
In a U-tube in which the cross-sectional area of the limb on the left is one-quarter, the limb on the right contains mercury (density 13.6 g/cc), and the level of mercury in the narrow limb is at a distance of 36cm from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water?
A. 1.2 cm
B. 2.35 cm
C. 0.5 cm
D. 0.8 cm
Answer
560.1k+ views
Hint: In this question we have been given that in a u tube the arms are filled with mercury. Later, water is added to the left arm which is one quarter the right arm. We have been asked to calculate the rise in mercury level in the right arm due to water added in the left arm. We know that pressure at a point in the u tube filled with a certain liquid is equal to the pressure at the point on the same horizontal plane in the other arm. This is due to Pascal's law.
Complete answer:
We know that when water is added in the u tube, the level of mercury in the arm will decrease by a certain height, say ‘y’ as shown in the figure. Therefore, water added will be (36+y) cm. now, as both the arms have different area say \[{{A}_{1}}\]and \[{{A}_{2}}\], the increase in mercury level in right arm will not be same as y.
Therefore, let us consider that the rise in mercury level in right arm is ‘x’
We know that,
\[{{A}_{2}}=\dfrac{1}{4}{{A}_{1}}\] ………… (1)
We know that the volume of mercury will remain constant.
Therefore,
\[{{A}_{2}}y={{A}_{1}}x\] ……….. (2)
From (1) and (2)
We get,
\[y=4x\] ……………. (3)
Now,
From Pascal's law for pressure at a point we know that the pressure at point A will be equal to the pressure at point b.
\[{{P}_{A}}={{P}_{B}}\] ……………. (4)
Now, pressure at point A is given by,
\[{{P}_{A}}={{P}_{0}}+(36+y){{\rho }_{w}}g\] ……………….. (5)
Similarly, pressure at point B,
\[{{P}_{B}}={{P}_{0}}+(x+y){{\rho }_{m}}g\]…………………. (6)
Now, from (4), (5) and (6)
We get,
\[{{P}_{0}}+(36+y){{\rho }_{w}}g={{P}_{0}}+(x+y){{\rho }_{m}}g\]
On solving,
\[(36+4x)\times 1=5x\times 13.6\] …………… (from (3))
Therefore,
\[x=0.56cm\]
Therefore, the correct answer is option C.
Note:
Pascal's law also known as Pascal's principle states that, in a fluid at rest in a closed container, pressure change at one point is transmitted through the entire container such that same changes occur everywhere. The pressure in the container will increase as the length of the liquid in the container is increased.
Complete answer:
We know that when water is added in the u tube, the level of mercury in the arm will decrease by a certain height, say ‘y’ as shown in the figure. Therefore, water added will be (36+y) cm. now, as both the arms have different area say \[{{A}_{1}}\]and \[{{A}_{2}}\], the increase in mercury level in right arm will not be same as y.
Therefore, let us consider that the rise in mercury level in right arm is ‘x’
We know that,
\[{{A}_{2}}=\dfrac{1}{4}{{A}_{1}}\] ………… (1)
We know that the volume of mercury will remain constant.
Therefore,
\[{{A}_{2}}y={{A}_{1}}x\] ……….. (2)
From (1) and (2)
We get,
\[y=4x\] ……………. (3)
Now,
From Pascal's law for pressure at a point we know that the pressure at point A will be equal to the pressure at point b.
\[{{P}_{A}}={{P}_{B}}\] ……………. (4)
Now, pressure at point A is given by,
\[{{P}_{A}}={{P}_{0}}+(36+y){{\rho }_{w}}g\] ……………….. (5)
Similarly, pressure at point B,
\[{{P}_{B}}={{P}_{0}}+(x+y){{\rho }_{m}}g\]…………………. (6)
Now, from (4), (5) and (6)
We get,
\[{{P}_{0}}+(36+y){{\rho }_{w}}g={{P}_{0}}+(x+y){{\rho }_{m}}g\]
On solving,
\[(36+4x)\times 1=5x\times 13.6\] …………… (from (3))
Therefore,
\[x=0.56cm\]
Therefore, the correct answer is option C.
Note:
Pascal's law also known as Pascal's principle states that, in a fluid at rest in a closed container, pressure change at one point is transmitted through the entire container such that same changes occur everywhere. The pressure in the container will increase as the length of the liquid in the container is increased.
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