SATHEE: Gravitation 2 Question 3

Gravitation 2 Question 3

6. From a solid sphere of mass $M$ and radius $R$ , a spherical portion of radius $\frac{R}{2}$ is removed as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$ , the potential at the centre of the cavity thus formed is ( $G$ = gravitational constant)

(2015 Main)

(a) $\frac{- G M}{R}$

(b) $\frac{- G M}{2 R}$

(d) $\frac{- 2 G M}{R}$

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Answer:

Correct Answer: 6. (a)

Solution:

$V_{R} = V_{T} - V_{C}$

$V_{R} =$ Potential due to remaining portion

$V_{T} =$ Potential due to total sphere

$V_{C} =$ Potential due to cavity

Radius of cavity is $\frac{R}{2}$ . Hence, volume and mass is $\frac{M}{8}$ .

$∴ V_{R} = - \frac{G M}{R^{3}} 1.5 R^{2} - 0.5 \frac{R^{2}}{2} + \frac{G \frac{M}{8}}{\frac{R}{2}} \frac{3}{2} = - \frac{G M}{R}$

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Gravitation 2 Question 3

6. From a solid sphere of mass M and radius R, a spherical portion of radius R2 is removed as shown in the figure. Taking gravitational potential V=0 at r=∞, the potential at the centre of the cavity thus formed is ( G = gravitational constant)

Answer:

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6. From a solid sphere of mass $M$ and radius $R$ , a spherical portion of radius $\frac{R}{2}$ is removed as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$ , the potential at the centre of the cavity thus formed is ( $G$ = gravitational constant)