Gravitation - NEET Physics Questions
← All Chapters

Gravitation

Practice Questions:
Question 81: easy

An iron sphere and an aluminium sphere, both of same radius are dropped from the top of a tower \(100\text{ m}\) high. At a height \(40\text{ m}\) above the ground, both of them will have same:

1. momentum
2. kinetic energy
3. potential energy
4. acceleration
View Answer

In the absence of air resistance, all falling bodies have the same acceleration due to gravity \(g\) independent of their mass. Since their masses are different due to different densities, other quantities will differ.

💬 Discuss this Question
Question 82: 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 \(T_e approx 2 times 1.0025 = 2.005\text{ s}\).

💬 Discuss this Question
Question 83: easy

A geostationary satellite has an orbital period of

1. 2 hours
2. 6 hours
3. 12 hours
4. 24 hours
View Answer

A geostationary satellite remains stationary relative to the Earth's surface, meaning its orbital period must equal the rotation period of the Earth, which is 24 hours.

💬 Discuss this Question
Question 84: easy

Imagine a light planet revolving around a very massive star in a circular orbit of radius \(r\) with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to \(r^{-5/2}\), then the square of the time period will be proportional to

1. \(r^3\)
2. \(r^2\)
3. \(r^{2.5}\)
4. \(r^{3.5}\)
View Answer

The centripetal force is \(F = m\omega^2 r = m\frac{4\pi^2}{T^2} r \propto \frac{r}{T^2}\). Given \(F \propto r^{-5/2}\), we get \(\frac{r}{T^2} \propto r^{-5/2} \Rightarrow T^2 \propto r^{3.5}\).

💬 Discuss this Question
Question 85: easy

Satellites A and B are orbiting around the earth in orbits of ratio R and 4R respectively. The ratio of their areal velocities is :

1. 1 : 2
2. 1 : 4
3. 1 : 8
4. 1 : 16
View Answer

Areal velocity is given by \(\frac{dA}{dt} = \frac{L}{2m} = \frac{vr}{2} = \frac{\sqrt{GMr}}{2}\). Since areal velocity is proportional to \(\sqrt{r}\), the ratio is \(\sqrt{\frac{R}{4R}} = \frac{1}{2}\) or 1 : 2.

💬 Discuss this Question
Question 86: difficult

A satellite is launched in the equatorial plane in such a way that it can transmit signals upto \(60^\circ\) latitude on the earth. Then the angular velocity of the satellite is :

1. \(\sqrt{\frac{GM}{8R^3}}\)
2. \(\sqrt{\frac{GM}{2R^3}}\)
3. \(\sqrt{\frac{GM}{4R^3}}\)
4. \(\sqrt{\frac{3\sqrt{3}GM}{8R^3}}\)
View Answer

For a transmission coverage up to latitude \(\theta = 60^\circ\), the orbital radius \(r\) satisfies \(\cos\theta = \frac{R}{r}\). This gives \(r = 2R\). The angular velocity is \(\omega = \sqrt{\frac{GM}{r^3}} = \sqrt{\frac{GM}{8R^3}}\).

💬 Discuss this Question
Question 87: easy

A satellite is seen after each 8 hours over equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth & satellite is same will be :

1. 8 hours
2. 12 hours
3. 24 hours
4. 6 hours
View Answer

When rotating oppositely, \(\frac{1}{T_{\text{rel}}} = \frac{1}{T_s} + \frac{1}{T_e} \Rightarrow \frac{1}{8} = \frac{1}{T_s} + \frac{1}{24}\), which gives \(T_s = 12\text{ hours}\). When rotating in the same direction, \(\frac{1}{T_{\text{rel}}'} = \frac{1}{T_s} - \frac{1}{T_e} = \frac{1}{12} - \frac{1}{24} = \frac{1}{24}\), so \(T_{\text{rel}}' = 24\text{ hours}\).

💬 Discuss this Question
Question 88: easy

A satellite of mass \(m\) is in a circular orbit of radius \(2R\) about the earth. How much energy is required to transfer it to a circular orbit of radius \(4R\) :  (\(R =\) Radius of earth)

1. \(\frac{mgR}{8}\)
2. \(\frac{mgR}{4}\)
3. \(\frac{mgR}{2}\)
4. None of these
View Answer

The total energy of a satellite is \(E = -\frac{GMm}{2r}\). The required energy is \(\Delta E = E_f - E_i = -\frac{GMm}{8R} - \left(-\frac{GMm}{4R}\right) = \frac{GMm}{8R}\). Since \(g = \frac{GM}{R^2}\), we get \(\Delta E = \frac{mgR}{8}\).

💬 Discuss this Question
Question 89: easy

If the earth be at one half its present distance from the sun, number of days in the year will be nearly

1. 129
2. 30
3. 200
4. 60
View Answer

According to Kepler's Third Law, \(T^2 \propto R^3\). Thus, \(\left(\frac{T'}{T}\right)^2 = \left(\frac{R'}{R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\), which gives \(T' = \frac{365}{\sqrt{8}} \approx 129\text{ days}\).

💬 Discuss this Question
Question 90: easy

The period of a satellite in a circular orbit of radius \(R\) is \(T\). What is the period of another satellite in a circular orbit of radius \(4R\) ?

1. 4T
2. T/8
3. T/4
4. 8 T
View Answer

By Kepler's Third Law, \(T^2 \propto R^3\). Therefore, \(\frac{T'}{T} = \left(\frac{4R}{R}\right)^{3/2} = 8\), which gives \(T' = 8T\).

💬 Discuss this Question