NEET Foundation Physics Chapter 3 Gravitation Long Answer Questions

Chapter 3 Gravitation Long Answer Type Question And Answers

Question 1. Interconnected vessel as shown in figure is filled with an ideal liquid P₁ and P₂ are airtight pistons having areas A₁ = 2 cm² and A₂ = 5 cm² respectively. A weight of 3 kg is kept on piston P₁. Calculate

1. pressure acting on piston P₁
2. pressure on piston P₂
3. force with which P₂ moves up.

Answer.

Given:

Interconnected vessel as shown in figure is filled with an ideal liquid P₁ and P₂ are airtight pistons having areas A₁ = 2 cm² and A₂ = 5 cm² respectively.

A weight of 3 kg is kept on piston P₁.

Pressure P₁ exerted by the piston P₁ on the confined liquid is given by

P₁ = \(\frac{\text { Force }}{\text { Area }}=\frac{\text { Weight }}{\text { Area }}=\frac{M_g}{A_1}\)

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

P₁ = 15 × 10⁴ Pa, (downwards)

By Pascal’s law, this pressure is communicated to the piston P₂.

∴ Upward pressure P₂ acting on piston P₂

= P₁

= 15 × 10⁴ Pa, (upwards)

We know that,

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

∴ F = PA

= P₂A₂

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

= 75 N

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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.

1. Take ρoceanwater = 1.02 × 10³ kgm^-3
2. 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 m³

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 × 10⁵ N

= 4.59 × 10⁶ 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

\(\frac{v_p}{v_e}=\sqrt{\left(\frac{M_p}{R_p} \times \frac{R_e}{M_r}\right)}=\sqrt{\left(\frac{1}{2} \times \frac{2}{1}\right)}=1\)

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/R²

Therefore,

GMm/R² = mg

Acceleration due to gravity g = GM/R²

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

1. reduce to 36%
2. 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

1. gravitational force between earth and the  sun and
2. gravitational force between the moon and  the earth.

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

Mass of earth, m₁ = 6 × 10²⁴ kg

Mass of sun, m₂ = 2 × 10³⁰kg

Distance between sun and earth, R = 1.5 × 10¹¹m

Gravitational force between the sun and the earth,

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

F = (6.67 × 10^-11 Nm² kg^-2 × 6 × 10²⁴ kg × 2 × 10³⁰kg)/(1.5 × 10¹¹m) ²

F = 3.6 × 10²² N

(2) Gravitational force between the moon and the earth

Mass of earth, m₁ = 6 × 10²⁴ kg

Mass of moon, m₂ = 7.4 × 10²² kg

Gravitational force between earth and the moon is

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

F = (6.67 × 10^-11Nm²kg^-2× 6 × 10²⁴kg × 7.4 × 10²² kg)/(3.8 × 10⁸m)²

F = 2.05 × 10²⁰ 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 = \(\frac{\sqrt{2 G M}}{R}\)

\(2 v_e^2-\frac{v_e^2}{2}=\frac{1}{2} v^{\prime 2}\)

Or, v’² = 3 v_e²

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.

1. How long does it take to fall 4.9 m?
2. How fast does it move at the end of the fall?
3. How fast does it move at the end of 7.9 m?
4. 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

v² – u² = 2ah

v² – 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.