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JEE Main 2020 (4 Sep, Shift 1) · Gravitational Potential & Energy

Gravitational Potential & Energy question from JEE Main 2020 (4 Sep, Shift 1)

Q21JEE Main 2020 · 4 SEP · SHIFT 1Medium

Gravitational Potential & Energy

On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is given by Ax(x2+a2)3/2\frac{Ax}{(x^2+a^2)^{3/2}} in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is:

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Answer: (c)

  1. V(x)=−∫∞xE dx=−A∫∞xx dx(x2+a2)3/2V(x) = -\int_\infty^x E\,dx = -A\int_\infty^x \frac{x\,dx}{(x^2+a^2)^{3/2}}
  2. ∫x dx(x2+a2)3/2=−1(x2+a2)1/2\int \frac{x\,dx}{(x^2+a^2)^{3/2}} = -\frac{1}{(x^2+a^2)^{1/2}}
  3. V=A(x2+a2)1/2V = \frac{A}{(x^2+a^2)^{1/2}}, so ∣V∣=A(x2+a2)1/2|V| = \frac{A}{(x^2+a^2)^{1/2}}
  4. So the answer is (c).
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