## Chapter 3 Gravitation Short Answer Type Question And Answers

**Question 1. Calculate the gravitational force of attraction between two small gold balls weighing 500 g and 2 kg kept on a horizontal table, with their centres 40 cm apart. Take G = 6.7 × 10 ^{-11} Nm^{2} kg^{2-2}**

**Answer.**

**Given:**

Two small gold balls weighing 500 g and 2 kg kept on a horizontal table, with their centres 40 cm apart.

Given: m_{1} = 500g = 0.5 kg

m_{2} = 2 kg

r = 40 cm = 0.40 m

and G = 6.7 × 10^{-11} Nm^{2} kg^{-2}

By Newton’s law of gravitation,

F = \(\frac{G m_1 m_2}{r^2}\)

= \(6.7 \times 10^{-11} \frac{\mathrm{Nm}^2}{\mathrm{~kg}^2} \times \frac{0.5 \mathrm{~kg} \times 2 \mathrm{~kg}}{(0.4 \mathrm{~m})^2}\)

= \(6.7 \times 10^{-11} \times \frac{0.5 \times 2}{0.16} \mathrm{~N}\)

= \(6.7 \times 10^{-11} \times \frac{100}{16} \mathrm{~N}\)

∴ F = 4.1875 × 10^{-10} N

**Question 2. With what force earth attracts moon? With what force moon attracts earth?**

**Answer.**

Take M_{E} = 6 × 10^{24} kg

M_{m} = 7.5 × 10^{22} kg

r = 3,80,000 km = 3.8 × 10^{8} m

According to Newton’s law of gravitation,

F = \(\frac{G M_E M_m}{r^2}\)

= \(6.7 \times 10^{-12} \frac{\mathrm{Nm}^2}{\mathrm{~kg}^2} \times \frac{6 \times 10^{24} \mathrm{~kg} \times 7.5 \times 10^{22} \mathrm{~kg}}{\left(3.8 \times 10^5 \mathrm{~m}\right)^2}\)

= \(\frac{6.7 \times 6 \times 7.5 \times 10^{35}}{3.8 \times 3.8 \times 10^{16}} \mathrm{~N}\)

Therefore, F = 20.9 × 10^{19} N

According to Newton’s III law, moon will also attract earth towards itself with the same force i.e., 20.9 × 10^{19} N.

**Question 3. A body is thrown vertically upwards with a velocity of 49 ms ^{-1}. Find**

**how long it takes to reach the highest point****maximum height attained by it.**

**Take g = 9.8 ms ^{-1} and neglect air-friction.**

**Answer.** Given: u = 49 ms^{-1}

v = 0 (at the highest point)

a = −9.8 ms^{-2}

t = ?

h = S = ?

(1) We know that

v = u + at

0 = 49 + (−9.8) × t

9.8 t = 49

∴ t = 49

9 8. = 5 s

(2) ∵ v^{2} = u^{2} + 2ah

0 = (49)^{2} + 2 (−9.8) h

19.6 h = 49 × 49

∴ h = \(\frac{49 \times 49}{19.6}\)

∴ h = 122.5 m

**Question 4. A body is dropped from the top of a tower 200 m tall. Simultaneously, another body is thrown vertically upwards from its foot with a velocity of 100 ms ^{-1}. Find when and where they meet or cross each other.
**

**Take g = 10 ms**

^{-2}and ignore air friction.**Answer.**

**Given:**

A body is dropped from the top of a tower 200 m tall. Simultaneously, another body is thrown vertically upwards from its foot with a velocity of 100 ms^{-1}.

Suppose the two bodies meet at point C after ‘x’ seconds.

Consider downward motion AC of first body.

∵ S = \(u t+\frac{1}{2} g t^2\)

∴ AC = \(0+\frac{1}{2} \times 10 \times x^2\)

∴ AC = 5x^{2} (1)

Consider upward motion BC of second body.

∴ S = \(u t+\frac{1}{2} g t^2\)

∴ BC = \(100 x+\frac{1}{2}(-10) x^2\)

∴ BC = 100x − 5x^{2} (2)

Adding equations (1) and (2), we get

AC + BC = 5x^{2} + 100x − 5x^{2}

∴ AB = 100x

i.e., 200 = 100x

∴ x = \(\frac{200}{100}=2 s\)

Substituting x = 25 in equation (i) we get,

AC = 5x^{2 }= 5 × 2^{2} = 20 m

Therefore, the two bodies meet at a point 20 m below the top of the tower or (200 − 20) m = 180 m above the foot of the tower, 2 s after they embark on their journey.

**Question 5. A girl weighing 40 kg is wearing pointed sandals having a total area of 2 × 10 ^{-4} m^{2}. How much pressure she will exert on ground? In which case, the ground is more likely to be damaged? Explain. (Take g = 10 ms^{-2})**

**Answer.**

**Given:**

A girl weighing 40 kg is wearing pointed sandals having a total area of 2 × 10^{-4} m^{2}.

A boy of same weight wears shoe having area of 100 × 10^{-4} m.

Pressure P_{g} exerted by girl on the ground due to her weight is given by

P_{g} = \(\frac{\text { Force }}{\text { Area }}=\frac{\text { Weight }}{\text { Area }}=\frac{M g}{A}\)

= \(\frac{40 \times 10 \mathrm{~N}}{2 \times 10^{-4} \mathrm{~m}^2}\)

P_{g} = 2 × 10^{6} Nm^{-2}or P_{a }Similarly pressure exerted by the boy on the ground due to his weight

P_{b} = \(\frac{\mathrm{Mg}_g}{A}=\frac{40 \times 10 \mathrm{~N}}{100 \times 10^{-4} \mathrm{~m}^2}\)

P_{b} = 4 × 10^{4} Nm^{-2} or Pa

Since P_{g} > P_{b}, the girl is more likely to damage the ground than the boy

**Question 6. A tall cylinder contains water column of height 3 m. Calculate the pressure it exerts on the bottom of the cylinder. Take g = 10 ms ^{-2}. Density of water = 1000 kg.**

**Answer.**

**Given:**

A tall cylinder contains water column of height 3 m.

We know that, pressure exerted by a liquid column is given by,

P = h ρ g

= \(3 \mathrm{~m} \times 1000 \frac{\mathrm{kg}}{\mathrm{m}^3} \times 10 \frac{\mathrm{m}}{\mathrm{s}^2}\)

∴ P = 3 × 10^{4} Nm^{2} or Pa

**Question 7. It is often said that atmospheric pressure is 76cm of mercury column. What does it mean? Express this pressure in Pascal.**

**Take ρmercury = 13.6 × 10 ^{3} kgm^{-3} and g = 9.8 ms^{-2}.**

**Answer.**

The given statement means that the earth’s atmosphere exerts some pressure as exerted by a mercury column of height 76 cm.

∴ Atmospheric pressure = Pressure of 76 cm of mercury

= h ρ g

= 0.76 m × 13.6 × 10^{3} kgm^{-3} × 9.8 ms^{-2}

= 1.013 × 10^{5} Pa

**Question 8. A ship having a total weight of 20,000 tonnes is floating on water. How much upthrust acts on the ship?**

**Take g = 10 ms ^{-2}**

**Answer.**

Ship is in equilibrium (i.e., at rest)

So, Upward force acting on the ship = Downward force acting on the ship.

U = W

= Mg

= \(20,000 \text { tonnes } \times 10^s \frac{\mathrm{kg}}{\text { tonne }} \times 10 \mathrm{~ms}^{-2}\)

= 2 × 10^{8} N