Unit Test (Solutions): Gravitation

Time: 1 hour 
M.M. 30 
Attempt all questions. 
Question numbers 1 to 5 carry 1 mark each. 
Question numbers 6 to 8 carry 2 marks each. 
Question numbers 9 to 11 carry 3 marks each. 
Question numbers 12 & 13 carry 5 marks each.

Q1. What is the force that keeps the moon in its orbit around the Earth? (1 Mark)

Ans: Gravitational force

Q2. Define centripetal force. (1 Mark)

Ans: It is the force that acts towards the center of a circular path and keeps a body moving along that path.

Q3. What is the SI unit of the universal gravitational constant (G)? (1 Mark)

Ans: N m² kg^-2

Q4. What is the acceleration due to gravity on Earth's surface (approximate value)? (1 Mark)

Ans: 9.8 m s^-2

Q5. Why do objects float in a liquid? (1 Mark)

Ans: Objects float when their density is less than the liquid's density.

Q6. State the universal law of gravitation. (2 Marks)

Ans: According to the universal law of gravitation, every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Q7. Why does a sheet of paper fall slower than a crumpled ball? (2 Marks)

Ans: The sheet of paper has a larger surface area, so it experiences greater air resistance than the crumpled ball, which falls faster due to lesser air resistance.

Q8. Define thrust and pressure. (2 Marks)

Ans: 

• Thrust is the force acting normally on a surface.

• Pressure is the thrust per unit area. Its SI unit is Pascal (Pa).

Q9. Calculate the gravitational force between the Earth and a 1 kg object on its surface. Given: Mass of Earth = 6 × 10²⁴ kg, Radius of Earth = 6.4 × 10⁶ m, G = 6.7 × 10^-11 N m² kg^-2.  Explain why this force is significant. (3 Marks)

Ans: Using F = G × (M × m) / d²
F = (6.7 × 10^-11 × 6 × 10²⁴ × 1) / (6.4 × 10⁶)²
F = (6.7 × 6 × 10¹³) / (4.096 × 10¹³)
F = 40.2 / 4.096 × 10⁰ = 9.81 N
Significance:
This is the gravitational force acting on a 1 kg object on Earth, also called its weight. It causes the object to accelerate downward at 9.8 m/s², which keeps everything grounded on Earth and governs the motion of falling objects.

Q10. Describe Archimedes' principle with an example. (3 Marks)

Ans: Archimedes' principle states that when a body is immersed fully or partially in a fluid, it experiences an upward force (buoyant force) equal to the weight of the fluid displaced by it.
Example: A ship floats in water because the upward buoyant force balances its weight.

Q11. How does the gravitational force between two objects change if:
(i) The mass of one object is doubled?
(ii) the distance between them is halved?
(iii) Both masses are doubled?   (3 Marks)

Ans: (i) The gravitational force doubles.
(ii) The gravitational force becomes four times (since F ∝ 1/d²).
(iii) The gravitational force becomes four times (F ∝ m₁ × m₂).

Q12.  Explain why the weight of an object on the moon is 1/6th its weight on Earth.  (5 Marks)

Ans: 

The weight of an object is the force with which it is attracted towards a celestial body and is given by the formula:
W = m × g,
where m is the mass of the object and g is the acceleration due to gravity.

The mass of the Moon is much smaller (7.36 × 10²² kg) and its radius is also smaller (1.74 × 10⁶ m) compared to the Earth (5.98 × 10²⁴ kg, 6.37 × 10⁶ m).
Using the formula for gravitational acceleration,
g = G × M / R²,
we find that the Moon's gravitational acceleration is approximately 1/6th that of Earth's.

Hence, the weight of an object on the Moon is 1/6th of its weight on Earth, because weight depends on gravitational acceleration.

Q13. A stone is dropped from a 100 m tower, and another is projected upward from the ground at 25 m/s simultaneously. Taking g = 10 m/s², calculate when and where they meet. (5 Marks)

Ans: Let time of meeting be t seconds.

For stone dropped from the tower:
Initial velocity, u₁ = 0
Distance covered = s₁ = ½ g t² = 5t²

For stone thrown upward:
Initial velocity, u₂ = 25 m/s
Distance covered = s₂ = ut - ½ g t² = 25t - 5t²

Since s₁ + s₂ = 100 (total height of the tower):
5t² + (25t - 5t²) = 100
25t = 100
t = 4 seconds

Now, distance covered by first stone:
s₁ = 5 × (4)² = 80 m from the top
So, height from ground = 100 - 80 = 20 m

Final Answer:
The stones meet after 4 seconds, at a height of 20 metres above the ground.