Four equal masses (each of mass M) are placed at the corners of a square of side a. The escape velocity of a body from the centre O of the square is

\(4\sqrt{\frac{2GM}{a}}\)
\(\sqrt{\frac{8\sqrt{2}GM}{a}}\)
\(\frac{4GM}{a}\)
\(\sqrt{\frac{4\sqrt{2}GM}{a}}\)

Solution

The earth is assumed to be sphere of radius R. A platform is arranged at a height R from the surface of Earth.The escape velocity of a body from this platform is kv, where v is its escape velocity from the surface of the earth. The value of k is

¹⁄_√2
¹⁄₃
¹⁄₂
√2

Solution

A (nonrotating) star collapses onto itself from an initial radius R_i with its mass remaining unchanged. Which curve in figure best gives the gravitational acceleration a_g on the surface of the star as a function of the radius of the star during the collapse

Solution

g ∝ ¹⁄_R²

R decreasing g increase hence, curve b represents correct variation.

The escape velocity from a planet is v_e. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

v_e
v_e/√2
v_e/2
zero

Solution

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1/2 of g at poles. The escape velocity of a particle on the pole of planet in terms of V is

V_e= 2V
V_e = V
V_e = V/2
V_e = √3V

Solution

A cavity of radius R/2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R/2 from the centre of the sphere. The gravitational force on a particle of mass ‘m’ at a distance R/2 from the centre of the sphere on the line joining both the centres of sphere and cavity is – (opposite to the centre of gravity)

[Here g = GM/R², where M is the mass of the sphere]

^mg⁄₂
^3mg⁄₈
^mg⁄₁₆
None of these

Solution

A satellite is revolving round the earth in an orbit of radius r with time period T. If the satellite is revolving round the earth in an orbit of radius r + Δr (Δr << r) with time period T+ ΔT then,

\(\frac{\Delta T}{T}=\frac{3}{2}\frac{\Delta r}{r}\)
\(\frac{\Delta T}{T}=\frac{2}{3}\frac{\Delta r}{r}\)
\(\frac{\Delta T}{T}=\frac{\Delta r}{r}\)
\(\frac{\Delta T}{T}=-\frac{\Delta r}{r}\)

Solution

Since, T²=kr³

Differentiating the above equation

\(\Rightarrow 2\frac{\Delta T}{T}=3\frac{\Delta r}{r}\Rightarrow \frac{\Delta T}{T}=\frac{3}{2}\frac{\Delta r}{r}\)

The percentage change in the acceleration of the earth towards the sun from a total eclipse of the sun to the point where the moon is on a side of earth directly opposite to the sun is

\(\frac{M_{s}}{M_{m}}\frac{r_{2}}{r_{1}}\times 100\)
\(\frac{M_{s}}{M_{m}}\left ( \frac{r_{2}}{r_{1}} \right )^{2}\times 100\)
\(2\left ( \frac{r_{1}}{r_{2}} \right )^{2}\frac{M_{m}}{M_{s}}\times 100\)
\(\left ( \frac{r_{1}}{r_{2}} \right )^{2}\frac{M_{m}}{M_{s}}\times 100\)

Solution

Four similar particles of mass m are orbiting in a circle of radius r in the same angular direction because of their mutual gravitational attractive force. Velocity of a particle is given by

\(\left [ \frac{Gm}{r}\left ( \frac{1+2\sqrt{2}}{4} \right ) \right ]^{1/2}\)
\(\sqrt[3]{\frac{GM}{r}}\)
\(\sqrt{\frac{GM}{r}(1+2\sqrt{2})}\)
\(\left [ \frac{1}{2}\frac{GM}{r}\left ( \frac{1+\sqrt{2}}{2} \right ) \right ]^{1/2}\)

Solution

In a region of only gravitational field of mass ‘M’ a particle is shifted from A to B via three different paths in the figure.The work done in different paths are W₁, W₂, W₃ respectively then

W₁ = W₂ = W₃
W₁ > W₂ > W₃
W₁ = W₂ > W₃
W₁ < W₂ < W₃

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