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A think tube sealed at both ends is 100 cm long. It lies horizontally, the middle 20 cm containing mercury and two equal ends containing air at standard atmospheric pressure. if the tube is now turned to a vertical position, by what amount will the mercury be displaced? 

(Given: cross section of the tube can be assumed to be uniform)

Option: 1

2.95 cm


Option: 2

5.18 cm


Option: 3

8.65 cm


Option: 4

0.0 cm


Answers (1)

best_answer

Let A and B are two chamber 

When the tube is horizontally placed

P_A=76cm \ Hg \ and \ P_B=76cm \ Hg

V_A=40a \ and \ V_B=40a

Where a is an area of chambers

When the tube is vertically placed

let by d amount will the mercury be displaces

So new pressure in chambers areP'_A \ and \ P'_B

and V'_A=(40+d)a \ and \ V'_B=(40-d)a

From

Boyle's law:P_1V_1=P_2V_2

For chamber A

(76 \times 40 a)=P'_{A} \times(40+d) a

P'_{A}=\frac{76 \times 40}{40+d}

For the chamber B

(76 \times 40 a)=P'_{B} \times(40-d) a

P'_{B}=\frac{76 \times 40}{40-d}

\text { Also } P'_{A}+20=P'_{B}

 on Solving we get

d=5.18 cm

 

Posted by

Gautam harsolia

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