### JEE Mains Previous Years Questions with Solutions

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### AIEEE 2008

MCQ (Single Correct Answer)
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

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

$\therefore$ ${v_t} = \sqrt {{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}}$
2

### AIEEE 2006

MCQ (Single Correct Answer)
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.
3

### AIEEE 2006

MCQ (Single Correct Answer)
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
4

### AIEEE 2005

MCQ (Single Correct Answer)
A $20$ $cm$ long capillary tube is dipped in water. The water rises up to $8$ $cm.$ If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be
A
$10$ $cm$
B
$8$ $cm$
C
$20$ $cm$
D
$4$ $cm$

## Explanation

In freely falling elevator $g$ = 0

Water fills the tube entirely in gravity less condition. Hence, length of water column in the capillary tube is 20 cm.

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