Properties Of Solids And Liquids Question 284
Question: The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ratio of density of mercury to that of air is 104. The height of the hill is
Options:
A)250 m
B)2.5 km
C)1.25 km
D)750 m
Show Answer
Answer:
Correct Answer: B
Solution:
Difference of pressure between sea level and the top of hill DP $ =(h _{1}-h _{2})\times {\rho _{Hg}}\times g $
$ =(75-50)\times {{10}^{-2}}\times {\rho _{Hg}}\times g $ -(i)
and pressure difference due to h meter of airDP =$ h\times {\rho _{air}}\times g $ -(ii)
By equating (i) and (ii) we get $ h\times {\rho _{air}}\times g=(75-50)\times {{10}^{-2}}\times {\rho _{Hg}}\times g $
$ \therefore \ h=25\times {{10}^{-2}}\left( \frac{{\rho _{Hg}}}{{\rho _{air}}} \right) $
$ =25\times {{10}^{-2}}\times 10^{4}=2500m $
Height of the hill = 2.5 km.
