Explanation: \(\text{Hints: } W = E \Delta A\) \(\text{Solve: ধরি, বড় ফোঁটার ব্যাসার্ধ, } R = 1 \, \text{m}\) \(\text{প্রমাণত, } 125 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies (5r)^3 = R^3\) \(\implies 5r = R \implies r = \frac{R}{5} = \frac{1}{5} = 0.2\) \(\implies W = 4\pi (Nr^2 - R^2) \times T = 16\pi T\) \(\text{Ans. (C)}\)