Question

When a vertical capillary of length \( l \) with the sealed upper end was brought in contact with the surface of a liquid, the level of this liquid rise to the height \( h \). The liquid density is \( \rho \), the inside diameter of the capillary is \( d \), the contact angle is \( \theta \), the atmospheric pressure is \( P_{0} \). Find the surface tension of the liquid.

(A) \( \left[\rho g h+\frac{P_{0} h}{l+h}\right] \frac{d}{4 \cos \theta} \)

(B) \( \left[\rho g h+\frac{P_{0} h}{l-h}\right] \frac{d}{4 \cos \theta} \)

(C) \( \left[\rho g h+\frac{l+h}{P_{0} h}\right] \frac{d}{4 \cos \theta} \)

(D) \( \left[\rho g h+\frac{P_{0} h}{l+h}\right] \frac{4 \cos \theta}{d} \)

Answer 1

Correct option is (B) \(\left[\rho g h+\frac{p_0 h}{l-h}\right] \frac{d}{4 \cos \theta}\)

We have the Boyle's law

\( \left[p_0-\rho g h+\frac{4 \alpha \cos \theta}{d}\right](l-h)=p_0 l\)

\( \text {or } \frac{4 \alpha \cos \theta}{d}=\rho g h+\frac{p_0 h}{l-h}\)

Hence, \(\alpha=\left[\rho g h+\frac{p_0 h}{l-h}\right] \frac{d}{4 \cos \theta}\).