrm{R}}=\mathrm{B}_{\text {total }}-\mathrm{B}_{\text {cavity }}$
<br/> <br/>$\begin{array}{l}
<br/> <br/>\frac{\mu_{0} I_{t}}{2 \pi a}-\frac{\mu_{0} I_{c}}{2 \pi\left(\frac{3}{2} a\right)} \\
<br/> <br/>\quad=\frac{\mu_{0}}{\pi a}\left[\frac{I_{t}}{2}-\frac{I_{c}}{3}\right]\left(\text { here } I_{t}=J\left(\pi a^{2}\right) I_{c}=J\left(\frac{\pi a^{2}}{4}\right)\right) \\
<br/> <br/>=\frac{\mu_{0}}{\pi a}\left[\frac{\pi a^{2} J}{2}-\frac{\pi a^{2} J}{12}\right] \\
<br/> <br/>\text { or, } \mathrm{B}_{\mathrm{R}}=\frac{5 \mu_{0} a J}{12}
<br/> <br/>\end{array}$
<br/> <br/>Comparing it with $\frac{N}{12} \mu_{0} a J$ We get $N=5$</div> | MarksBatch0.db |
39 | A cylindrical furnace has height H and diameter D both 1 m . It is maintained at temperature 360 K . The air gets heated inside the furnace at constant pressure P a and its temperature becomes T = 360 K . The hot air with density Ï rises up a vertical chimney of diameter d = 0 . 1 m and height h = 9 m above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density Ï a = 1 . 2 kg m - 3 , pressure P a and temperature T a = 300 K enters the furnace. Assume air as an ideal gas, neglect the variations in Ï and T inside the chimney and the furnace. Also ignore the viscous effects. [Given: The acceleration due to gravity g = 10 m s - 2 and Ï = 3 . 14 ] When the chimney is closed using a cap at the top, a pressure difference â P develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of â P is _____ N m â 2 . | a-cylindrical-furnace-has-height-h-and-diameter-d-both-1-m-it-is-maintained-at-temperature-360-k-78555 | <div class="question"><p>A cylindrical furnace has height <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>H</mi></mfenced></math> and diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>D</mi></mfenced></math> both <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math>. It is maintained at temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>360</mn><mo> </mo><mi mathvariant="normal">K</mi></math>. The air gets heated inside the furnace at constant pressure <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub></math> and its temperature becomes <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi><mo>=</mo><mn>360</mn><mo> </mo><mi mathvariant="normal">K</mi></math>. The hot air with density <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ρ</mi></math> rises up a vertical chimney of diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mo>=</mo><mn>0</mn><mo>.</mo><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math> and height <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>h</mi><mo>=</mo><mn>9</mn><mo> </mo><mi mathvariant="normal">m</mi></math> above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>ρ</mi><mi>a</mi></msub><mo>=</mo><mn>1</mn><mo>.</mo><mn>2</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>-</mo><mn>3</mn></mrow></msup></math>, pressure <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub></math> and temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>T</mi><mi>a</mi></msub><mo>=</mo><mn>300</mn><mo> </mo><mi mathvariant="normal">K</mi></math> enters the furnace. Assume air as an ideal gas, neglect the variations in <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ρ</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi></math> inside the chimney and the furnace. Also ignore the viscous effects.</p><br/><br/><p>[Given: The acceleration due to gravity <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>2</mn></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>π</mi><mo>=</mo><mn>3</mn><mo>.</mo><mn>14</mn></math>]</p><br/><br/><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3fbf9ea5-3f88-41c1-b37d-775e24fecac2.png" style="width: 357px; height: 652px;"/></p><br/><br/>When the chimney is closed using a cap at the top, a pressure difference <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∆</mo><mi>P</mi></math> develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∆</mo><mi>P</mi></math> is _____ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">N</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>−</mo><mn>2</mn></mrow></msup></math>.