### JEE Mains Previous Years Questions with Solutions

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1

### 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
${A_2}{v_2} = {A_1}{v_1}$
$\pi {{D_2^2} \over 4} \times {v_2} = \pi {{D_1^2} \over 4}{v_1}$
$\Rightarrow \,\,{D_1} = {D_2}\sqrt {{{{v_1}} \over {{v_2}}}} = 3.55 \times {10^{ - 3}}m$
2

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

### 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.
4

### 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 $(2)$
${{{F_e}} \over {mg}} = {{2{F_e}'} \over {mg}} \Rightarrow {F_e} = 2{F_e}' \Rightarrow {F_e}' = {{{F_e}} \over 2}$