Q. Three masses 200 kg, 300 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 \( G = 6.7 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \))
(A) \(4.77 \times 10^{-7}\)
(B) \(1.74 \times 10^{-7}\)
(C) \(9.86 \times 10^{-6}\)
(D) \(2.85 \times 10^{-7}\)
Explanation (Complete Step by Step Calculation)
The work done in rearranging the masses is equal to the change in gravitational potential energy of the system.
For three masses placed at the vertices of an equilateral triangle, the total gravitational potential energy is:
\[
U = -G\left(\frac{m_1 m_2}{r} + \frac{m_2 m_3}{r} + \frac{m_3 m_1}{r}\right)
\]
Given masses:
\[
m_1 = 200\,\text{kg}, \quad m_2 = 300\,\text{kg}, \quad m_3 = 400\,\text{kg}
\]
Initial side length:
\[
r_1 = 20\,\text{m}
\]
Final side length:
\[
r_2 = 25\,\text{m}
\]
Sum of pairwise products of masses:
\[
m_1 m_2 + m_2 m_3 + m_3 m_1 = (200 \times 300) + (300 \times 400) + (400 \times 200)
\]
\[
= 60000 + 120000 + 80000 = 260000
\]
Initial potential energy:
\[
U_1 = -G \frac{260000}{20}
\]
Final potential energy:
\[
U_2 = -G \frac{260000}{25}
\]
Work done is:
\[
W = U_2 - U_1
\]
\[
W = -G \times 260000\left(\frac{1}{25} - \frac{1}{20}\right)
\]
\[
\frac{1}{25} - \frac{1}{20} = \frac{4 - 5}{100} = -\frac{1}{100}
\]
\[
W = G \times \frac{260000}{100}
\]
\[
W = 6.7 \times 10^{-11} \times 2600
\]
\[
W = 1.74 \times 10^{-7}\,\text{J}
\]
Thus, the work done in this process is \(1.74 \times 10^{-7}\,\text{J}\).