Question

Calculate the amount of energy evolved when 343 droplets of mercury each of radius. 0.05 mm, combine to form one drop. The surface tension of mercury is 50 x 10^-2 N/m.

Answer 1

Let R = radius of big drop

r = radius of small drop

T = 50 x 10^-2 N/m

r = 0 .05 x l0^-3 m ,

Energy evolved = ?

∴ Volume of one drop = Volume of 343 droplets

`4/3 piR^3 = 343 xx 4/3 pir^3`

∴ R³ = 343 r³

∴ R = 7r

Energy evolved

 = Surface tension × Change in area

 = T. dA

= T  × [343 × 4πr² - 4πR²]

= T x [343 x 4πr² - 4π(7r)²]

= T x 4πr² [343 - 49]

= T x 4πr² x 294

= 50 x10^-2 x 4π (0.05 x 10^-3)² x 294

= 46158 x 10^-10

∴ Energy evolved= 4 .6158 x 10^-6 J