Numerical

A raindrop of diameter 4 mm is about to fall on the ground. Calculate the pressure inside the raindrop. [Surface tension of water T = 0.072 N/m, atmospheric pressure = 1.013 x 10^{5} N/m^{2} ]

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#### Solution

Given:

T = 0.072 N/m,

d = 4mm ∴ r = 2 * 10^{-3} m,

P_{o }= 1.013 * 10^{5} N/m^{2}

To find: Pressure inside the raindrop (P_{i})

Formula: Pi = Po + 2T/r

Calculation: From formula,

`P_i=1.013xx10^5+(2xx0.072)/(2xx10^(-3))`

`P_i=1.013xx10^5+0.072xx10^3`

`=1.013xx10^5+0.00072xx10^5`

`thereforeP_i=1.01372xx10^5Pa`

**The pressure inside the raindrop is 1.01372 * 10 ^{5} Pa.**

Concept: Surface Tension

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