## Chapter 3 Gravitation Long Answer Type Question And Answers

**Question 1. Interconnected vessel as shown in figure is filled with an ideal liquid P _{1} and P_{2} are airtight pistons having areas A_{1} = 2 cm^{2} and A_{2} = 5 cm^{2} respectively. A weight of 3 kg is kept on piston P_{1}. Calculate**

**pressure acting on piston P**_{1}**pressure on piston P**_{2}**force with which P**_{2}moves up.

**Answer.**

**Given:**

Interconnected vessel as shown in figure is filled with an ideal liquid P_{1} and P_{2} are airtight pistons having areas A_{1} = 2 cm^{2} and A_{2} = 5 cm^{2} respectively.

A weight of 3 kg is kept on piston P_{1}.

Pressure P_{1} exerted by the piston P_{1} on the confined liquid is given by

P_{1} = \(\frac{\text { Force }}{\text { Area }}=\frac{\text { Weight }}{\text { Area }}=\frac{M_g}{A_1}\)

= \(\frac{3 \times 10 \mathrm{~N}}{2 \times 10^{-1} \mathrm{~m}^2}\)

P_{1} = 15 × 10^{4} Pa, (downwards)

By Pascal’s law, this pressure is communicated to the piston P_{2}.

∴ Upward pressure P_{2} acting on piston P_{2}

= P_{1}

= 15 × 10^{4} Pa, (upwards)

We know that,

P = \(\frac{F}{A}\)

∴ F = PA

= P_{2}A_{2}

= \(15 \times 10^4 \frac{\mathrm{N}}{\mathrm{m}^2} \times 5 \times 10^{-4} \mathrm{~m}^2\)

= 75 N

**Read and Learn More NEET Foundation Long Answer Questions**

**Question 2. Nine tenth of an iceberg having a total volume of 500 m3 is submerged in ocean. Calculate the buoyant force acting on the iceberg.**

**Take ρoceanwater = 1.02 × 10**^{3}kgm^{-3}**g = 10 ms**^{-2}

**Answer.**

**Given:**

Nine tenth of an iceberg having a total volume of 500 m3 is submerged in ocean

Buoyant force or upthrust is given by

U = Vρ g

where

V = Volume of immersed part of solid.

= \(\left(\frac{9}{10} \times 500\right) \mathrm{m}^3\)

= 450 m^{3}

So U = V ρ g

= \(450 \mathrm{~m}^3 \times 1.02 \times 10^3 \frac{\mathrm{kg}}{\mathrm{m}^3} \times 10 \mathrm{~m}\)

= 45 × 1.02 × 10^{5} N

= 4.59 × 10^{6} N

**Question 3. A piece of metal weighs 200 gf in air and 180 gf in water. Finds its relative density.**

**Answer.**

**Given:**

A piece of metal weighs 200 gf in air and 180 gf in water.

\(\text { R. D. of a solid }=\frac{\text { Weight in air }}{\text { Loss of weight in water }}\)= \(\frac{W_1}{W_1-W_2}\)

= \(\frac{200 \mathrm{gf}}{(200-180) g f}\)

= \(\frac{200}{20}\)

∴ RD of metal = 10 (No unit).

**Question 4. Imagine a planet whose both diameter and mass are one half of the Earth. The day’s temperature of this planet’s surface reaches up to 800 K. Find whether oxygen molecules are possible in the atmosphere of this planet.**

**Answer.**

**Given:**

Imagine a planet whose both diameter and mass are one half of the Earth. The day’s temperature of this planet’s surface reaches up to 800 K.

Escape velocity,

v_{e} = \(\sqrt{2} G M / R\)

Let v_{p} = escape velocity on the planet

v_{e} = escape velocity on the earth

v_{p} = v_{e} = 11.2 km/s

From kinetic theory of gases

v_{rms} = \(\sqrt{3 R T / M})=\sqrt{3 N K T / M}=\sqrt{3 N K T / N m}\)

where N = Avogadro’s number

m = mass of oxygen molecule

K = Boltzmann constant

v_{rms} = \(\sqrt{3 R T / m}\)

= \(\sqrt{\left(\frac{\left(3 \times 1.38 \times 10^{-23} \times 800\right)}{5.3 \times 10^{-2 h}}\right)}\)

(m = 5.3 × 10−26 kg)

v_{rms} = 0.79 km/s

As v_{rms} is very small compared to the escape velocity on the planet, molecules cannot escape from the surface of the planet’s atmosphere.

**Question 5. Deduce the relationship between g and G.**

**Answer.**

** Relationship between g and G:**

Let g = acceleration due to gravity at a planet

M = mass

R = radius

m = mass of object

By Newton’s law of motion, force on a body due to gravity on its surface

F = mass × acceleration due to gravity

= m g

By Newton’s gravitational law, force is

F = GMm/R^{2}

Therefore,

GMm/R^{2} = mg

Acceleration due to gravity g = GM/R^{2}

**Question 6. How much below the surface of earth does the acceleration due to gravity**

**reduce to 36%****reduce by 36%, of its value on the surface of earth, (radius of earth = 6400 km).**

**Answer.** Case (1)

We have,

g_{h} = g (1 – d/R)

Or d = (g – g_{h}) R/g (1)

Here

g_{h} = 36/100 g (2)

Using (1) and (2)

g_{h} = 36/100 g

d = (1 – 36/100) R

d = (100 – 36)/100 × 6400

d = 4096 km

Case (2)

