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JEE Mains Previous Years Questions with Solutions

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1

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.

Questions Asked from Properties of Matter

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