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JEE Mains Previous Years Questions with Solutions

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1

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

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

AIEEE 2011

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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
$$W = 2T4\pi \left[ {{{\left( 5 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}}$$
$$ = 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}\,J$$
$$ = 0.4\pi \times {10^{ - 3}}\,J$$
$$ = 0.4\pi mJ$$

Questions Asked from Properties of Matter

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