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Two tubes of same length and diameters \(4 \mathrm{~mm}\) and \(8 \mathrm{~mm}\) are joined together to form a U-shaped tube open at both the ends. If the U-tube contains water, then the difference between the levels of water in the two limbs of the tube is
(Surface tension of water at the temperature of experiment is \(7.3 \times 10^{-2} \mathrm{Nm}^{-1}\), angle of contact \(=0^{\circ}\), density of water \(=1.0 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\) and acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
Options:
(Surface tension of water at the temperature of experiment is \(7.3 \times 10^{-2} \mathrm{Nm}^{-1}\), angle of contact \(=0^{\circ}\), density of water \(=1.0 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\) and acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
Solution:
1275 Upvotes
Verified Answer
The correct answer is:
\(3.65 \mathrm{~mm}\)
A figure of U-shaped tube of different diameter is given below,
Given, radius of first tube, \(R_1=2 \mathrm{~mm}\) and radius of second tube, \(R_2=4 \mathrm{~mm}\). density of water, \(\rho=1.0 \times 10^3 \mathrm{~kg}-\mathrm{m}^{-3}\)
Since, the tube is open at both the ends. So, \(p_A=p_B=p_0\) (atmospheric pressure) Pressure in capillary tube \(B\),
\(p_B=\frac{2 T}{R_2}=p_0\)
[Here, \(T\) = Surface tension of water]
Similarly pressure at capillary tube \(A\),
\(p_A=\frac{2 T}{R_1}=p_0\)
Hence, for same height \(O O^{\prime}\),
\(\begin{aligned}
-\rho g x+\frac{2 T}{R_1} & =\frac{2 T}{R_2} \\
x & =\frac{2 T}{\rho g}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
\end{aligned}\)
Putting the given values, we get
\(x=\frac{2 \times 7.3 \times 10^{-2}}{10^3 \times 10 \times 10^{-3}}\left[\frac{1}{2}-\frac{1}{4}\right]=3.65 \mathrm{~mm}\)
Hence, the difference between the levels of water in the two limbs of the tube is \(3.65 \mathrm{~mm}\).
So, the correct option is (a).
Given, radius of first tube, \(R_1=2 \mathrm{~mm}\) and radius of second tube, \(R_2=4 \mathrm{~mm}\). density of water, \(\rho=1.0 \times 10^3 \mathrm{~kg}-\mathrm{m}^{-3}\)
Since, the tube is open at both the ends. So, \(p_A=p_B=p_0\) (atmospheric pressure) Pressure in capillary tube \(B\),
\(p_B=\frac{2 T}{R_2}=p_0\)
[Here, \(T\) = Surface tension of water]
Similarly pressure at capillary tube \(A\),
\(p_A=\frac{2 T}{R_1}=p_0\)
Hence, for same height \(O O^{\prime}\),
\(\begin{aligned}
-\rho g x+\frac{2 T}{R_1} & =\frac{2 T}{R_2} \\
x & =\frac{2 T}{\rho g}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
\end{aligned}\)
Putting the given values, we get
\(x=\frac{2 \times 7.3 \times 10^{-2}}{10^3 \times 10 \times 10^{-3}}\left[\frac{1}{2}-\frac{1}{4}\right]=3.65 \mathrm{~mm}\)
Hence, the difference between the levels of water in the two limbs of the tube is \(3.65 \mathrm{~mm}\).
So, the correct option is (a).
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