1

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

A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time is - A
${{m{g^2}{t^2}} \over 8}$
B
${{3m{g^2}{t^2}} \over 8}$
C
$-$ ${{m{g^2}{t^2}} \over 8}$
D
0

## Explanation

N $-$ mg = ${{mg} \over 2}$ $\Rightarrow$ N = ${{3mg} \over 2}$

The distance travelled by the system in time t is

S = ut + ${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$

Now, work done

W = N.S = $\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$

$\Rightarrow$  W = ${{3m{g^2}{t^2}} \over 8}$
2

### 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
3

### 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
4

### 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