InterviewSolution
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InterviewSolution
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The gravitational field due to a mass distribution is I = k/r3 in the x - direction (k is a constant). The gravitational potential is taken to be zero at infinity, then its value at a distance x is(a) k/x(b) k/2x(c) k/x2(d) k/2x2 |
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Answer» Answer is : (d) \(\frac{k}{2x^2}\) As, I = \(\frac{-dV}{dr}\) ⇒ dV = - ldr So, \(\int\limits^v_0 dV\) = \(\int\limits^x_0\) - ldr = \(\int\limits^x_0\) - kr-3 dr ⇒ V = - k\(\left(\cfrac{r^{-3+1}}{-3+1}\right)^x_0\) = \(\frac{k}{2x^2}\) |
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