AP EAMCET · PHYSICS · Mechanical Properties of Fluids
A capillary tube of radius \(0.1 \mathrm{~mm}\) is dipped in water. The water rises to a height of \(2 \mathrm{~cm}\) in the tube. If the surface tension of water is \(0.072 \mathrm{Nm}^{-1}\), the contact angle between water and wall of the tube is
- A \(\theta=\cos ^{-1}\left(\frac{1}{3.6}\right)\)
- B \(\theta=\cos ^{-1}\left(\frac{1}{7.2}\right)\)
- C \(\theta=\cos ^{-1}\left(\frac{1}{1.8}\right)\)
- D \(\theta=\cos ^{-1}\left(\frac{1}{6.2}\right)\)
Answer & Solution
Correct Answer
(B) \(\theta=\cos ^{-1}\left(\frac{1}{7.2}\right)\)
Step-by-step Solution
Detailed explanation
Radius of capillary tube, \(\mathrm{r}=0.1 \mathrm{~mm}=0.1 \times 10^{-3} \mathrm{~m}\) Height of water rises, \(\mathrm{h}=2 \mathrm{~cm}=02 \mathrm{~m}\) Surface tension, \(T=0.072 \mathrm{~N} / \mathrm{m}\) Rise in water is given by…
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