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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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,
$$2Tl = mg$$
$$T = {{mg} \over {2l}} = {{1.5 \times {{10}^{ - 2}}} \over {2 \times 30 \times {{10}^{ - 2}}}} = {{1.5} \over {60}}$$
$$ = 0.025\,N/m = 0.025Nm$$

Questions Asked from Properties of Matter

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