1

### JEE Main 2016 (Online) 10th April Morning Slot

A neutron moving with a speed ‘v’ makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is :
A
10.2 eV
B
16.8 eV
C
12.1 eV
D
20.4 eV

## Explanation

Let, velocity offer collision = v1

$\therefore$   From conservation of momentum,

mv = (m + m) v1

$\Rightarrow$    v1 = ${v \over 2}$

$\therefore$   Loss in kinetic energy

= ${1 \over 2}$ mv2 $-$ ${1 \over 2}$ (2m) $\times$ ${\left( {{v \over 2}} \right)^2}$

= ${1 \over 4}\,$mv2

lost kinetic energy is used by the electron to jump from first orbit to second orbit.

$\therefore$   ${1 \over 4}$mv2 = (13.6 $-$ 3.4) eV = 10.2 eV

$\Rightarrow$  ${1 \over 2}$mv2 = 20.4 eV
2

### JEE Main 2016 (Online) 10th April Morning Slot

Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on the body in first two seconds of the motion is : A
12000 J
B
$-$ 12000 J
C
$-$ 4500 J
D
$-$ 9300 J

## Explanation Here u = 50 m/s , what t = 0

$\alpha$  =  ${{\Delta v} \over {\Delta t}}$  =  ${{50 - 0} \over {0 - 10}}$  =  $-$5 m/s2

Speed of the body at t = 2 s

v   =   u + at

=  50 + ($-$ 5) $\times$ 2

=  40 m/s

From work energy theorem,

$\Delta$w  =  ${1 \over 2}m{v^2} - {1 \over 2}m{u^2}$

=  ${1 \over 2}$ m(v2 $-$u2)

= ${1 \over 2}$ $\times$ 10 $\times$ (402 $-$ 502)

=  5 $\times$ ($-$10)(90)

=  $-$ 4500 J
3

### JEE Main 2017 (Offline)

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is ${1 \over 8}mv_0^2$, the value of k will be:
A
10-1 kg m-1 s-1
B
10-3 kg m-1
C
10-3 kg s-1
D
10-4 kg m-1

## Explanation

According to the question, final kinetic energy = ${1 \over 8}mv_0^2$

Let final speed of the body = Vf

So final kinetic energy = ${1 \over 2}mv_f^2$

According to question,

${1 \over 2}mv_f^2$ = ${1 \over 8}mv_0^2$

$\Rightarrow {v_f} = {{{v_0}} \over 2}$ = ${{10} \over 2}$ = 5 m/s

Given that, F = –kv2

$\Rightarrow$ $m\left( {{{dv} \over {dt}}} \right)$$= - k{v^2}$

$\Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}$

$\Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt}$

$\Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10$

$\Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}$
4

### JEE Main 2017 (Offline)

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be:
A
18 J
B
4.5 J
C
22 J
D
9 J

## Explanation

Given that, F = 6t

We know, F = ma = $m{{dv} \over {dt}}$

$\therefore$ $m{{dv} \over {dt}} = 6t$

$\Rightarrow$ $1.{{dv} \over {dt}} = 6t$ [as m = 1]

$\Rightarrow$ $\int\limits_0^v {dv} = \int {6t} dt$

$\Rightarrow$ $v = 6\left[ {{{{t^2}} \over 2}} \right]_0^1$

$\Rightarrow$ $v = {6 \over 2} = 3$ m/s [ as given t = 1 sec ]

Work done by the body during the first 1 form work-energy theorem,

W = $\Delta$K.E = ${1 \over 2}m\left( {{V^2} - {v^2}} \right)$

= ${1 \over 2}.1.\left( {{3^2} - {0^2}} \right)$ = 4.5 J