 JEE Mains Previous Years Questions with Solutions

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1

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 ]
$\Rightarrow k{x_0} = Mg - \sigma {L \over 2}Ag$
$\Rightarrow {x_0} = {{Mg - {{\sigma LAg} \over 2}} \over k}$
$= {{Mg} \over k}\left( {1 - {{LA\sigma } \over {2M}}} \right)$
2

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,
$2Tl = mg$
$T = {{mg} \over {2l}} = {{1.5 \times {{10}^{ - 2}}} \over {2 \times 30 \times {{10}^{ - 2}}}} = {{1.5} \over {60}}$
$= 0.025\,N/m = 0.025Nm$
3

AIEEE 2011

Water is flowing continuously from a tap having an internal diameter $8 \times {10^{ - 3}}\,\,m.$ The water velocity as it leaves the tap is $0.4\,\,m{s^{ - 1}}$ . The diameter of the water stream at a distance $2 \times {10^{ - 1}}\,\,m$ below the tap is close to :
A
$7.5 \times {10^{ - 3}}m$
B
$9.6 \times {10^{ - 3}}m$
C
$3.6 \times {10^{ - 3}}m$
D
$5.0 \times {10^{ - 3}}m$

Explanation

From Bernoulli's theorem,
${P_0} + {1 \over 2}\rho v_1^2\rho gh = {P_0} + {1 \over 2}\rho v_2^2 + 0$
${v_2} = \sqrt {v_1^2 + 2gh}$
$= \sqrt {0.16 + 2 \times 10 \times 0.2}$
$= 2.03\,m/s$
From equation of continuity
${A_2}{v_2} = {A_1}{v_1}$
$\pi {{D_2^2} \over 4} \times {v_2} = \pi {{D_1^2} \over 4}{v_1}$
$\Rightarrow \,\,{D_1} = {D_2}\sqrt {{{{v_1}} \over {{v_2}}}} = 3.55 \times {10^{ - 3}}m$
4

AIEEE 2011

Work done in increasing the size of a soap bubble from a radius of $3$ $cm$ to $5$ $cm$ is nearly (Surface tension of soap solution $= 0.03N{m^{ - 1}},$
A
$0.2\pi mJ$
B
$2\pi mJ$
C
$0.4\pi mJ$
D
$4\pi mJ$

Explanation

$W = T \times \,\,$ change in surface area
$W = 2T4\pi \left[ {{{\left( 5 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}}$
$= 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}\,J$
$= 0.4\pi \times {10^{ - 3}}\,J$
$= 0.4\pi mJ$