Question

if value of acceleration due to gravity at a place decreases by 3% then find the change in the height of the mercury in a barometer at that place

Answer 1

density remains same
Graviry get low by 3 percent so new gravity will be 97/100g now pressure must remain same in barometer and pressure= hdg but now pressure becomes hd97/100g so hdg=hd97/100g so new height will be100/97 of intial height

Answer 2

The height of mercury in barometer increases by 3%, if the acceleration due to gravity decreased by 3%

Explanation:

The pressure in the mercury barometer is defined by the formula,

$P=\rho h g \rightarrow(1)$

So,

$h=\frac{P}{\rho g} \rightarrow(2)$

Where,

P is the pressure in the barometer

ρ is mercury’s density

g is the acceleration due to gravity

The term P/ρ will remain constant in barometer. So it states that height of mercury will be inversely proportional to the acceleration due to gravity.

The change in height of the mercury in barometer can be found by differentiating eqn (2), we get,

$\Delta h=\frac{P \Delta g}{\rho g^{2}} \rightarrow(3)$

Now, divide eqn (3) with eqn (2)

$\frac{\Delta h}{h}=-\frac{P \Delta g}{\rho g^{2}} \times \frac{\rho g}{P}$

So,  

$\frac{\Delta h}{h}=-\frac{\Delta g}{g} \rightarrow(4)$

As it is given in question,

$\Delta \frac{g}{g}=-3 \%$

On substituting this in eqn (4), we get,

$\frac{\Delta h}{h}=3 \% \rightarrow(5)$

So the height of the mercury in barometer increases by 3%