1

### JEE Main 2021 (Online) 1st September Evening Shift

A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of 0.8$\sqrt {gh}$. The value of workdone by the air-friction is :
A
$-$0.68 mgh
B
mgh
C
1.64 mgh
D
0.64 mgh

## Explanation

Work done = Chang in kinetic energy

${W_{mg}} + {W_{air - friction}} = {1 \over 2}m{\left( {.8\sqrt {gh} } \right)^2} - {1 \over 2}m{(0)^2}$

${W_{air - friction}} = {{.64} \over 2}$ mgh $-$ mgh = $-$0.68 mgh

Option (a)
2

### JEE Main 2021 (Online) 31st August Evening Shift

A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1 : 2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is :-
A
${{1 \over 3}}$
B
${{2 \over 3}}$
C
${{1 \over 8}}$
D
${{1 \over 4}}$

## Explanation 3MV0 = 2MV2 + MV1

3V0 = 2V2 + V1

120 = 2V2 + 60 $\Rightarrow$ V2 = 30 m/s

${{\Delta K.E.} \over {K.E.}} = {{{1 \over 2}MV_1^2 + {1 \over 2}2MV_2^2 - {1 \over 2}3MV_0^2} \over {{1 \over 2}3MV_0^2}}$

$= {{V_1^2 + 2V_2^2 - 3V_0^2} \over {3V_0^2}}$

$= {{3600 + 1800 - 4800} \over {4800}} = {1 \over 8}$
3

### JEE Main 2021 (Online) 27th July Evening Shift

An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by :
A
${\left( {{{9P} \over {8m}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}$
B
${\left( {{{8P} \over {9m}}} \right)^{{1 \over 2}}}{t^{{2 \over 3}}}$
C
${\left( {{{9m} \over {8P}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}$
D
${\left( {{{8P} \over {9m}}} \right)^{{1 \over 2}}}{t^{{3 \over 2}}}$

## Explanation

P = const.

$P = Fv = {{m{v^2}dv} \over {dx}}$

$\int\limits_0^x {{P \over m}dx} = \int\limits_0^v {{v^2}dv}$

${{Px} \over m} = {{{v^3}} \over 3}$

${\left( {{{3Px} \over m}} \right)^{1/3}} = v = {{dx} \over {dt}}$

${\left( {{{3P} \over m}} \right)^{1/3}}\int\limits_0^t {dt} = \int\limits_0^x {{x^{ - 1/3}}} dx$

$\Rightarrow x = {\left( {{{8P} \over {9m}}} \right)^{1/2}}{t^{3/2}}$
4

### JEE Main 2021 (Online) 27th July Evening Shift

Given below is the plot of a potential energy function U(x) for a system, in which a particle is in one dimensional motion, while a conservative force F(x) acts on it. Suppose that Emech = 8 J, the incorrect statement for this system is : [ where K.E. = kinetic energy ]
A
at x > x4 K.E. is constant throughout the region.
B
at x < x1, K.E. is smallest and the particle is moving at the slowest speed.
C
at x = x2, K.E. is greatest and the particle is moving at the fastest speed.
D
at x = x3, K.E. = 4 J.

## Explanation

Emech. = 8J

(A) at x > x4

U = constant = 6J

K = Emech. $-$ U = 2J = constant

(B) at x < x1

U = constant = 8J

K = Emech $-$ U = 8 $-$ 8 - 0 J

Particle is at rest.

(C) At x = x2,

U = 0 $\Rightarrow$ Emech. = K = 8 J

KE is greatest, and particle is moving at fastest speed.

(D) At x = x3

U = 4J

U + K = 8J

K = 4J