The surface tension and vapour pressure of water at 20° C are 7.28× 10- 2 Nm- 1 and 2.33× 103Pa respectively. At this temperature, the radius of the smallest possible drop, which can form without evaporating is

Question

The surface tension and vapour pressure of water at 20° C are 7.28× 10- 2 Nm- 1 and 2.33× 103Pa respectively. At this temperature, the radius of the smallest possible drop, which can form without evaporating is (A) 1.25× 10-5 m (B) 6.25× 10-5 m (C) 4.3× 10-5 m (D) 3.4× 10-5 m

Tardigrade Admin · Accepted Answer

Here, S=7.28× 10- 2Nm- 1,P=2.33× 103Pa The drop will evaporate if the water pressure is greater than the vapour pressure P As P=(2 S/R) where R is the radius of drop ∴ R=(2 S/P)=(2 × 7.28 × 10- 2/2.33 × 103)=6.25× 10- 5

Q. The surface tension and vapour pressure of water at $2 0^{\circ} C$ are $7.28 \times 1 0^{- 2} N m^{- 1}$ and $2.33 \times 1 0^{3} P a$ respectively. At this temperature, the radius of the smallest possible drop, which can form without evaporating is

$1.25 \times 1 0^{- 5} m$

$6.25 \times 1 0^{- 5} m$

$4.3 \times 1 0^{- 5} m$

$3.4 \times 1 0^{- 5} m$

Q. The surface tension and vapour pressure of water at 20∘C are 7.28×10−2Nm−1 and 2.33×103Pa respectively. At this temperature, the radius of the smallest possible drop, which can form without evaporating is

1.25×10−5m

6.25×10−5m

4.3×10−5m

3.4×10−5m

Here, S=7.28×10−2Nm−1,P=2.33×103Pa The drop will evaporate if the water pressure is greater than the vapour pressure P As P=R2S​ where R is the radius of drop ∴R=P2S​=2.33×1032×7.28×10−2​=6.25×10−5

Q. The surface tension and vapour pressure of water at $2 0^{\circ} C$ are $7.28 \times 1 0^{- 2} N m^{- 1}$ and $2.33 \times 1 0^{3} P a$ respectively. At this temperature, the radius of the smallest possible drop, which can form without evaporating is

$1.25 \times 1 0^{- 5} m$

$6.25 \times 1 0^{- 5} m$

$4.3 \times 1 0^{- 5} m$

$3.4 \times 1 0^{- 5} m$