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Coulomb's Law (NCERT) - Free MCQ Practice Test with solutions, NEET NCERT


MCQ Practice Test & Solutions: Test: Coulomb's Law (NCERT) (10 Questions)

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Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

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Test: Coulomb's Law (NCERT) - Question 1

The constant k in Coulomb's law depends on

Detailed Solution: Question 1

The value of k = (1/4πε0) = 8.854 x 10-12 C2 N-1 m-2 where εis permittivity of free space.

Test: Coulomb's Law (NCERT) - Question 2

Which of the following statement is not a similarity between electrostatic and gravitational forces?

Detailed Solution: Question 2

Electrostatic forces are both attractive and repulsive depending upon the type of charge, but gravitational forces is always attractive.

Test: Coulomb's Law (NCERT) - Question 3

SI unit of permittivity of free space is

Test: Coulomb's Law (NCERT) - Question 4

The force between two small charged spheres having charges of 1 x 10-7 C and 2 x 10-7 C placed 20 cm apart in air is

Detailed Solution: Question 4

Test: Coulomb's Law (NCERT) - Question 5

The nucleus of helium atom contains two proton that are separated by distance 3.0 x 10-15 m. The magnitude of the electrostatic force that each proton exerts on the other is

Detailed Solution: Question 5

Charge of proton is qP = 1.6 x 10-19 C

Distance between the protons is, r = 3 x 10-15m
The magnitude of electrostatic force between protons is 

Test: Coulomb's Law (NCERT) - Question 6

Two insulated charged metallic spheres P and Q have their centres separated by a distance of 60 cm. The radii of P and Q are negligible compared to the distance of separation. The mutual force of electrostatic repulsion if the charge on each is 3.2 x 10-7 C is

Detailed Solution: Question 6

Here, q1 = q2 = 3.2 x 10-7 C, r = 60cm = 0.6 m
Electrostatic force,

Test: Coulomb's Law (NCERT) - Question 7

Two point charges of +3 μC and +4 μC repel each other with a force of 10 N. If each is given an additional charge of -6 μC, the new force is

Detailed Solution: Question 7

The force between two charges q1 and q2 at a distance r is given by Coulomb's law:

F = k*q1*q2/r^2

where k is Coulomb's constant (9*10^9 N*m^2/C^2).

Given that F = 10 N, q1 = 3 μC = 3*10^-6 C, and q2 = 4 μC = 4*10^-6 C, we can solve for r^2:

10 = 9*10^9 * 3*10^-6 * 4*10^-6 / r^2
r^2 = 9*10^9 * 3*10^-6 * 4*10^-6 / 10
r^2 = 1.08*10^-2 m^2

Now, if each charge is given an additional -6 μC, the new charges are q1' = -3 μC = -3*10^-6 C and q2' = -2 μC = -2*10^-6 C. The new force F' is:

F' = k*q1'*q2'/r^2
F' = 9*10^9 * -3*10^-6 * -2*10^-6 / 1.08*10^-2
F' = 5 N

So, the new force is 5N. Hence, option

Test: Coulomb's Law (NCERT) - Question 8

The ratio of magnitude of electrostatic force and gravitational force for an electron and proton is

Detailed Solution: Question 8

Here, for an electron and a proton 

Test: Coulomb's Law (NCERT) - Question 9

The electrostatic attracting force on a small sphere of charge 0.2μC due to another small sphere of charge -0.4μC in air is 0.4N. The distance between the two spheres is

Detailed Solution: Question 9

Here, q1 = 0.2 μC = 0.2 x 10-6 C
q2 = 0.4 μC = -0.4 x 10-6 C, F = -0.4 N
As F = 

= 1.8 x 10-3
∴ r = (1.8 x 10-3)1/2 = 0.0424 m = 42.4 x 10-3 m

Test: Coulomb's Law (NCERT) - Question 10

Under the action of a given coulombic force the acceleration of an electron is 2.5 x 1022 ms-2. Then the magnitude of the acceleration of a proton under the action of same force is nearly

Detailed Solution: Question 10

The acceleration due to given coulombic force F is 
a = (F/m) or a ∝ (1/m) ...(1)
∴  where mc and mp are masses of electron and proton respectively

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