</div> | ['Physics', 'Mechanical Properties of Fluids', 'JEE Advanced', 'JEE Advanced 2023 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7850</span> </div> | <div class="solution"><blockquote><p>Applying Bernoulli’s theorem between top and bottom points of furnace :</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>ρ</mi><mi>a</mi></msub><msup><mi>V</mi><mn>2</mn></msup><mo>=</mo><msub><mi>P</mi><mi>a</mi></msub><mo>+</mo><mi>ρ</mi><mi>g</mi><mi>H</mi><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>ρ</mi><msup><mi>V</mi><mn>2</mn></msup><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mfenced><mn>1</mn></mfenced></math></p><p>Also, since <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mi>M</mi><mo>=</mo><mi>ρ</mi><mi>R</mi><mi>T</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><msub><mi>ρ</mi><mi>a</mi></msub><mo>×</mo><mn>300</mn><mo>=</mo><mi>ρ</mi><mo>×</mo><mn>360</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>ρ</mi><mo>=</mo><mn>1</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>-</mo><mn>3</mn></mrow></msup><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mfenced><mn>2</mn></mfenced></math></p><p>From equation(1), we can write</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mfrac><msup><mi>V</mi><mn>2</mn></msup><mn>2</mn></mfrac><mfenced><mrow><mn>0</mn><mo>.</mo><mn>2</mn></mrow></mfenced><mo>=</mo><mn>1</mn><mo>×</mo><mn>10</mn><mo>×</mo><mn>1</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>V</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p>Now, let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mo>'</mo></math> be the speed of air in chimney.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> By continuity equation,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi mathvariant="normal">π</mi><msup><mi>D</mi><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mo>·</mo><mi>V</mi><mo>=</mo><mfrac><mrow><mi mathvariant="normal">π</mi><msup><mi>d</mi><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mo>·</mo><mi>V</mi><mo>'</mo></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>V</mi><mo>'</mo><mo>=</mo><mn>100</mn><mi>V</mi><mo>=</mo><mn>1000</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p>Therefore, pressure difference <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∆</mo><mi>p</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mi>ρ</mi><mi>V</mi><msup><mo>'</mo><mn>2</mn></msup><mo>·</mo></math>area</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>1</mn><mo>×</mo><msup><mfenced><mn>1000</mn></mfenced><mn>2</mn></msup><mo>×</mo><mfrac><mrow><mi>π</mi><msup><mi>d</mi><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mspace linebreak="newline"></mspace><mo>=</mo><mn>1</mn><mo>×</mo><msup><mfenced><mn>1000</mn></mfenced><mn>2</mn></msup><mo>×</mo><mfrac><mrow><mn>3</mn><mo>.</mo><mn>14</mn><mo>×</mo><msup><mfenced><mrow><mn>0</mn><mo>.</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mspace linebreak="newline"></mspace><mo>=</mo><mn>7850</mn><mo> </mo><mi mathvariant="normal">N</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mn>2</mn></msup></math></p></blockquote></div> | MarksBatch0.db |
40 | A cylindrical furnace has height H and diameter D both 1 m . It is maintained at temperature 360 K . The air gets heated inside the furnace at constant pressure P a and its temperature becomes T = 360 K . The hot air with density Ï rises up a vertical chimney of diameter d = 0 . 1 m and height h = 9 m above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density Ï a = 1 . 2 kg m - 3 , pressure P a and temperature T a = 300 K enters the furnace. Assume air as an ideal gas, neglect the variations in Ï and T inside the chimney and the furnace. Also ignore the viscous effects. [Given: The acceleration due to gravity g = 10 m s - 2 and Ï = 3 . 14 ] Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is _____ gm s â 1 . | a-cylindrical-furnace-has-height-h-and-diameter-d-both-1-m-it-is-maintained-at-temperature-360-k-53309 | <div class="question"><p>A cylindrical furnace has height <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>H</mi></mfenced></math> and diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>D</mi></mfenced></math> both <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math>. It is maintained at temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>360</mn><mo> </mo><mi mathvariant="normal">K</mi></math>. The air gets heated inside the furnace at constant pressure <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub></math> and its temperature becomes <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi><mo>=</mo><mn>360</mn><mo> </mo><mi mathvariant="normal">K</mi></math>. The hot air with density <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ρ</mi></math> rises up a vertical chimney of diameter <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mo>=</mo><mn>0</mn><mo>.