An inverted bell lying at the bottom of a lak | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Boyle's law,for a constant temperature,the pressure and volume of a gas are inversely proportional: $P_1V_1 = P_2V_2$.At the bottom o

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(B) According to Boyle's law,for a constant temperature,the pressure and volume of a gas are inversely proportional: $P_1V_1 = P_2V_2$.At the bottom of the lake,the total pressure $P_1$ is the sum of atmospheric pressure and the pressure due to the water column: $P_1 = P_{atm} + h ho_w g$.Given: $P_{atm} = 70 \, cm$ of $Hg = 70 	imes 13.6 \, g/cm^2 	imes g$,$h = 47.6 \, m = 4760 \, cm$,$ho_w = 1 \, g/cm^3$,$V_1 = 50 \, cm^3$.At the surface,$P_2 = P_{atm} = 70 	imes 13.6 \, g/cm^2 	imes g$.Substituting the values into $V_2 = V_1 	imes (P_1 / P_2)$:$V_2 = 50 	imes \frac{(70 	imes 13.6 	imes g) + (4760 	imes 1 	imes g)}{70 	imes 13.6 	imes g}$$V_2 = 50 	imes \frac{952 + 4760}{952} = 50 	imes \frac{5712}{952} = 50 	imes 6 = 300 \, cm^3$.

An inverted bell lying at the bottom of a lake $47.6 \, m$ deep has $50 \, cm^3$ of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be ......... $cm^3$ (atmospheric pressure $= 70 \, cm$ of $Hg$ and density of $Hg = 13.6 \, g/cm^3$).

Similar Questions

$A$ body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{mv^2}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle?

If $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$,then $(x, y)$ is equal to

The average force applied on the walls of a closed container depends on $T^{x}$ where $T$ is the temperature of an ideal gas. The value of '$x$' is

Which of the following statements is true for transition elements?

The total number of nitrogenous bases in the human genome is estimated to be about

Vedclass Test Series

Exam Paper Generator

Online Exam Module