A manometer tube contanins a liquid of density `3 xx 10^(3) kg m^(-3)`. When connected to a vessel containing a gas, the liquid level in the other arm of the tube is higher by 10 cm. When connected to another sample of enclosed gas, the liquid level in the other arm of the manometer tube falls 7 cm below the liquid level in the first arm. Which of the two samples exerts more pressureand by what amount ?

Explanation

For Sample 1
`h_(1) = 10 cm = 0.1 m`
`P_(1) = P_(a) + rho gh_(1)" "...(i)`

For Sample 2
In this case, level of the liquid in the left arm is higher than that in the right arm by 7 cm.
`therefore` Atmospheric pressure `P_(a)` is greater than the pressure exerted by the sample
`rArr P_(a) = P_(2) + rho gh_(2)`
`rArr P_(2) = P_(a) - rho gh_(2)" "...(ii)`

Comparing equaitons (i) and (ii), it is clear that `P_(1) gt P_(2)`. Therefore the gas in sample 1 exerts greater pressure than that in sample 2.
The difference in the two pressures is
`P_(1) - P_(2)= (P_(a) + rho gh_(1)) - (P_(a) - rho gh_(2))`
`= rho g(h_(1) + h_(2))`
`=rho g(17 cm)`
`= (3 xx 10^(3) kg m^(-3)) xx (9.8 ms^(-2))(0.17 m)`
`4.99 xx 10^(3) Pa`
= 5 kPa