</mo><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math> and height <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>h</mi><mo>=</mo><mn>9</mn><mo> </mo><mi mathvariant="normal">m</mi></math> above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>ρ</mi><mi>a</mi></msub><mo>=</mo><mn>1</mn><mo>.</mo><mn>2</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>-</mo><mn>3</mn></mrow></msup></math>, pressure <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub></math> and temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>T</mi><mi>a</mi></msub><mo>=</mo><mn>300</mn><mo> </mo><mi mathvariant="normal">K</mi></math> enters the furnace. Assume air as an ideal gas, neglect the variations in <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ρ</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi></math> inside the chimney and the furnace. Also ignore the viscous effects.</p><br/><br/><p>[Given: The acceleration due to gravity <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>2</mn></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>π</mi><mo>=</mo><mn>3</mn><mo>.</mo><mn>14</mn></math>]</p><br/><br/><p><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/3fbf9ea5-3f88-41c1-b37d-775e24fecac2.png" style="width: 357px; height: 652px;"/></p><br/><br/>Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is _____ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>gm</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>−</mo><mn>1</mn></mrow></msup></math>.</div> | ['Physics', 'Mechanical Properties of Fluids', 'JEE Advanced', 'JEE Advanced 2023 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">7850</span> </div> | <div class="solution"><blockquote><p>Applying Bernoulli’s theorem between top and bottom points of furnace :</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>P</mi><mi>a</mi></msub><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>ρ</mi><mi>a</mi></msub><msup><mi>V</mi><mn>2</mn></msup><mo>=</mo><msub><mi>P</mi><mi>a</mi></msub><mo>+</mo><mi>ρ</mi><mi>g</mi><mi>H</mi><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>ρ</mi><msup><mi>V</mi><mn>2</mn></msup><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mfenced><mn>1</mn></mfenced></math></p><p>Also, since <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mi>M</mi><mo>=</mo><mi>ρ</mi><mi>R</mi><mi>T</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><msub><mi>ρ</mi><mi>a</mi></msub><mo>×</mo><mn>300</mn><mo>=</mo><mi>ρ</mi><mo>×</mo><mn>360</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>ρ</mi><mo>=</mo><mn>1</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">m</mi><mrow><mo>-</mo><mn>3</mn></mrow></msup><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mfenced><mn>2</mn></mfenced></math></p><p>From equation(1), we can write</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mfrac><msup><mi>V</mi><mn>2</mn></msup><mn>2</mn></mfrac><mfenced><mrow><mn>0</mn><mo>.</mo><mn>2</mn></mrow></mfenced><mo>=</mo><mn>1</mn><mo>×</mo><mn>10</mn><mo>×</mo><mn>1</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>V</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p>Required mass flow rate <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mi>V</mi><mo>×</mo><mfrac><mrow><mi>π</mi><msup><mi>D</mi><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mo>×</mo><mi>ρ</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>10</mn><mo>×</mo><mfrac><mrow><mn>3</mn><mo>.</mo><mn>14</mn></mrow><mn>4</mn></mfrac><mo>×</mo><msup><mn>1</mn><mn>2</mn></msup><mo>×</mo><mn>1</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>7</mn><mo>.</mo><mn>85</mn><mo> </mo><mi>kg</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>7850</mn><mo> </mo><mi>gm</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p></blockquote></div> | MarksBatch0.db |
41 | A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it upto height H . Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm . Find the fall in height (in mm ) of water level due to opening of the orifice. [Take atmospheric pressure = 1.0 Ã 1 0 5 Nm â 2 , density of water = 1000 kg m â 3 and g = 10 ms â 2 . Neglect any effect of surface tension.] | a-cylindrical-vessel-of-height-500-mm-has-an-orifice-small-hole-at-its-bottom-the-orifice-is-init-59533 | <div class="question">A cylindrical vessel of height $500 \mathrm{~mm}$ has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it upto height $H$. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 $\mathrm{mm}$. Find the fall in height (in $\mathrm{mm}$ ) of water level due to opening of the orifice.