Gravitation

JEE Main level, single correct, medium

105 of 223

Question 105 of 223

Three masses 200 kg,300 kg200 \mathrm{~kg}, 300 \mathrm{~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/kg2\mathrm{G}=6.7 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2})

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Answer: (C) 1.74×10−71.74 \times 10^{-7}

Work done equals negative change in gravitational potential energy.
For three masses in an equilateral triangle, gravitational PE is U=−G(m1m2r12+m2m3r23+m1m3r13)U = -G\left(\frac{m_1m_2}{r_{12}} + \frac{m_2m_3}{r_{23}} + \frac{m_1m_3}{r_{13}}\right).
Initial configuration (side 20 m):
Ui=−G20(200×300+300×400+200×400)U_i = -\frac{G}{20}(200 \times 300 + 300 \times 400 + 200 \times 400)
=−260000G20=−13000G= -\frac{260000G}{20} = -13000G
Final configuration (side 25 m):
Uf=−G25(260000)=−10400GU_f = -\frac{G}{25}(260000) = -10400G
Work done:
W=−(Uf−Ui)=−10400G+13000GW = -(U_f - U_i) = -10400G + 13000G
=2600G=2600×6.7×10−11= 2600G = 2600 \times 6.7 \times 10^{-11}
=1.742×10−7= 1.742 \times 10^{-7} J ≈1.74×10−7\approx 1.74 \times 10^{-7} J

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