Question

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

• A. $9.86 \times 10^{-6}$
• B. $2.85 \times 10^{-7}$
• C. $4.77 \times 10^{-7}$
• D. $1.74 \times 10^{-7}$

Hint:

Work done in rearranging masses against gravity equals the change in gravitational potential energy of the system.

Answer 1

The Correct Option is D

To determine the work done in rearranging the masses on the vertices of a larger equilateral triangle, we have to calculate the change in gravitational potential energy. The formula for gravitational potential energy between two masses is given by:

\(U = -\frac{G \cdot m_1 \cdot m_2}{r}\)

where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses, and \(r\) is the distance between the masses.

First, we need to calculate the initial potential energy when masses are placed at the vertices of the smaller triangle with side \(20 \, \text{m}\). The initial gravitational potential energy, \(U_i\), can be calculated by summing the potential energies for each unique pair of masses:

• Masses at vertices: \(200 \, \text{kg}, 300 \, \text{kg}, 400 \, \text{kg}\)
• Distance between each pair: \(r = 20 \, \text{m}\)

The initial potential energy is:

\(U_i = - \left( \frac{G \times 200 \times 300}{20} + \frac{G \times 300 \times 400}{20} + \frac{G \times 400 \times 200}{20} \right)\)

Substitute the values:

\(U_i = - \left( \frac{6.7 \times 10^{-11} \times 200 \times 300}{20} + \frac{6.7 \times 10^{-11} \times 300 \times 400}{20} + \frac{6.7 \times 10^{-11} \times 400 \times 200}{20} \right)\)

Calculating the individual terms:

• \(\text{First pair: } \frac{G \times 200 \times 300}{20} = 2.01 \times 10^{-7} \, \text{J}\)
• \(\text{Second pair: } \frac{G \times 300 \times 400}{20} = 4.02 \times 10^{-7} \, \text{J}\)
• \(\text{Third pair: } \frac{G \times 400 \times 200}{20} = 2.68 \times 10^{-7} \, \text{J}\)

Sum of the initial potential energies:

\(U_i = -(2.01 \times 10^{-7} + 4.02 \times 10^{-7} + 2.68 \times 10^{-7}) = -8.71 \times 10^{-7} \, \text{J}\)

Now calculate the potential energy when the masses are on a larger triangle with side \(25 \, \text{m}\). The new gravitational potential energy, \(U_f\), is given by:

\(U_f = - \left( \frac{G \times 200 \times 300}{25} + \frac{G \times 300 \times 400}{25} + \frac{G \times 400 \times 200}{25} \right)\)

Substitute the values:

\(U_f = - \left( \frac{6.7 \times 10^{-11} \times 200 \times 300}{25} + \frac{6.7 \times 10^{-11} \times 300 \times 400}{25} + \frac{6.7 \times 10^{-11} \times 400 \times 200}{25} \right)\)

Calculating the individual terms:

• \(\text{First pair: } \frac{G \times 200 \times 300}{25} = 1.61 \times 10^{-7} \, \text{J}\)
• \(\text{Second pair: } \frac{G \times 300 \times 400}{25} = 3.22 \times 10^{-7} \, \text{J}\)
• \(\text{Third pair: } \frac{G \times 400 \times 200}{25} = 2.15 \times 10^{-7} \, \text{J}\)

Sum of the final potential energies:

\(U_f = -(1.61 \times 10^{-7} + 3.22 \times 10^{-7} + 2.15 \times 10^{-7}) = -6.98 \times 10^{-7} \, \text{J}\)

The work done in rearranging the masses is the difference in potential energies:

\(\Delta U = U_f - U_i = -6.98 \times 10^{-7} + 8.71 \times 10^{-7} = 1.73 \times 10^{-7} \, \text{J}\)

Thus, the correct answer is \(1.73 \times 10^{-7} \, \text{J}\).