Initial Volume = final volume (D) \(T \times 4 \pi R^2\)
\(\frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3\)
\(R^3 = 8r^3\)
\(R = 2r\)
\(r = \frac{R}{2}\)
Work Done W = T.d5
W = \(T . [8(4 \pi r ^2) - 4 \pi R^2]\)
Correct Option is (D)
Work done = \(T \times 4 \pi[8r^2 - R^2]\)
\(\because r = \frac{R}{2}\)
\(W = T \times 4 \pi[8 (\frac{R}{2})^2-R^2]\)
\(W = T \times 4 \pi [28 \times \frac{R^2}{4} - R^2]\)
\(W = T \times 4 \pi R^2\)
