 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

An open glass tube is immersed in mercury in such a way that a length of $8$ $cm$ extends above the mercury level. The open end of the tube is then closed and scaled and the tube is raised vertically up by additional $46$ $cm$. What will be length of the air column above mercury in the tube now? (Atmospheric pressure $=76$ $cm$ of $Hg$)
A
$16$ $cm$
B
$22$ $cm$
C
$38$ $cm$
D
$6$ $cm$

Explanation Length of the air column above mercury in the tube is,
$P + x = {P_0}$
$\Rightarrow P = \left( {76 - x} \right)$
$\Rightarrow 8 \times A \times 76 = \left( {76 - x} \right) \times A \times \left( {54 - x} \right)$
$\therefore$ $x=38$
Thus, length of air column $=54-38=16cm.$
2

JEE Main 2013 (Offline)

Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible ? The surface tension is $T,$ density of liquid is $\rho$ and $L$ is its latent heat of vaporization.
A
$\rho L/T$
B
$\sqrt {T/\rho L}$
C
$T/\rho L$
D
$2T/\rho L$

Explanation

When radius is decrease by $\Delta R,$
$4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right]$
$\Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right]$
$\Rightarrow \rho {R^2}\Delta RL = T2R\Delta R\,\,$ [ $\Delta R$ is very small ]
$\Rightarrow R = {{2T} \over {\rho L}}$
3

JEE Main 2013 (Offline)

A uniform cylinder of length $L$ and mass $M$ having cross-sectional area $A$ is suspended, with its length vertical, from a fixed point by a mass-less spring such that it is half submerged in a liquid of density $\sigma$ at equilibrium position. The extension ${x_0}$ of the spring when it is in equilibrium is:
A
${{Mg} \over k}$
B
${{Mg} \over k}\left( {1 - {{LA\sigma } \over M}} \right)$
C
${{Mg} \over k}\left( {1 - {{LA\sigma } \over {2M}}} \right)$
D
${{Mg} \over k}\left( {1 + {{LA\sigma } \over M}} \right)$

Explanation From figure, $k{x_0} + {F_B} = Mg$
$k{x_0} + \sigma {L \over 2}Ag = Mg$
[ as mass $=$ density $\times$ volume ]
$\Rightarrow k{x_0} = Mg - \sigma {L \over 2}Ag$
$\Rightarrow {x_0} = {{Mg - {{\sigma LAg} \over 2}} \over k}$
$= {{Mg} \over k}\left( {1 - {{LA\sigma } \over {2M}}} \right)$
4

AIEEE 2012

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of $1.5 \times {10^{ - 2}}\,\,N$ (see figure). The length of the slider is $30$ $cm$ and its weight negligible. The surface tension of the liquid film is A
$0.0125\,\,N{m^{ - 1}}$
B
$0.1\,\,N{m^{ - 1}}$
C
$0.05\,\,N{m^{ - 1}}$
D
$0.025\,\,N{m^{ - 1}}$

Explanation

At equilibrium,
$2Tl = mg$
$T = {{mg} \over {2l}} = {{1.5 \times {{10}^{ - 2}}} \over {2 \times 30 \times {{10}^{ - 2}}}} = {{1.5} \over {60}}$
$= 0.025\,N/m = 0.025Nm$