JEE Mains Previous Years Questions with Solutions

4.5
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1

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

AIEEE 2010

A ball is made of a material of density $\rho$ where ${\rho _{oil}}\, < \rho < {\rho _{water}}$ with ${\rho _{oil}}$ and ${\rho _{water}}$ representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?
A
B
C
D

Explanation

Oil will float on water so, $(2)$ or $(4)$ is the correct option, But density of ball is more than that of oil, hence it will sinkin oil.
3

AIEEE 2010

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of ${30^ \circ }$ with each other. When suspended in a liquid of density $0.8g$ $c{m^{ - 3}},$ the angle remains the same. If density of the material of the sphere is $1.6$ $g$ $c{m^{ - 3}},$ the dielectric constant of the liquid is
A
$4$
B
$3$
C
$2$
D
$1$

Explanation

${F_e} = T\sin {15^ \circ }\,\,;$
$mg = T\cos {15^ \circ }$
$\Rightarrow \tan {15^ \circ } = {{{F_e}} \over {mg}}$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$
In liquid, ${F_e}' = T'\sin {15^ \circ }$ $\,\,\,\,\,\,...(ii)$

$mg = {F_B} + T'\cos {15^ \circ }$
${F_B}' = V\left( {d - \rho } \right)g = V\left( {1.6 - 0.8} \right)g = 0.8\,Vg$
$= 0.8{m \over d}g = {{0.8mg} \over {1.6}} = {{mg} \over 2}$
$\therefore$ $mg = {{mg} \over 2} + T'\cos {15^ \circ }$
$\Rightarrow {{mg} \over 2} = T'\cos {15^ \circ }$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( B \right)$
From $(A)$ and $(B),$ $\tan \,{15^ \circ } = {{2{F_e}'} \over {mg}}\,\,\,\,\,\,\,\,\,...\left( 2 \right)$
From $(1)$ and $(2)$
${{{F_e}} \over {mg}} = {{2{F_e}'} \over {mg}} \Rightarrow {F_e} = 2{F_e}' \Rightarrow {F_e}' = {{{F_e}} \over 2}$

4

AIEEE 2009

Two wires are made of the same material and have the same volume. However wire $1$ has cross-sectional area $A$ and wire $2$ has cross-sectional area $3A.$ If the length of wire $1$ increases by $\Delta x$ on applying force $F,$ how much force is needed to stretch wire $2$ by the same amount?
A
$4F$
B
$6F$
C
$9F$
D
$F$

Explanation

As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of cross-section $3A$ will be $\ell /3$ (same volume as wire $1$).
For wire $1,$
$y = {{F/A} \over {\Delta x/\ell }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$
For wire $2.$
$Y = {{F'/3A} \over {\Delta x/\left( {\ell /3} \right)}}........(ii)$
From $(i)$ and $(ii),$ ${F \over A} \times {\ell \over {\Delta x}} = {{F'} \over {3A}} \times {\ell \over {3\Delta x}} \Rightarrow F' = 9F$