Test: Basic Properties & Coulomb’s Law (October 24) - NEET MCQ

Test: Basic Properties & Coulomb’s Law (October 24) - NEET MCQ

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10 Questions MCQ Test Daily Test for NEET Preparation - Test: Basic Properties & Coulomb’s Law (October 24)

Test: Basic Properties & Coulomb’s Law (October 24) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Basic Properties & Coulomb’s Law (October 24) questions and answers have been prepared according to the NEET exam syllabus.The Test: Basic Properties & Coulomb’s Law (October 24) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Properties & Coulomb’s Law (October 24) below.
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Test: Basic Properties & Coulomb’s Law (October 24) - Question 1

The constant k in Coulomb's law depends on

Detailed Solution for Test: Basic Properties & Coulomb’s Law (October 24) - 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: Basic Properties & Coulomb’s Law (October 24) - Question 2

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 for Test: Basic Properties & Coulomb’s Law (October 24) - Question 2

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

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Test: Basic Properties & Coulomb’s Law (October 24) - Question 3

If 109 electrons move out of a body to another body every second, then the time required to get charge of 1 C on the other body is

Detailed Solution for Test: Basic Properties & Coulomb’s Law (October 24) - Question 3

The charge given out in one second = 1.6 x 10-19 C x 109 = 1.6 x 10-10C
Time required to accumulate a charge of 1 C
= 1/(1.6 x 10-10) = 6.25 x 109 s = 198 years

Test: Basic Properties & Coulomb’s Law (October 24) - Question 4

The number of electrons present in -1C of charge is

Detailed Solution for Test: Basic Properties & Coulomb’s Law (October 24) - Question 4

By quantisation of charge, q = ne
or

Test: Basic Properties & Coulomb’s Law (October 24) - Question 5

SI unit of permittivity of free space is

Test: Basic Properties & Coulomb’s Law (October 24) - Question 6

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

Detailed Solution for Test: Basic Properties & Coulomb’s Law (October 24) - Question 6

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

Test: Basic Properties & Coulomb’s Law (October 24) - Question 7

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 for Test: Basic Properties & Coulomb’s Law (October 24) - Question 7

Here, q1 = 1 x 10-7 C, q2 = 2 x 10-7 C, r = 20cm = 20 x 10-2 m

Test: Basic Properties & Coulomb’s Law (October 24) - Question 8

The number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of +2.4 C is

Detailed Solution for Test: Basic Properties & Coulomb’s Law (October 24) - Question 8

Total charge, q = 2.4 C
Then by quantization of charge, q = ne
∴ number of electrons, n = q/e = 2.4C/1.6 x 10-19 C = 1.5 x 1019

Test: Basic Properties & Coulomb’s Law (October 24) - Question 9

A polythene piece rubbed with wool is found to have a negative charge of 6 x 10-7 C. The number of electrons transferred to polythene from wool is

Detailed Solution for Test: Basic Properties & Coulomb’s Law (October 24) - Question 9

Here, q = -6 x 10-7 C
∴ Number of electrons transferred from wool to polythene, n =

Test: Basic Properties & Coulomb’s Law (October 24) - Question 10

A cup contains 250 g of water. Find the total positive charges present in the cup of water.

Detailed Solution for Test: Basic Properties & Coulomb’s Law (October 24) - Question 10

Mass of water = 250 g,
Molecular mass of water = 18 g
Number of molecules in 18 g of water = 6.02 x 1023
Number of molecules in one cup of water = (250/18) x 6.02 x 1023
Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons.
∴ Total positive charge present in one cup of water
= (250/18) x 6.02 x 1023 x 10 x 1.6 x 10-19 C = 1.34 x 107C.

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