Here

g_{h} = g – 36/100 g

i.e., g_{h} = 64/100 g (3)

Using equation (1) and (3)

d = (g – 64/100 g) 6400/g

d = 100 – 64/100 × 6400

d = 2304 km

**Question 7. Deduce gravitational force between**

**gravitational force between earth and the sun and****gravitational force between the moon and the earth.**

**Answer.** (1) Gravitational force between earth and the sun

Mass of earth, m_{1} = 6 × 10^{24} kg

Mass of sun, m_{2} = 2 × 10^{30}kg

Distance between sun and earth, R = 1.5 × 10^{11}m

Gravitational force between the sun and the earth,

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

F = (6.67 × 10^{-11} Nm^{2} kg^{-2} × 6 × 10^{24} kg × 2 × 10^{30}kg)/(1.5 × 10^{11}m) ^{2}

F = 3.6 × 10^{22} N

(2) Gravitational force between the moon and the earth

Mass of earth, m_{1} = 6 × 10^{24} kg

Mass of moon, m_{2} = 7.4 × 10^{22} kg

Gravitational force between earth and the moon is

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

F = (6.67 × 10^{-11}Nm^{2}kg^{-2}× 6 × 10^{24}kg × 7.4 × 10^{22} kg)/(3.8 × 10^{8}m)^{2}

F = 2.05 × 10^{20} N

**Question 8. Explain why a sheet of paper falls slower than one that is crumpled into a ball?**

**Answer.**

The sheet of paper falls slower than the one that is crumpled into a ball because the air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the sheet of paper is more than the resistance offered by air to the paper ball because the sheet has larger area.

**Question 9. The escape velocity of a body on the surface of the earth is 11.2 km/s. A body is projected away with twice this speed. What is the speed of the body at infinity? Ignore the presence of other heavenly bodies.**

**Answer.**

If v is the velocity of projection and v’ is the velocity at infinity, then we have by energy conservation principle.

\(\frac{1}{2} m v^2-\frac{G M m}{R}=\frac{1}{2} m v^{\prime 2}+0\)Here v = 2v_{e}

Thus,

\(\left(\frac{1}{2}\right) \cdot 4 v_e^2-\frac{G M}{R}=\frac{1}{2} v^{\prime 2}\) \(2 v_e^2-\frac{G M}{R}=\frac{1}{2} v^{\prime 2}\)Now, v_{e} = \(\frac{\sqrt{2 G M}}{R}\)

Or, v’^{2} = 3 v_{e}^{2}

Or, v’ = \(\sqrt{3} v_e=\sqrt{3} \times 11.2 \mathrm{~km} / \mathrm{s}=19.4 \mathrm{~km} / \mathrm{s}\)

**Question 10. Deduce the equation showing the variation in the value of g with depth below the surface of the earth.**

**Answer.**

Let us consider a body of mass m at a depth h below the surface of earth. Then, radius of the inner solid sphere of the earth = R – h.

Volume of the inner solid sphere of the earth = \(\frac{4}{3} \pi(R-h)^3 d\)

If d is the average density of the earth, then

Mass of inner solid sphere of earth = \(\frac{4}{3} \pi(R-h)^3 d\).

According to the law of gravitation,

mg_{d} = \(G \times \frac{4}{3} \pi(R-h)^3 d \times m /(R-h)^2\)

This gives

g_{d} = \(G \times \frac{4}{3} \pi(R-h) d (1)\)

On the surface of earth,

g = \(\frac{G M}{R^2}\)

= \(G \times \frac{4}{3} \pi R^3 \frac{d}{R^2}\)

= \(G \times \frac{4}{3} \pi R d (2)\)

From equation (1) and (2)

g_{d}/g = \(G \times \frac{4}{3} \pi(R-h) \times d / G \times \frac{4}{3} \pi R d\)

= (R – h)/R

Or g_{d} = g (1 – h/R) or (R – h)/R < 1

So, g_{d} < g

Thus, the value of g at a depth inside the earth is less than that on the surface of the earth.

The value (R – h)/R decreases with the value of h, i.e. depth below the surface of the earth. So the value of g decreases as we go down below the surface of earth.

**Question 11. A stone is dropped from the edge of the roof.**

**How long does it take to fall 4.9 m?****How fast does it move at the end of the fall?****How fast does it move at the end of 7.9 m?****What is its acceleration after 1 s and after 2 s?**

**Answer.** (1) As the stone is dropped, its initial velocity,

u = 0, h = 4.9 m

a = g = 9.8 ms^{-2}, time, t =?

From

h = \(u t+\frac{1}{2} g \mathrm{t}^2\)

4.9 = \(0+\frac{1}{2} \times 9.8 t^2=4.9 t^2\)

Or \(f^2=\frac{4.9}{4.9}=1, t=\sqrt{1}=1 \mathrm{~s}\)

(2) Final velocity, v = ? at t = 1 s

From

v = u + gt

v = 0 + 9.8 × 1 = 9.8 m s^{-1}

(3) Let v be the final velocity when h = 7.9 m

From

v^{2} – u^{2} = 2ah

v^{2} – 0 = 2 (9.8) 7.9

Or v = \(\sqrt{2 \times 9.8 \times 7.9}=\sqrt{154.84}\)

v = 12.4 m s^{–}^{1}

(4) The acceleration of a freely falling body remains the same at all times, i.e., a = g = 9.8 ms^{-1} after 1 s and 2 s.