First slide

Surface tension

Question

A mercury drop of radius I cm is broken into $10^{6}$ droplets of equal size. The work done is ( $T = 35 \times 10^{- 2}$ N/m)

Moderate

A

$4.35 \times 10^{- 2} J$
B

$4.35 \times 10^{- 3} J$
C

$4.35 \times 10^{- 6} J$
D

$4.35 \times 10^{- 8} J$

Solution

Initial radius of drop,
R = 1 cm = $1 \times 10^{- 2} m$
Suppose, r is the radius of droplets.
Since, volume remain constant.
$\frac{4}{3} {πR}^{3} = 10^{6} \times \frac{4}{3} {πr}^{3}$
[Since, number of droplets = $10^{6}$ ]
$∴ r = \frac{1 \times 10^{- 2}}{10^{2}} = 10^{- 4} [∵ R = 1 \times 10^{- 2} m]$
Increase in surface area,
$∆ S = 10^{6} \times 4 {πr}^{2} - 4 {πR}^{2}$
= $10^{6} \times 4 \times π \times 10^{- 8} - 4 π \times 10^{- 4}$
= $4 \times 9.9 \times π \times 10^{- 3}$
$∴ Work done = Increase in surface area \times Surface tension$
= $4 \times 9.9 \times π \times 10^{- 3} \times 35 \times 10^{- 2}$
= $4.35 \times 10^{- 2} J$

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