Energy needed in breaking a drop of radius R into n identical drops of radii r is given by

Question

Energy needed in breaking a drop of radius R into n identical drops of radii r is given by (A) 4 π T (. n r2 - R2 .) (B) (4/3) π (. r3 n - R2 .) (C) 4 π T (. R2 - n r2 .) (D) 4 π T (. n r2 + R2 .)

Tardigrade Admin · Accepted Answer

Energy needed = Increment in surface energy = (surface energy of n small drops) - (surface energy of one big drop) = n 4 π r2 T - 4 π R2 T = 4 π T (. n r2 - R2 .)

Q. Energy needed in breaking a drop of radius $R$ into $n$ identical drops of radii $r$ is given by

$4 π T (n r^{2} - R^{2})$

$\frac{4}{3} π (r^{3} n - R^{2})$

$4 π T (R^{2} - n r^{2})$

$4 π T (n r^{2} + R^{2})$

Energy needed = Increment in surface energy = (surface energy of n small drops) - (surface energy of one big drop) $= n 4 π r^{2} T - 4 π R^{2} T = 4 π T (n r^{2} - R^{2})$

Q. Energy needed in breaking a drop of radius R into n identical drops of radii r is given by

4πT(nr2−R2)

34​π(r3n−R2)

4πT(R2−nr2)

4πT(nr2+R2)