1

### JEE Main 2019 (Online) 10th January Evening Slot

A particle which is experiencing a force, given by $\overrightarrow F = 3\widehat i - 12\widehat j,$ undergoes a displacement of $\overrightarrow d = 4\overrightarrow i$ particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?
A
9 J
B
10 J
C
12 J
D
15 J

## Explanation

Work done = $\overrightarrow F \cdot \overrightarrow d$

$=$ 12 J

work energy theorem

wnet $=$ $\Delta$K.E.

12 $=$ Kf $-$ 3

Kf = 15 J
2

### JEE Main 2019 (Online) 11th January Morning Slot

A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25 $\times$ 106 N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms–2 , the value of x will be close to :
A
8 cm
B
4 cm
C
40 cm
D
80 cm

## Explanation

velocity of 1 kg block just before it collides with 3kg block

= $\sqrt {2gh} = \sqrt {2000}$ m/s

Applying momentum conversation just before and just after collision.

1 $\times$ $\sqrt {2000}$ = 4v $\Rightarrow$ v = ${{\sqrt {2000} } \over 4}$ m/s initial compression of spring

1.25 $\times$ 106 x0 = 30 $\Rightarrow$ x0 $\approx$ 0

applying work energy theorem,

Wg + Wsp = $\Delta$KE

$\Rightarrow$   40 $\times$ x + ${1 \over 2}$ $\times$ 1.25 $\times$ 106 (02 $-$ x2)

= 0 $-$ ${1 \over 2}$ $\times$ 4 $\times$ v2

solving x $\approx$ 4 cm
3

### JEE Main 2019 (Online) 12th January Morning Slot

For the given cyclic process CAB as shown for a gas, the work done is : A
1 J
B
10 J
C
5 J
D
30 J

## Explanation

Since P$-$V indicator diagram is given, so work done by gas is area under the cyclic diagram.

$\therefore$  $\Delta$W = Work done by gas = ${1 \over 2}$ $\times$ 4 $\times$ 5 J

= 10 J
4

### JEE Main 2019 (Online) 12th January Morning Slot

There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R (t) is :
A B C D ## Explanation

At any instant 't'

Total energy of charge distribution is constant

i.e.     ${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$

$\therefore$  ${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$

$\therefore$  V = $\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} \right)}$

$\therefore$  V = $\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} \right)}$ = C$\sqrt {{1 \over {{R_0}}} - {1 \over R}}$

Also the slope of v-s curve will go on decreasing