<br/>[Take atmospheric pressure $=1.0 \times 10^5 \mathrm{Nm}^{-2}$, density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ and $g=10$ $\mathrm{ms}^{-2}$. Neglect any effect of surface tension.]</div> | ['Physics', 'Mechanical Properties of Fluids', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">6</span> </div> | <div class="solution">In this question we will have to assume that temperature of enclosed air about water is constant (or $p V=$ constant)<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/8dM8QYJxDmaHvc7TZMSYMcDZb9J5Qzf03BYUgRzDGn4.original.fullsize.png"/><br/><br/><br/>$$<br/>\begin{gathered}<br/>p=p_0-\rho g h \\<br/>p_0[A(500-H)]=p[A(200)]<br/>\end{gathered}<br/>$$<br/>Solving these two equations, we get<br/>$$<br/>\begin{aligned}<br/>H & =206 \mathrm{~mm} \\<br/>\therefore \quad \text { Level fall } & =(206-200) \mathrm{mm} \\<br/>& =6 \mathrm{~mm}<br/>\end{aligned}<br/>$$</div> | MarksBatch0.db |
42 | A decapeptide (molecular weight 796) on complete hydrolysis gives glycine (molecular weight 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is | a-decapeptide-molecular-weight-796-on-complete-hydrolysis-gives-glycine-molecular-weight-75-ala-21183 | <div class="question">A decapeptide (molecular weight 796) on complete hydrolysis gives glycine (molecular weight 75), alanine and phenylalanine. Glycine contributes $47.0 \%$ to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is</div> | ['Chemistry', 'Biomolecules', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">6</span> </div> | <div class="solution">$$<br/>\text { A decapeptide has nine peptide (amide) linkage as }<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/QRsCQ14li6NT9gp0NOovuoPzTGcmi68dMG0a74dUtUg.original.fullsize.png"/><br/><br/>Therefore, no hydrolysis, it will absorb nine water molecules.<br/>Hence, total mass of hydrolysis product $=796+18 \times 9=958$<br/>$\Rightarrow$ mass of glycine in hydrolysis product $=\frac{958 \times 47}{100}=450$<br/>$\Rightarrow$ number of glycine molecule in one molecule of decapeptide $=\frac{450}{75}=6$</div> | MarksBatch0.db |
43 | A diatomic ideal gas is compressed adiabatically to 32 1 â of its initial volume. If the initial temperature of the gas is T i â T i â (in kelvin) and the final temperature is a T i â , the value of a is | a-diatomic-ideal-gas-is-compressed-adiabatically-to-32-1-of-its-initial-volume-if-the-initial-t-29573 | <div class="question">A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_i$ $T_i$ (in kelvin) and the final temperature is $a T_i$, the value of $a$ is</div> | ['Physics', 'Thermodynamics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)'] | None | <div class="correct-answer">
The correct answer is:
<span class="option-value">4</span> </div> | <div class="solution">In adiabatic process,<br/>$$<br/>\begin{aligned}<br/>& T V^{\gamma-1}=\text { constant } \\<br/>& \therefore T_i V_i^{0.4}=T_f V_f^{0.4} \\<br/>& \text { (as } \gamma=1.4 \text { for diatomic gas) } \\<br/>& \text { or } \quad T_i V_i^{0.4}=\left(\alpha T_i\right)\left(\frac{V_i}{32}\right)^{0.4} \\<br/>& \text { or } \quad \alpha(32)^{0.4}=4 \\<br/>&<br/>\end{aligned}<br/>$$<br/>$\therefore$ The correct answer is 4 .</div> | MarksBatch0.db |
44 | A disk of radius 4 a â having a uniformly distributed charge 6 C is placed in the x â y plane with its centre at ( 2 â a â , 0 , 0 ) . A rod of length a carrying a uniformly distributed charge 8 C is placed on the x -axis from x = 4 a â to x = 4 5 a â . Two point charges â 7 C and 3 C are placed at ( 4 a â , 4 â a â , 0 ) and ( 4 â 3 a â , 4 3 a â , 0 ) , respectively. Consider a cubical surface formed by six surfaces x = ± 2 a â , y = ± 2 a â , z = ± 2 a â . The electric flux through this cubical surface is | a-disk-of-radius-4-a-having-a-uniformly-distributed-charge-6-c-is-placed-in-the-x-y-plane-wi-97430 | <div class="question">A disk of radius $\frac{a}{4}$ having a uniformly distributed charge $6 \mathrm{C}$ is placed in the $x-y$ plane with its centre at $\left(\frac{-a}{2}, 0,0\right)$. A rod of length a carrying a uniformly distributed charge $8 \mathrm{C}$ is placed on the $x$-axis from $x=\frac{a}{4}$ to $x=\frac{5 a}{4}$. Two point charges $-7 \mathrm{C}$ and $3 \mathrm{C}$ are placed at $\left(\frac{a}{4}, \frac{-a}{4}, 0\right)$ and $\left(\frac{-3 a}{4}, \frac{3 a}{4}, 0\right)$, respectively. Consider a cubical surface formed by six surfaces $x=\pm \frac{a}{2}, y=\pm \frac{a}{2}, z=\pm \frac{a}{2}$.