### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### 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.

2

### 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

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

$\therefore$ ${v_t} = \sqrt {{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}}$
3

### 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.
4

### AIEEE 2006

If the terminal speed of a sphere of gold (density $= 19.5\,\,kg/{m^3}$) is $0.2$ $m/s$ in a viscous liquid (density $= 1.5\,\,kg/{m^3}$, find the terminal speed of a sphere of silver (density $= 10.5\,\,kg/{m^3}$) of the same size in the same liquid
A
$0.4$ $m/s$
B
$0.133$ $m/s$
C
$0.1$ $m/s$
D
$0.2$ $m/s$

## Explanation

Let Terminal velocity = vt

Upward viscous force = downward weight of sphere

$\Rightarrow 6\pi \eta r{v_t} = \left( {{4 \over 3}\pi {r^3}} \right)\left( {\rho - \sigma } \right)g$

$\Rightarrow {v_t} = {{2{r^2}\left( {\rho - \sigma } \right)g} \over {9\eta }}$ ........ (1)

where, $\rho$ = density of substance of a body

$\sigma$ = density of liquid

Now let the terminal velocity of gold = vg and silver = vs.

From equation (1), we can write

${{{v_g}} \over {{v_s}}} = {{{\rho _g} - \sigma } \over {{\rho _s} - \sigma }}$ $= {{19.5 - 1.5} \over {10.5 - 1.5}}$ = ${{18} \over 9}$ $= {2 \over 1}$

$\therefore$ ${v_s} = {{{v_g}} \over 2}$ = ${{0.2} \over 2}$ = 0.1