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

Step 1: Expression for gravitational potential energy.
For three masses placed at the vertices of an equilateral triangle of side $r$, total gravitational potential energy is:
\[ U = -G\left(\frac{m_1 m_2 + m_2 m_3 + m_3 m_1}{r}\right) \] Step 2: Calculating initial potential energy.
\[ U_i = -G\left(\frac{(200)(300) + (300)(400) + (400)(200)}{20}\right) \] \[ U_i = -6.7 \times 10^{-11} \times \frac{260000}{20} \] Step 3: Calculating final potential energy.
\[ U_f = -G\left(\frac{260000}{25}\right) \] Step 4: Work done in rearrangement.
\[ W = U_f - U_i \] \[ W = 6.7 \times 10^{-11} \times 260000\left(\frac{1}{20} - \frac{1}{25}\right) \] \[ W = 1.74 \times 10^{-7}\,\text{J} \] Step 5: Final conclusion.
The work done in rearranging the masses is $1.74 \times 10^{-7}\,\text{J}$.