JEE Mains Previous Years Questions with Solutions

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1

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
2

AIEEE 2005

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.
3

AIEEE 2005

If $S$ is stress and $Y$ is young's modulus of material of a wire, the energy stored in the wire per unit volume is
A
${{{S^2}} \over {2Y}}$
B
$2{S^2}Y$
C
${S \over {2Y}}$
D
${{2Y} \over {{S^2}}}$

Explanation

Energy stored per unit volume of wire,

$E = {1 \over 2} \times \,stress\, \times \,strain$

$\therefore$ $E = {1 \over 2} \times \,stress\, \times \,{{stress} \over Y} = {1 \over 2}{{{S^2}} \over Y}$

[ As Young's modulus(Y) = ${{Stress} \over {Strain}}$

$\therefore$ Strain = ${{Stress} \over Y}$ ]
4

AIEEE 2004

A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
A
$Ec$
B
$2E/c$
C
$E/c$
D
$E/{c^2}$

Explanation

Momentum of photon $= {E \over c}$

Change in momentum $= {{2E} \over c}$

$=$ momentum transferred to the surface

(the photon will reflect with same magnitude of momentum in opposite direction)