WBJEE · Physics · Mechanical Properties of Fluids
A uniform capillary tube of length \(l\) and inner radius \(r\) with its upper end sealed is submerged vertically into water. The outside pressure is \(p_{0}\) and surface tension of water is \(\gamma\). When a length \(x\) of the capillary is submerged into water, it is found that water levels inside and outside the capillary coincide. The value of \(x\) is
- A \(\frac{l}{\left(1+\frac{p_{0} r}{4 \gamma}\right)}\)
- B \(l\left(1-\frac{p_{0} r}{4 \gamma}\right)\)
- C \(l\left(1-\frac{p_{0} r}{2 \gamma}\right)\)
- D \(\frac{l}{\left(1+\frac{p_{0} r}{2 \gamma}\right)}\)
Answer & Solution
Correct Answer
(D) \(\frac{l}{\left(1+\frac{p_{0} r}{2 \gamma}\right)}\)
Step-by-step Solution
Detailed explanation
When the sealed capillary tube is submerged vertically into water, the pressure inside the tube changes For the inside capillary, \[ p_{1} V_{1}=p_{2} v_{2} \] \(\therefore \quad P_{0}(l A)=p^{\prime}(l-x) A\) where \(p^{\prime}\) is pressure in capillary after being submerged…
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