MHT CET · Physics · Mechanical Properties of Fluids
Two narrow tube of diameters \(d_1\) and \(d_2\) are joined together to form a Utube open at both ends. If U-tube contains water, the difference in water levels in the limbs is ( \(T\) is the surface tension of water, angle of contact is zero and density of water is \(\rho, g\) is the acceleration due to gravity)
- A \(\frac{4 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]\)
- B \(\frac{4 T}{\rho g}\left[\frac{d_1 d_2}{d_1+d_2}\right]\)
- C \(\frac{2 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]\)
- D \(\frac{2 T}{\rho g}\left[\frac{d_1+d_2}{d_1 d_2}\right]\)
Answer & Solution
Correct Answer
(A) \(\frac{4 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]\)
Step-by-step Solution
Detailed explanation
Consider the situation below:
When angle of contact is zero degree, the radius of meniscus equals the radius of tube.
Excess pressure in the first tube \(P_1=\frac{4 T}{d_1}\)
Excess pressure in the second tube, \(P_2=\frac{4 T}{d_2}\)
Hence, the pressure difference in the two limbs of the tube is given by,
\(\Delta P=P_1-P_2=h \rho g\)
\(\Rightarrow h=\frac{P_1-P_2}{\rho g}=\frac{4 T}{\rho g}\left(\frac{1}{d_1}-\frac{1}{d_2}\right)=\frac{4 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]\)
When angle of contact is zero degree, the radius of meniscus equals the radius of tube.
Excess pressure in the first tube \(P_1=\frac{4 T}{d_1}\)
Excess pressure in the second tube, \(P_2=\frac{4 T}{d_2}\)
Hence, the pressure difference in the two limbs of the tube is given by,
\(\Delta P=P_1-P_2=h \rho g\)
\(\Rightarrow h=\frac{P_1-P_2}{\rho g}=\frac{4 T}{\rho g}\left(\frac{1}{d_1}-\frac{1}{d_2}\right)=\frac{4 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]\)
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