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Question: The length of the air column in a capillary glass tube closed at one end containing mercury pellet \...

The length of the air column in a capillary glass tube closed at one end containing mercury pellet 10  cm10\;{\text{cm}} long are 43  cm43\;{\text{cm}} and 33  cm33\;{\text{cm}} respectively with the close end up and down. The atmospheric pressure is:
A) 75  cm75\;{\text{cm}} of Hg
B) 77  cm77\;{\text{cm}} of Hg
C) 70  cm70\;{\text{cm}} of Hg
D) 100  cm100\;{\text{cm}} of Hg

Explanation

Solution

The above problem can be solved by using the principle of balancing the pressure in both the pellet of the mercury. The density of the mercury is the same for both pellets, so the volume of the mercury should be equal to balance the pressure in both pellets.

Complete step by step solution:
Given: The length of the air column in capillary glass tube closed at one end is l=10  cml = 10\;{\text{cm}}, the length of the one pellet is l1=43  cm{l_1} = 43\;{\text{cm}}, the length of the other pellet is l2=33  cm{l_2} = 33\;{\text{cm}}.
Let the density of the mercury in both pellets is ρ\rho , the length of the atmospheric column of the mercury is y and the area of both the pellets is A.

The pressure in the one pellet is given by the equation as:
P1=ρl1A(y−l)......(1){P_1} = \rho {l_1}A\left( {y - l} \right)......\left( 1 \right)
The pressure in the other pellet is given by the equation as:
P2=ρl2A(y+l)......(2){P_2} = \rho {l_2}A\left( {y + l} \right)......\left( 2 \right)
Equate the equation (1) and equation (2) to find the equation for the length of the atmospheric column.
P1=P2{P_1} = {P_2}
ρl1A(y−l)=ρl2A(y+l)\rho {l_1}A\left( {y - l} \right) = \rho {l_2}A\left( {y + l} \right)
l1(y−l)=l2(y+l)......(3){l_1}\left( {y - l} \right) = {l_2}\left( {y + l} \right)......\left( 3 \right)
Substitute 43  cm43\;{\text{cm}} for l1{l_1}, 33  cm33\;{\text{cm}} for l2{l_2}and 10  cm10\;{\text{cm}} in the equation (3) to find the length of atmospheric column.
(43  cm)(y−10  cm)=(33  cm)(y+10  cm)\left( {43\;{\text{cm}}} \right)\left( {y - 10\;{\text{cm}}} \right) = \left( {33\;{\text{cm}}} \right)\left( {y + 10\;{\text{cm}}} \right)
y−10  cm=(0.77)(y+10  cm)y - 10\;{\text{cm}} = \left( {0.77} \right)\left( {y + 10\;{\text{cm}}} \right)
y−10  cm=0.77y+7.7  cmy - 10\;{\text{cm}} = 0.77y + 7.7\;{\text{cm}}
0.23y=17.7  cm0.23y = 17.7\;{\text{cm}}
y=76.96  cmy = 76.96\;{\text{cm}}
y≈77  cmy \approx 77\;{\text{cm}}

Thus, the atmospheric pressure in glass tubes is 77  cm77\;{\text{cm}} of Hg and the option (B) is the correct answer.

Note: Be careful in putting the values of the length of both pellets in the appropriate equation to find the atmospheric pressure in the pellet.