1
Numerical

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

The electric field in an electromagnetic wave is given by E = (50 NC$-$1) sin$\omega$ (t $-$ x/c)

The energy contained in a cylinder of volume V is 5.5 $\times$ 10$-$12 J. The value of V is _____________ cm3. (given $\in$0 = 8.8 $\times$ 10$-$12C2N$-$1m$-$2)

## Explanation

$E = 50\sin \left( {\omega t - {\omega \over c}.\,x} \right)$

Energy density = ${1 \over 2}{ \in _0}E_0^2$

Energy of volume $V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}$

${1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}$

$V = {{5.5 \times 2} \over {2500 \times 8.8}} = .0005{m^3}$

= .0005 $\times$ 106 (c.m)3

= 500 (c.m)3
2
Numerical

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

Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T1 is connected to T2. As the current in R = 6$\Omega$ attains a maximum value of steady state level, T1 is disconnected from T2 and immediately connected to T3. Potential drop across r = 3$\Omega$ resistor immediately after T1 is connected to T3 is __________ V. (Round off to the Nearest Integer)

## Explanation

What T1 and T2 are connected, then the steady state current in the inductor $I = {6 \over 6} = 1A$

When T1 and T3 are connected then current through inductor remains same. So potential difference across 3$\Omega$

V = Ir = 1 $\times$ 3 = 3 Volt
3
Numerical

### JEE Main 2021 (Online) 25th July Morning Shift

A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance '2 m' when each is carrying same charge 'q'. If the free charged particle is displaced from its equilibrium position through distance 'x' (x < < 1 m). The particle executes SHM. Its angular frequency of oscillation will be ____________ $\times$ 105 rad/s if q2 = 10 C2.

## Explanation

Net force on free charged particle

$F = {{k{q^2}} \over {{{(d + x)}^2}}} - {{k{q^2}} \over {{{(d - x)}^2}}}$

$F = - k{q^2}\left[ {{{4dx} \over {{{({d^2} - {x^2})}^2}}}} \right]$

$a = - {{4k{q^2}d} \over m}\left( {{x \over {{d^4}}}} \right)$

$a = - \left( {{{4k{q^2}} \over {m{d^3}}}} \right)x$

So, angular frequency

$\omega = \sqrt {{{4k{q^2}} \over {m{d^3}}}}$

$\omega = \sqrt {{{4 \times 9 \times {{10}^9} \times 10} \over {1 \times {{10}^{ - 6}} \times {1^3}}}}$

$\omega = 6 \times {10^8}$ rad/sec
4
Numerical

### JEE Main 2021 (Online) 20th July Morning Shift

A body having specific charge 8 $\mu$C/g is resting on a frictionless plane at a distance 10 cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V/m is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be _______________ s.