### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2008

A jar is filled with two non-mixing liquids $1$ and $2$ having densities ${\rho _1}$ and ${\rho _2}$ respectively. A solid ball, made of a material of density ${\rho _3}$, is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for ${\rho _1}$ , ${\rho _1}$ and ${\rho _3}$ ?
A
${\rho _3} < {\rho _1} < \rho {}_2$
B
${\rho _1} > {\rho _3} > \rho {}_2$
C
${\rho _1} < {\rho _2} < \rho {}_3$
D
${\rho _1} < {\rho _3} < \rho {}_2$

## Explanation

From the figure it is clear that liquid $1$ floats on liquid $2.$ The lighter liquid floats over heavier liquid. Therefore we can conclude that ${\rho _1} < {\rho _2}$

Also ${\rho _3} < {\rho _2}$ otherwise the ball would have sink to the bottom of the jar.

Also ${\rho _3} > {\rho _1}$ otherwise the ball would have floated in liquid $1.$ From the above discussion we conclude that ${\rho _1} < {\rho _2} < {\rho _3}.$

2

### AIEEE 2008

A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?
A
B
C
D

## Explanation

In case of water, the meniscus shape is concave upwards. Also according to ascent formula $h = {{2T\,\cos \,\theta } \over {r\rho g}}$

The surface tension $(I)$ of soap solution is less than water. Therefore rise of soap solution in the capillary tube is less as compared to water. As in the case of water. the meniscus shape of soap solution is also concave upwards.

3

### AIEEE 2008

A spherical solid ball of volume $V$ is made of a material of density ${\rho _1}$. It is falling through a liquid of density ${\rho _2}\left( {{\rho _2} < {\rho _1}} \right)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v,$ i.e., ${F_{viscous}} = - k{v^2}\left( {k > 0} \right).$ The terminal speed of the ball is
A
$\sqrt {{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}}$
B
${{{Vg{\rho _1}} \over k}}$
C
$\sqrt {{{Vg{\rho _1}} \over k}}$
D
${{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}$

## Explanation

The forces acting on the ball -

(1) mg = $V{\rho _1}g$ downward direction

(2) Thrust upward direction ( By Archimedes principle )

(3) Force of friction ( Buoynat force) upward direction

The ball reaches to its terminal speed $\left( {{v_t}} \right)$ when acceleration = 0.

So, weight $=$ Buoyant force $+$ Viscous force

$\therefore$ $V\rho {}_1g = V{\rho _2}g + kv_t^2$

$\therefore$ ${v_t} = \sqrt {{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}}$
4

### AIEEE 2006

A wire elongates by $l$ $mm$ when a LOAD $W$ is hanged from it. If the wire goes over a pulley and two weights $W$ each are hung at the two ends, the elongation of the wire will be (in $mm$)
A
$l$
B
$2l$
C
zero
D
$l/2$

## Explanation

Case $(i)$
At equilibrium, $T=W$
$Y = {{W/A} \over {\ell /L}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ...\left( 1 \right)$
Case $(ii)$ At equilibrium $T=W$
$\therefore$ $Y = {{W/A} \over {{{\ell /2} \over {L/2}}}} \Rightarrow Y = {{W/A} \over {\ell /L}}$
$\Rightarrow$ Elongation is the same.