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)
Your Input ________

Answer

Correct Answer is 500

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)

Your Input ________

Answer

Correct Answer is 3

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.
Your Input ________

Answer

Correct Answer is 6

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.

Your Input ________

Answer

Correct Answer is 1

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