Gravitation - NEET Physics Questions
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Gravitation

Practice Questions:
Question 41: moderate

The acceleration due to gravity increases by \(0.5\%\) when we go from the equator to the poles. What will be the time period of the pendulum at the equator which beats seconds at the poles?

1. 1.950 s
2. 1.995 s
3. 2.050 s
4. 2.005 s
View Answer

The time period of a seconds pendulum at the poles is \(T_p = 2\text{ s}\). Since \(T propto g^{-1/2}\), we have \(frac{T_e}{T_p} = sqrt{frac{g_p}{g_e}} = sqrt{1 + 0.005} approx 1 + 0.0025\). This gives \(T_e approx 2 times 1.0025 = 2.005\text{ s}\).

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Question 42: moderate

Two equal masses \(m\) and \(m\) are hung from a balance whose scale pans differ in vertical height by \(h\). The error in weighing in terms of density of the earth \(\rho\) is :

1. \(\pi G \rho m h\)
2. \(\frac{1}{3} \pi G \rho m h\)
3. \(\frac{8}{3} \pi G \rho m h\)
4. \(\frac{4}{3} \pi G \rho m h\)
View Answer

The change in gravity over height \(h\) is \(dg = \frac{2g}{R} h\). Using \(g = \frac{4}{3} \pi G \rho R\), we get \(dg = \frac{8}{3} \pi G \rho h\). Thus, the difference in weight (error) is \(m \cdot dg = \frac{8}{3} \pi G \rho m h\).

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Question 43: moderate

Two concentric shells have mass \(M\) and \(m\) and their radii are \(R\) and \(r\) respectively, where \(R > r\). What is the gravitational potential at their common centre ?

1. \(-\frac{GM}{R}\)
2. \(-\frac{GM}{r}\)
3. \(-G \left[ \frac{M}{R} - \frac{m}{r} \right]\)
4. \(-G \left[ \frac{M}{R} + \frac{m}{r} \right]\)
View Answer

The gravitational potential inside any spherical shell is constant and equals the potential at its surface. Therefore, the total potential at the common centre is the sum of the potentials: \(V = -\frac{GM}{R} - \frac{Gm}{r} = -G\left[\frac{M}{R} + \frac{m}{r}\right]\).

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Question 44: moderate

The gravitational force between two particles with masses \(m\) and \(M\), initially at rest at great separation, pulls them together. When their separation becomes \(d\), then speed of either particle relative to the other will be :

1. \(\sqrt{G(M+m)/2d}\)
2. \(\sqrt{G(M+m)/d}\)
3. \(\sqrt{4G(M+m)/d}\)
4. \(\sqrt{2G(M+m)/d}\)
View Answer

By conservation of mechanical energy, the relative speed is found using the reduced mass \(\mu = \frac{mM}{m+M}\). Thus, \(\frac{1}{2} mu v_{\text{rel}}^2 = \frac{GMm}{d}\), which simplifies to \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).

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Question 45: moderate

The escape velocity for a planet is \(v_e\). A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be

1. \(\sqrt{1.5} v_e\)
2. \(\frac{v_e}{\sqrt{2}}\)
3. \(v_e\)
4. zero
View Answer

Using conservation of mechanical energy from infinity to the centre: \(0 = \frac{1}{2}mv^2 - \frac{3GmM}{2R}\). Since escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\), we get \(v^2 = \frac{3GM}{R} = 1.5 v_e^2⇒ v = \sqrt{1.5} v_e\).

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Question 46: moderate

The mass of a spaceship is \(1000\text{ kg}\). It is to be launched from the earth’s surface out into free space. The value of \(‘g’\) and \(‘R’\) (radius of earth) are \(10\text{ m/s}^2\) and \(6400\text{ km}\) respectively. The required energy for this work will be :

1. \(6.4 \times 10^{10}\text{ Joules}\)
2. \(6.4 \times 10^{11}\text{ Joules}\)
3. \(6.4 \times 10^8\text{ Joules}\)
4. \(6.4 \times 10^9\text{ Joules}\)
View Answer

The minimum energy required to escape the earth's gravitational pull from the surface is \(E = \frac{GMm}{R} = mgR\). Substituting \(m = 1000\text{ kg}\), \(g = 10\text{ m/s}^2\), and \(R = 6.4 \times 10^6\text{ m}\), we find \(E = 6.4 \times 10^{10}\text{ Joules}\).

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Question 47: moderate

Two bodies, each of mass \(M\), are kept fixed with a separation \(2L\). A particle of mass \(m\) is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is \(G\). The correct statement(s) is (are) :

(a) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(4\sqrt{\frac{GM}{L}}\)

(b) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(2\sqrt{\frac{GM}{L}}\)

(c) The minimum initial velocity of the mass \(m\) to escape the gravitational field of the two bodies is \(\sqrt{\frac{2GM}{L}}\)

(d) The energy of the mass \(m\) remains constant

 

1. a, b
2. b, d
3. a, c
4. a, d
View Answer

At the midpoint, potential energy is \(U = -\frac{2GMm}{L}\). For escaping to infinity, total mechanical energy must be at least 0:

\(\frac{1}{2}mv^2 - \frac{2GMm}{L} = 0 ⇒ v = 2\sqrt{\frac{GM}{L}}\). Mechanical energy remains conserved as only gravity acts.

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Question 48: moderate

The energy required to put a satellite of mass \(m\) from earth surface into a orbit of radius \(2R\) is \(E_1\). The energy further needed to change the orbit of this satellite from its present orbit to radius \(4R\) is \(E_2\). The ratio \(\frac{E_1}{E_2}\) is (where \(R\) is radius of earth:

1. \(4 : 1\)
2. \(1 : 4\)
3. \(6 : 1\)
4. \(1 : 6\)
View Answer

The energy required to put a satellite in orbit from earth's surface is \(E_1 = -\frac{GMm}{2(2R)} - \left(-\frac{GMm}{R}\right) = \frac{3GMm}{4R}\). The energy to change orbit from \(2R\) to \(4R\) is \(E_2 = -\frac{GMm}{2(4R)} - \left(-\frac{GMm}{2(2R)}\right) = \frac{GMm}{8R}\). Thus, \(\frac{E_1}{E_2} = 6\), which gives the ratio \(6 : 1\).

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