<br/>The electric flux through this cubical surface is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/AvcnvTT5sePtQoutuhERbdjweJ91m1yd8m1dkYFaLNw.original.fullsize.png"/><br/></div> | ['Physics', 'Electrostatics', 'JEE Main'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{-2 C}{\varepsilon_0}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2 C}{\varepsilon_0}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{10 C}{\varepsilon_0}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{12 C}{\varepsilon_0}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{-2 C}{\varepsilon_0}$<br/></span> </div> | <div class="solution">Total enclosed charge as already shown is<br/>$$<br/>q_{\text {net }}=\frac{6 C}{2}+\frac{8 C}{4}-7 C=-2 C<br/>$$<br/>From Gauss-theorem, net flux,<br/>$$<br/>\varphi_{\text {net }}=\frac{q_{\text {net }}}{\varepsilon_0}=\frac{-2 C}{\varepsilon_0}<br/>$$</div> | MarksBatch0.db |
45 | A disk of radius 4 a â having a uniformly distributed charge 6 C is placed in the x â y plane with its centre at ( 2 â a â , 0 , 0 ) . A rod of length a carrying a uniformly distributed charge 8 C is placed on the x -axis from x = 4 a â to x = 4 5 a â . Two point charges â 7 C and 3 C are placed at ( 4 a â , 4 â a â , 0 ) and ( 4 â 3 a â , 4 3 a â , 0 ) , respectively. Consider a cubical surface formed by six surfaces x = ± 2 a â , y = ± 2 a â , z = ± 2 a â . The electric flux through this cubical surface is | a-disk-of-radius-4-a-having-a-uniformly-distributed-charge-6-c-is-placed-in-the-x-y-plane-wi-33837 | <div class="question">A disk of radius $\frac{a}{4}$ having a uniformly distributed charge $6 \mathrm{C}$ is placed in the $x-y$ plane with its centre at $\left(\frac{-a}{2}, 0,0\right)$. A rod of length a carrying a uniformly distributed charge $8 \mathrm{C}$ is placed on the $x$-axis from $x=\frac{a}{4}$ to $x=\frac{5 a}{4}$. Two point charges $-7 \mathrm{C}$ and $3 \mathrm{C}$ are placed at $\left(\frac{a}{4}, \frac{-a}{4}, 0\right)$ and $\left(\frac{-3 a}{4}, \frac{3 a}{4}, 0\right)$, respectively. Consider a cubical surface formed by six surfaces $x=\pm \frac{a}{2}, y=\pm \frac{a}{2}, z=\pm \frac{a}{2}$.<br/>The electric flux through this cubical surface is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/AvcnvTT5sePtQoutuhERbdjweJ91m1yd8m1dkYFaLNw.original.fullsize.png"/><br/></div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{-2 C}{\varepsilon_0}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2 C}{\varepsilon_0}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{10 C}{\varepsilon_0}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{12 C}{\varepsilon_0}$</span> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$\frac{-2 C}{\varepsilon_0}$<br/></span> </div> | <div class="solution">Total enclosed charge as already shown is<br/>$$<br/>q_{\text {net }}=\frac{6 C}{2}+\frac{8 C}{4}-7 C=-2 C<br/>$$<br/>From Gauss-theorem, net flux,<br/>$$<br/>\varphi_{\text {net }}=\frac{q_{\text {net }}}{\varepsilon_0}=\frac{-2 C}{\varepsilon_0}<br/>$$</div> | MarksBatch0.db |
46 | A disk of radius R with uniform positive charge density Ï is placed on the x y plane with its center at the origin. The Coulomb potential along the z -axis is V z = Ï 2 ϵ 0 R 2 + z 2 - z A particle of positive charge q is placed initially at rest at a point on the z -axis with z = z 0 and z 0 > 0 . In addition to the Coulomb force, the particle experiences a vertical force F â = - c k ^ with c > 0 . Let β = 2 c ε 0 q Ï . Which of the following statement(s) is(are) correct? | a-disk-of-radius-r-with-uniform-positive-charge-density-is-placed-on-the-x-y-plane-with-its-cente-24247 | <div class="question"><p>A disk of radius <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> with uniform positive charge density <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>σ</mi></math> is placed on the <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mi>y</mi></math> plane with its center at the origin. The Coulomb potential along the <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi></math>-axis is<br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mfenced><mi>z</mi></mfenced><mo>=</mo><mfrac><mi>σ</mi><mrow><mn>2</mn><msub><mi>ϵ</mi><mn>0</mn></msub></mrow></mfrac><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></math></p><p>A particle of positive charge <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>q</mi></math> is placed initially at rest at a point on the <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi></math>-axis with <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><msub><mi>z</mi><mn>0</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>></mo><mn>0</mn></math>. In addition to the Coulomb force, the particle experiences a vertical force <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>F</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mi>c</mi><mover><mi>k</mi><mo>^</mo></mover></math> with <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>c</mi><mo>></mo><mn>0</mn></math>. Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mrow><mn>2</mn><mi>c</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mrow><mi>q</mi><mi>σ</mi></mrow></mfrac></math>. Which of the following statement(s) is(are) correct?</p></div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2022 (Paper 2)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mn>25</mn><mn>7</mn></mfrac><mi>R</mi></math>, the particle reaches the origin.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mn>3</mn><mn>7</mn></mfrac><mi>R</mi></math>, the particle reaches the origin.</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mi>R</mi><msqrt><mn>3</mn></msqrt></mfrac></math>, the particle returns back to <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><msub><mi>z</mi><mn>0</mn></msub></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>></mo><mn>1</mn></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>></mo><mn>0</mn></math>, the particle always reaches the origin.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value">For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mn>25</mn><mn>7</mn></mfrac><mi>R</mi></math>, the particle reaches the origin., For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>=</mo><mfrac><mi>R</mi><msqrt><mn>3</mn></msqrt></mfrac></math>, the particle returns back to <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><msub><mi>z</mi><mn>0</mn></msub></math>, For <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>β</mi><mo>></mo><mn>1</mn></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub><mo>></mo><mn>0</mn></math>, the particle always reaches the origin.</span> </div> | <div class="solution"><p>For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>c</mi><mi>z</mi><mo>≥</mo><mfrac><mrow><mi>σ</mi><mi>q</mi></mrow><mrow><mn>2</mn><msub><mi>ϵ</mi><mn>0</mn></msub></mrow></mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mspace linebreak="newline"></mspace><mo>⇒</mo><mfrac><mrow><mi>c</mi><mn>2</mn><msub><mi>ϵ</mi><mn>0</mn></msub></mrow><mrow><mi>σ</mi><mi>q</mi></mrow></mfrac><mo>≥</mo><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mspace linebreak="newline"></mspace><mo>⇒</mo><mi>β</mi><mo>≥</mo><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac></math></p><p>For option A</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mo>=</mo><mfrac><mrow><mi>R</mi><mo>-</mo><mi>R</mi><mfenced><mrow><msqrt><mn>1</mn><mo>+</mo><mstyle displaystyle="true"><mfrac><mn>625</mn><mn>49</mn></mfrac></mstyle></msqrt><mo>-</mo><mstyle displaystyle="true"><mfrac><mn>25</mn><mn>7</mn></mfrac></mstyle></mrow></mfenced></mrow><mrow><mi>R</mi><mstyle displaystyle="true"><mfrac><mn>25</mn><mn>7</mn></mfrac></mstyle></mrow></mfrac><mo>≃</mo><mn>0</mn><mo>.</mo><mn>242</mn></math></p><p>Since, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn><mo>.</mo><mn>25</mn><mo>></mo><mn>0</mn><mo>.</mo><mn>242</mn></math></p><p>Option A is correct.</p><p>For option B</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mo>=</mo><mfrac><mrow><mi>R</mi><mo>-</mo><mi>R</mi><mfenced><mrow><msqrt><mn>1</mn><mo>+</mo><mstyle displaystyle="true"><mfrac><mn>9</mn><mn>49</mn></mfrac></mstyle></msqrt><mo>-</mo><mstyle displaystyle="true"><mfrac><mn>3</mn><mn>7</mn></mfrac></mstyle></mrow></mfenced></mrow><mrow><mi>R</mi><mstyle displaystyle="true"><mfrac><mn>3</mn><mn>7</mn></mfrac></mstyle></mrow></mfrac><mo>≃</mo><mn>0</mn><mo>.</mo><mn>79</mn></math></p><p>Since, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn><mo>.</mo><mn>25</mn><mo><</mo><mn>0</mn><mo>.</mo><mn>79</mn></math></p><p>Option B is incorrect</p><p>For option C</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mo>=</mo><mfrac><mrow><mi>R</mi><mo>-</mo><mi>R</mi><mfenced><mrow><msqrt><mn>1</mn><mo>+</mo><mstyle displaystyle="true"><mfrac><mn>1</mn><mn>3</mn></mfrac></mstyle></msqrt><mo>-</mo><mstyle displaystyle="true"><mfrac><mn>1</mn><msqrt><mn>3</mn></msqrt></mfrac></mstyle></mrow></mfenced></mrow><mrow><mi>R</mi><mstyle displaystyle="true"><mfrac><mn>1</mn><msqrt><mn>3</mn></msqrt></mfrac></mstyle></mrow></mfrac><mo>≃</mo><mn>0</mn><mo>.