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 $$G=6.7 \times 10^{-11} Nm^{2}/kg^{2}$$)

Three point masses of 200 kg, 300 kg, and 400 kg originally occupy the vertices of an equilateral triangle of side 20 m, and they are then rearranged to the vertices of a larger equilateral triangle of side 25 m with the same center. We wish to find the work done in this rearrangement.

The gravitational potential energy of three point masses at the vertices of a triangle is given by $$U = -G\left(\frac{m_1 m_2}{r_{12}} + \frac{m_2 m_3}{r_{23}} + \frac{m_1 m_3}{r_{13}}\right).$$ Since the triangle is equilateral, all pairwise distances are equal to a common value $$r$$, so $$U = -\frac{G}{r}\bigl(m_1m_2 + m_2m_3 + m_1m_3\bigr)\,.$$

Calculating the sum of mass products gives $$m_1m_2 + m_2m_3 + m_1m_3 = 200\times300 + 300\times400 + 200\times400 = 60000 + 120000 + 80000 = 260000 \text{ kg}^2\,.$$

The work done is the change in potential energy, so $$W = U_f - U_i = -\frac{G \cdot 260000}{25} - \Bigl(-\frac{G \cdot 260000}{20}\Bigr).$$ This gives $$W = G \cdot 260000\Bigl(\frac{1}{20} - \frac{1}{25}\Bigr) = G \cdot 260000 \cdot \frac{1}{100} = G \cdot 2600\,.$$

Substituting the gravitational constant $$G = 6.7 \times 10^{-11}$$ yields $$W = 6.7 \times 10^{-11} \times 2600 = 6.7 \times 2.6 \times 10^{-8} = 17.42 \times 10^{-8} = 1.742 \times 10^{-7} \approx 1.74 \times 10^{-7} \text{ J}.\,$$

The correct answer is Option (4): $$1.74 \times 10^{-7}$$.