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JEE Physics 2026 Question with Solution

Three masses 200 kg200\,kg, 300 kg300\,kg and 400 kg400\,kg are placed at the vertices of an equilateral triangle of side 20 m20\,m. They are rearranged on the vertices of a bigger triangle of side 25 m25\,m with the same centre. The work done in this process is \hspace{1cm J.

(Gravitational constant G=6.7×10−11 Nm2kg−2G = 6.7 \times 10^{-11\,N m^2kg^{-2}})

  • A

    9.86×10−69.86 \times 10^{-6}

  • B

    2.85×10−72.85 \times 10^{-7}

  • C

    4.77×10−74.77 \times 10^{-7}

  • D

    1.74×10−71.74 \times 10^{-7}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Three masses 200 kg200\,kg, 300 kg300\,kg and 400 kg400\,kg are initially at the vertices of an equilateral triangle of side 20 m20\,m and finally at the vertices of an equilateral triangle of side 25 m25\,m. The gravitational constant is G=6.7×10−11 N m2kg−2G = 6.7 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}.

Find: The work done in rearranging the masses.

For three masses placed at the vertices of an equilateral triangle of side rr, the total gravitational potential energy is

U=−G(m1m2+m2m3+m3m1r)U = -G\left(\frac{m_1m_2 + m_2m_3 + m_3m_1}{r}\right)

Initial potential energy:

Ui=−G((200)(300)+(300)(400)+(400)(200)20)U_i = -G\left(\frac{(200)(300) + (300)(400) + (400)(200)}{20}\right) Ui=−6.7×10−11×26000020U_i = -6.7 \times 10^{-11} \times \frac{260000}{20}

Final potential energy:

Uf=−G(26000025)U_f = -G\left(\frac{260000}{25}\right)

Work done in rearrangement is the change in gravitational potential energy:

W=Uf−UiW = U_f - U_i W=6.7×10−11×260000(120−125)W = 6.7 \times 10^{-11} \times 260000\left(\frac{1}{20} - \frac{1}{25}\right) W=1.74×10−7 JW = 1.74 \times 10^{-7}\,\text{J}

Therefore, the work done in rearranging the masses is 1.74×10−7 J1.74 \times 10^{-7}\,\text{J}. The correct option is D.

Common mistakes

  • Using only one pair of masses in gravitational potential energy is incorrect because the system has three interacting pairs: m1m2m_1m_2, m2m3m_2m_3 and m3m1m_3m_1. Add all three pair contributions before dividing by the common side length.

  • Taking work done as Ui−UfU_i - U_f here gives the wrong sign. Since the solution defines work done in rearrangement against gravity as the change in potential energy, use W=Uf−UiW = U_f - U_i.

  • Using the wrong separation distance in the final state is a common error. After rearrangement, every pair is separated by the new side length 25 m25\,m, not 20 m20\,m.

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