</mo><mn>73</mn></math></p><p>Since, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn><mo>.</mo><mn>25</mn><mo><</mo><mn>0</mn><mo>.</mo><mn>73</mn></math></p><p>Therefore, the particle returns to <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mn>0</mn></msub></math></p><p>Hence, option C is correct.</p><p>For option D,</p><p>For any value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>></mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mfenced close="]" open="["><mrow><mi>R</mi><mo>-</mo><mfenced><mrow><msqrt><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup></msqrt><mo>-</mo><mi>z</mi></mrow></mfenced></mrow></mfenced><mi>z</mi></mfrac><mo><</mo><mn>1</mn></math></p><p>Hence, option D is correct.</p></div> | MarksBatch0.db |
47 | A double star system consists of two stars A and B which have time period T A â and T B â . Radius R A â and R B â and mass M A â and M B â . Choose the correct option. | a-double-star-system-consists-of-two-stars-a-and-b-which-have-time-period-t-a-and-t-b-radi-89463 | <div class="question">A double star system consists of two stars $A$ and $B$ which have time period $T_A$ and $T_B$. Radius $R_A$ and $R_B$ and mass $M_A$ and $M_B$. Choose the correct option.</div> | ['Physics', 'Gravitation', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>If $T_A>T_B$, then $R_A>R_B$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>If $T_A>T_B$, then $M_A>M_B$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\left(\frac{T_A}{T_B}\right)^2=\left(\frac{R_A}{R_B}\right)^3$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$T_A=T_B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$T_A=T_B$</span> </div> | <div class="solution">In case of binary star system angular velocity and hence the time period of both the stars are equal.</div> | MarksBatch0.db |
48 | A few electric field lines for a system of two charges Q 1 â and Q 2 â fixed at two different points on the x -axis are shown in the figure. These lines suggest that | a-few-electric-field-lines-for-a-system-of-two-charges-q-1-and-q-2-fixed-at-two-different-po-67686 | <div class="question">A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/R5lzBNcidnS01bSprkXrbMp3l9bUBkdMt6YycbuULb8.original.fullsize.png"/><br/></div> | ['Physics', 'Electrostatics', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\left|Q_1\right|>\left|Q_2\right|$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\left|Q_1\right| < \left|Q_2\right|$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>at a finite distance to the left of $Q_1$ the electric field is zero<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>at a finite distance to the right of $Q_2$ the electric field is zero</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>$\left|Q_1\right|>\left|Q_2\right|$<br/>, <br/>at a finite distance to the right of $Q_2$ the electric field is zero</span> </div> | <div class="solution">From the behaviour of electric lines, we can say that $Q_1$ is positive and $Q_2$ is negative. Further, $\left|Q_1\right|>\left|Q_2\right|$<br/>At some finite distance to the right of $Q_2$, electric field will be zero. Because electric field due to $Q_1$ is towards right (away from $Q_1$ ) and due to $Q_2$ is towards left (towards $Q_2$ ). But since magnitude of $Q_1$ is more, the two fields may cancel each other because distance of that point from $Q_1$ will also be more.<br/>$\therefore$ The correct options are (a) and (d).</div> | MarksBatch0.db |
49 | A field line is shown in the figure. This field cannot represent | a-field-line-is-shown-in-the-figure-this-field-cannot-represent-34016 | <div class="question">A field line is shown in the figure. This field cannot represent<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/sBuAndXrVPaCS8wD88iMYjym52y85QgiPtevdojilzs.original.fullsize.png"/><br/></div> | ['Physics', 'Magnetic Effects of Current', 'JEE Advanced', 'JEE Advanced 2006'] | <ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>Magnetic field<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Electrostatic field<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Induced electric field<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>Gravitational field</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul> | <div class="correct-answer">
The correct answers are:
<span class="option-value"><br/>Magnetic field<br/>, <br/>Gravitational field</span> </div> | <div class="solution">Electrostatic and gravitational field do not make closed loops.</div> | MarksBatch0.db |
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