MHT CET · Physics · Mechanical Properties of Fluids

When capillary is dipped vertically in water, rise of water in capillary is ' h '. The angle of contact is zero. Now the tube is depressed so that its length above the water surface is \(\frac{\mathrm{h}}{3}\). The new apparent angle of contact is \(\left(\cos 0^{\circ}=1\right)\)

A \(\cos ^{-1}\left(\frac{1}{2}\right)\)
B \(\cos ^{-1}\left(\frac{1}{3}\right)\)
C \(\cos ^{-1}\left(\frac{1}{4}\right)\)
D \(\cos ^{-1}\left(\frac{1}{6}\right)\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\cos ^{-1}\left(\frac{1}{3}\right)\)

Step-by-step Solution

Detailed explanation

Surface tension in terms of capillary rise \(h\) is
\(\begin{array}{ll}
& \mathrm{T}=\frac{\text { rh } \rho g}{2 \cos \theta} \Rightarrow \frac{\mathrm{rh}^{\prime} \rho g}{2 \cos \theta^{\prime}} \\
\therefore & \frac{\cos \theta^{\prime}}{\cos \theta}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=\frac{1}{3} \\
\therefore & \cos \theta^{\prime}=\frac{1}{3} \cos 0^{\circ}=\frac{1}{3} \\
\therefore & \theta^{\prime}=\cos ^{-1}\left(\frac{1}{3}\right)
\end{array}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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