Answer
64.8k+ views
Hint: 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 $l = 10\;{\text{cm}}$, the length of the one pellet is ${l_1} = 43\;{\text{cm}}$, the length of the other pellet is ${l_2} = 33\;{\text{cm}}$.
Let the density of the mercury in both pellets is $\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:
${P_1} = \rho {l_1}A\left( {y - l} \right)......\left( 1 \right)$
The pressure in the other pellet is given by the equation as:
${P_2} = \rho {l_2}A\left( {y + l} \right)......\left( 2 \right)$
Equate the equation (1) and equation (2) to find the equation for the length of the atmospheric column.
${P_1} = {P_2}$
$\rho {l_1}A\left( {y - l} \right) = \rho {l_2}A\left( {y + l} \right)$
${l_1}\left( {y - l} \right) = {l_2}\left( {y + l} \right)......\left( 3 \right)$
Substitute $43\;{\text{cm}}$ for ${l_1}$, $33\;{\text{cm}}$ for ${l_2}$and $10\;{\text{cm}}$ in the equation (3) to find the length of atmospheric column.
$\left( {43\;{\text{cm}}} \right)\left( {y - 10\;{\text{cm}}} \right) = \left( {33\;{\text{cm}}} \right)\left( {y + 10\;{\text{cm}}} \right)$
$y - 10\;{\text{cm}} = \left( {0.77} \right)\left( {y + 10\;{\text{cm}}} \right)$
$y - 10\;{\text{cm}} = 0.77y + 7.7\;{\text{cm}}$
$0.23y = 17.7\;{\text{cm}}$
$y = 76.96\;{\text{cm}}$
$y \approx 77\;{\text{cm}}$
Thus, the atmospheric pressure in glass tubes is $77\;{\text{cm}}$ 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.
Complete step by step solution:
Given: The length of the air column in capillary glass tube closed at one end is $l = 10\;{\text{cm}}$, the length of the one pellet is ${l_1} = 43\;{\text{cm}}$, the length of the other pellet is ${l_2} = 33\;{\text{cm}}$.
Let the density of the mercury in both pellets is $\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:
${P_1} = \rho {l_1}A\left( {y - l} \right)......\left( 1 \right)$
The pressure in the other pellet is given by the equation as:
${P_2} = \rho {l_2}A\left( {y + l} \right)......\left( 2 \right)$
Equate the equation (1) and equation (2) to find the equation for the length of the atmospheric column.
${P_1} = {P_2}$
$\rho {l_1}A\left( {y - l} \right) = \rho {l_2}A\left( {y + l} \right)$
${l_1}\left( {y - l} \right) = {l_2}\left( {y + l} \right)......\left( 3 \right)$
Substitute $43\;{\text{cm}}$ for ${l_1}$, $33\;{\text{cm}}$ for ${l_2}$and $10\;{\text{cm}}$ in the equation (3) to find the length of atmospheric column.
$\left( {43\;{\text{cm}}} \right)\left( {y - 10\;{\text{cm}}} \right) = \left( {33\;{\text{cm}}} \right)\left( {y + 10\;{\text{cm}}} \right)$
$y - 10\;{\text{cm}} = \left( {0.77} \right)\left( {y + 10\;{\text{cm}}} \right)$
$y - 10\;{\text{cm}} = 0.77y + 7.7\;{\text{cm}}$
$0.23y = 17.7\;{\text{cm}}$
$y = 76.96\;{\text{cm}}$
$y \approx 77\;{\text{cm}}$
Thus, the atmospheric pressure in glass tubes is $77\;{\text{cm}}$ 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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)