JEE Main 2022 (Online) 26th June Morning Shift | Electrostatics Question 58 | Physics

JEE Main 2022 (Online) 26th June Morning Shift

MCQ (Single Correct Answer)

-1

Given below two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negative charge distribution.

In the light of above statements, choose the most appropriate answer from the options given below.

Both (A) and (R) are correct and (R) is the correct explanation of (A).

Both (A) and (R) are correct and (R) is not the correct explanation of (A).

(A) is correct but (R) is not correct.

(A) is not correct but (R) is correct.

JEE Main 2022 (Online) 25th June Morning Shift

MCQ (Single Correct Answer)

-1

In the figure, a very large plane sheet of positive charge is shown. P₁ and P₂ are two points at distance l and 2l from the charge distribution. If $$\sigma$$ is the surface charge density, then the magnitude of electric fields E₁ and E₂ at P₁ and P₂ respectively are :

JEE Main 2022 (Online) 25th June Morning Shift Physics - Electrostatics Question 52 English

$${E_1} = \sigma /{\varepsilon _0},\,{E_2} = \sigma /2{\varepsilon _0}$$

$${E_1} = 2\sigma /{\varepsilon _0},\,{E_2} = \sigma /{\varepsilon _0}$$

$${E_1} = {E_2} = \sigma /2{\varepsilon _0}$$

$${E_1} = {E_2} = \sigma /{\varepsilon _0}$$

JEE Main 2022 (Online) 24th June Evening Shift

MCQ (Single Correct Answer)

-1

Two identical charged particles each having a mass 10 g and charge 2.0 $$\times$$ 10^$$-$$7C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms^$$-$$2]

12 cm

10 cm

8 cm

5 cm

JEE Main 2022 (Online) 24th June Evening Shift

MCQ (Single Correct Answer)

-1

A long cylindrical volume contains a uniformly distributed charge of density $$\rho$$. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :

$${{\rho q{R^2}} \over {4{\varepsilon _0}}}$$

$${{\rho q{R^2}} \over {2{\varepsilon _0}}}$$

$${{q\rho } \over {4{\varepsilon _0}{R^2}}}$$

$${{4{\varepsilon _0}{R^2}} \over {q\rho }}$$

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