Class 11 Physics Chapter Gravitation MCQs are provided here with answers. These questions are designed as per the latest CBSE syllabus and NCERT curriculum. Solving these chapter-wise MCQs will help students to score good marks in the final exam. Gravitation Class 11 Physics MCQs are prepared for a better understanding of the concept. It allows students to test their knowledge and answering skills in the given time frame.
MCQs on Class 11 Chapter Gravitation
Check the multiple-choice questions for the 11th Class Physics Gravitation chapter. Each MCQ will have four options here, out of which only one is correct. Students have to pick the correct option and check the answer provided here.
1. The gravitational force between two objects is given by:
\begin{array}{l} F = G \frac{m_1 m_2}{r^2} \end{array}
If the mass of both objects is doubled, what happens to the force?
A) It remains the same
B) It becomes twice
C) It becomes four times
D) It becomes one-fourth
Answer: C) It becomes four times
Explanation:
The gravitational force is directly proportional to the product of the masses. If both masses are doubled, then:
\begin{array}{l} F’ = G \frac{(2m_1)(2m_2)}{r^2} = 4F \end{array}
Thus, the force increases four times.
2. The acceleration due to gravity at a height hh above Earth’s surface is given by:
\begin{array}{l} g_h = g \left( 1 – \frac{2h}{R} \right) \end{array}
where R is Earth’s radius. What happens to gh if h≪R ?
A) It remains unchanged
B) It slightly decreases
C) It slightly increases
D) It becomes zero
Answer: B) It slightly decreases
Explanation:
Since the formula shows that gh is slightly less than g, acceleration due to gravity decreases with height.
3. The weight of a body is maximum at:
A) Poles
B) Equator
C) Center of Earth
D) In space
Answer: A) Poles
Explanation:
Weight is given by W=mg. Since g is maximum at the poles due to the Earth’s rotation and shape (flattened at poles and bulged at equator), the weight of the body is maximum there.
4. If the distance between two masses is doubled, the gravitational force between them:
A) Remains the same
B) Becomes twice
C) Becomes one-fourth
D) Becomes one-half
Answer: C) Becomes one-fourth
Explanation:
The gravitational force is inversely proportional to the square of the distance:
\begin{array}{l} F = G \frac{m_1 m_2}{r^2} \end{array}
When r is doubled,
\begin{array}{l} F’ = G \frac{m_1 m_2}{(2r)^2} = \frac{F}{4} \end{array}
5. The orbital velocity of a satellite in a circular orbit of radius R around Earth is given by:
\begin{array}{l} v = \sqrt{\frac{GM}{R}} \end{array}
What happens to the orbital velocity if the orbit’s radius is increased?
A) Increases
B) Decreases
C) Remains the same
D) First increases, then decreases
Answer: B) Decreases
Explanation:
Since velocity is inversely proportional to the square root of R, increasing R results in a decrease in v.
6. The escape velocity of an object from Earth’s surface is given by:
\begin{array}{l} v_e = \sqrt{\frac{2GM}{R}} \end{array}
What happens if Earth’s radius were doubled while keeping the mass constant?
A) It increases
B) It decreases
C) It remains the same
D) It becomes zero
Answer: B) It decreases
Explanation:
Since escape velocity is inversely proportional to √R, doubling R, decreases ve by a factor of 1/√2.
7. The time period of a satellite orbiting close to the Earth’s surface is approximately:
A) 24 hours
B) 84 minutes
C) 365 days
D) 1 hour
Answer: B) 84 minutes
Explanation:
Using Kepler’s third law:
\begin{array}{l} T = 2\pi \sqrt{\frac{R^3}{GM}} \end{array}
For a satellite near Earth’s surface, T≈84 84 minutes.
8. If an object is taken to the center of the Earth, its weight will:
A) Remain the same
B) Increase
C) Decrease
D) Become zero
Answer: D) Become zero
Explanation:
Weight is given by W=mg, and g decreases linearly with depth. At the Earth’s center, g=0, so W=0.
9. The total energy of a satellite in a stable orbit around Earth is:
A) Positive
B) Negative
C) Zero
D) Infinite
Answer: B) Negative
Explanation:
Total energy E=−GMm/2R, which is always negative for a bound system.
10. The value of acceleration due to gravity at a depth dd below the Earth’s surface is given by:
\begin{array}{l} g_d = g \left(1 – \frac{d}{R} \right) \end{array}
What happens to g as d increases?
A) Increases
B) Decreases
C) Remains the same
D) First decreases, then increases
Answer: B) Decreases
Explanation:
Since the equation shows a decrease in g with d, acceleration due to gravity decreases as we go deeper into the Earth.
