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TS EAMCET · Physics · Gravitation

The bodies of masses \(100 \mathrm{~kg}\) and \(8100 \mathrm{~kg}\) are held at a distance of \(1 \mathrm{~m}\). The gravitational field at a point on the line joining them is zero. The gravitational potential at that point in \(\mathrm{J} / \mathrm{kg}\) is \(\left(G=6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2\right)\)

  1. A \(-6.67 \times 10^7\)
  2. B \(-6.67 \times 10^{10}\)
  3. C \(-13.34 \times 10^7\)
  4. D \(-6.67 \times 10^9\)
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Answer & Solution

Correct Answer

(A) \(-6.67 \times 10^7\)

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Detailed explanation

Mass of first body \(\left(m_1\right)=100 \mathrm{~kg}\) Mass of second body \(\left(m_2\right)=8100 \mathrm{~kg}\) Distance between two bodies \(=1 \mathrm{~m}\) Distance of null point from \(100 \mathrm{~kg}\)…
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