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Gravitation question

2015 · Shift 0 · Q61
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  5. /2015 · Shift 0 · Q61

Gravitation question

2015 · Shift 0 · Q61

JEE MainPhysicsGravitationMCQ+4 / −1
From a solid sphere of mass MMM and radius R,R,R, a spherical portion of radius R/2R/2R/2 is removed, as shown in the figure. Taking gravitational potential V=0V=0V=0 at r=∞,r = \infty ,r=∞, the potential at the center of the cavity thus formed is: (G=gravitationalconstantG=gravitational constantG=gravitationalconstant)JEE Main 2015 (Offline) Physics - Gravitation Question 178 English
  1. A
    −2GM3R{{ - 2GM} \over {3R}}3R−2GM​
  2. B
    −2GMR{{ - 2GM} \over R}R−2GM​
  3. C
    −GM2R{{ - GM} \over {2R}}2R−GM​
  4. D
    −GMR{{ - GM} \over R}R−GM​
View written solutionFree

Correct answer: D

  1. Use superposition

A cavity can be treated as:

  • a full solid sphere of mass MMM and radius RRR, plus
  • a negative mass sphere representing the removed part.

So, Vcavity center=Vfull sphere at that point+Vremoved part at that point.V_{\text{cavity center}} = V_{\text{full sphere at that point}} + V_{\text{removed part at that point}}.Vcavity center​=Vfull sphere at that point​+Vremoved part at that point​.

  1. Geometry of the figure

The removed spherical cavity has radius R/2R/2R/2 and is tangent to the outer surface, so the distance between the center of the original sphere and the center of the cavity is d=R−R2=R2.d = R - \frac{R}{2} = \frac{R}{2}.d=R−2R​=2R​.

Thus, the center of the cavity lies inside the original sphere at a distance r=R/2r=R/2r=R/2 from its center.

  1. Potential inside a uniform solid sphere

For a uniform solid sphere of mass MMM and radius RRR, the gravitational potential at distance rrr from its center is V(r)=−GM2R3(3R2−r2).V(r) = -\frac{GM}{2R^3}(3R^2-r^2).V(r)=−2R3GM​(3R2−r2).

At r=R/2r=R/2r=R/2,

= -\frac{GM}{2R^3}\cdot \frac{11R^2}{4} = -\frac{11GM}{8R}.$$ --- 4. **Potential due to the removed sphere** The removed sphere has radius $$a=\frac{R}{2}.$$ Since the original sphere is uniform, its density is $$\rho = \frac{M}{\frac{4}{3}\pi R^3}.$$ Hence mass of removed part: $$m = \rho \cdot \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = M\left(\frac{1}{8}\right)=\frac{M}{8}.$$ Potential at the center of a solid sphere of mass $m$ and radius $a$ is $$V_{\text{at center}} = -\frac{3Gm}{2a}.$$ So for the removed sphere, $$V_{\text{removed (if present)}} = -\frac{3G(M/8)}{2(R/2)} = -\frac{3GM}{8}\cdot \frac{1}{R} = -\frac{3GM}{8R}.$$ But this mass is **removed**, so its contribution must be subtracted: $$V_{\text{negative sphere}} = +\frac{3GM}{8R}.$$ --- 5. **Net potential at the cavity center** $$V = -\frac{11GM}{8R} + \frac{3GM}{8R} = -\frac{8GM}{8R} = -\frac{GM}{R}.$$ --- 6. **Match with options** The correct option is $$\boxed{-\frac{GM}{R}}$$ which is **Option D**. --- 7. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** So they agree.
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