Work done in splitting a drop of water of 1 mm radius into 10 droplets is (surface tension of water =72 × 10-3J/m2 )

Question

Work done in splitting a drop of water of 1 mm radius into 10 droplets is (surface tension of water =72 × 10-3J/m2 ) (A)  9.58 × 10-5J  (B)  8.95 × 10-5J  (C)  5.89 × 10-5J  (D)  5.98 × 10-6J

Tardigrade Admin · Accepted Answer

Work done in splitting a water drop of radius R into n drops of equal size =4π R2T(n1/3-1) =4π × (10-3)2× 72× 10-3× (106/3-1) =4π × 10-6× 72× 10-3× 99 =8.95× 10-5J

Q. Work done in splitting a drop of water of 1 mm radius into 10 droplets is (surface tension of water $= 72 \times 10^{- 3} J / m^{2}$ )

$9.58 \times 10^{- 5} J$

$8.95 \times 10^{- 5} J$

$5.89 \times 10^{- 5} J$

$5.98 \times 10^{- 6} J$

Work done in splitting a water drop of radius $R$ into $n$ drops of equal size $= 4 π R^{2} T (n^{1/3} - 1)$ $= 4 π \times (10^{- 3})^{2} \times 72 \times 10^{- 3} \times (10^{6/3} - 1)$ $= 4 π \times 10^{- 6} \times 72 \times 10^{- 3} \times 99$ $= 8.95 \times 10^{- 5} J$

Q. Work done in splitting a drop of water of 1 mm radius into 10 droplets is (surface tension of water =72×10−3J/m2 )

9.58×10−5J

8.95×10−5J

5.89×10−5J

5.98×10−6J