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JEE Main 2026 (24 Jan, Shift 1) · Gravitational Potential & Energy

Gravitational Potential & Energy question from JEE Main 2026 (24 Jan, Shift 1)

Q5JEE Main 2026 · 24 JAN · SHIFT 1Medium

Gravitational Potential & Energy

Three masses 200 kg,300 kg200\text{ kg}, 300\text{ kg} and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process ______ J. (Gravitational constant G=6.7×10−11 N m2/kg2G = 6.7 \times 10^{-11}\text{ N m}^2/\text{kg}^2)

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Answer: (b)

  1. U=−G(m1m2+m2m3+m3m1)aU = -\frac{G(m_1m_2 + m_2m_3 + m_3m_1)}{a}; ∑mimj=60000+120000+80000=2.6×105\sum m_im_j = 60000 + 120000 + 80000 = 2.6\times10^5
  2. W=Uf−Ui=G(2.6×105)(120−125)W = U_f - U_i = G(2.6\times10^5)\left(\frac{1}{20} - \frac{1}{25}\right)
  3. W=6.7×10−11×2.6×105×0.01W = 6.7\times10^{-11}\times2.6\times10^5\times0.01
  4. W=1.74×10−7 JW = 1.74\times10^{-7}\ \text{J}
  5. So the answer is (b).
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