Question

the hight of a mercury barometer is 75 cm at sea level and 50cm at the top of a hill. ratio of density of mercury to that of air is 100000.the height of tge hill is

Answer 1

Here the pressure difference between the sea level and the hill and that due to atmospheric pressure would be same.

So,

Pressure difference between sea level and the hill would be

dP1 = dh1 X ρm X g  (1)

where dh = h1 (sea level mercury height = 75 cm) - h2 (hill top mercury height = 50 cm)

 ρm is the density of mercury

 

and  the pressure difference in atmospheric pressure would be

dP2 = dh2 X ρa X g  (2)

where dh2 is the height of the hill, ρa is the density of air 

now, we know that (1) = (2), so

 

dh1 X ρm X g = dh2 X ρa X g

 

by substituting appropriate values and rearranging, we get

 

dh2 = 0.25 X  (ρm/ ρa) = 0.25 X 104 m

or dh2 = 2500 m 

 

thus, the height of the hill is 

dh2 = 2500m = 2.5 km

Answer 2

Answer:

Pressure differences between sea level and the top of hill

△p=(h1−h2)×pHg×g

=(75−50)×10−2×pHg×g   ...(i)

and pressure difference due to h metre of air

△p=h×pair×g    ...(ii)

By equating Eqs. (i) and (ii)

h×pair×g

=(75−50)×10−2×pHg×p

p=25×10−2(pairpHg)

=25×10−2×104=2500m

∴ Height of hill = 2.5 km