NEET Foundation Physics Chapter 3 Gravitation Short Answer Questions

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 Nm2 kg2-2
Answer.

Given:

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

Given:  m1 = 500g = 0.5 kg

m2 = 2 kg

r = 40 cm = 0.40 m

and G = 6.7 × 10-11 Nm2 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 ME = 6 × 1024 kg

Mm = 7.5 × 1022 kg

r = 3,80,000 km = 3.8 × 108 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 × 1019 N

According to Newton’s III law, moon will also attract earth towards itself with the same force i.e., 20.9 × 1019 N.

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

  1. how long it takes to reach the highest point
  2. 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) ∵ v2 = u2 + 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.

NEET Foundation Physics Gravitation Two bodies meet at point C

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 = 5x2 (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 − 5x2 (2)

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

AC + BC = 5x2 + 100x − 5x2

∴ AB = 100x

i.e., 200 = 100x

∴ x = \(\frac{200}{100}=2 s\)

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

AC = 5x2 = 5 × 22 = 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 m2. 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 m2.

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

Pressure Pg exerted by girl on the ground due to her weight is given by

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

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

Pg = 2 × 106 Nm-2or Pa Similarly pressure exerted by the boy on the ground due to his weight

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

Pb = 4 × 104 Nm-2 or Pa

Since Pg > Pb, 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.

NEET Foundation Physics Gravitation Pressure exerted liquid column

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 × 104 Nm2 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 × 103 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 × 103 kgm-3 × 9.8 ms-2

= 1.013 × 105 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 × 108 N

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