Question

The height of Mercury column in a barometer tube is 75 cm at sea level and 60 cm at the top of Hill. If density of Mercury stand per four times of density of Mercury is 10⁴ times of density of air, then height of the hill will be (assume density of air to be uniform)
a) 1.5 Km
b) 2.5 Km
c) 6 Km
d) 4.5 Km
​

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

The height of Mercury column in a barometer tube is 75 cm at sea level and 60 cm at the top of Hill. If density of Mercury stand per four times of density of Mercury is 10⁴ times of density of air, then height of the hill will be (assume density of air to be uniform)

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Height of Mercury column in a barometer tube at sea level ${\sf h_1}$ = 75cm

• Height of Mercury column in a barometer tube at top of hill ${\sf h_2}$= 60cm

• ${\sf \dfrac{\rho_{Hg}}{\rho_{air}}=10^{-4}}$

$\large\underline{\underline{\sf To\:Find:}}$

• Height of hill (H)

We know ⎯

❏$\large{\boxed{\bf \blue{Pressure (P)=h\rho g} }}$

Pressure difference between sea level and top of hill = Pressure difference due to air

$\implies{\sf (h_2-h_1)×\rho_{Hg}×g=(H-0)×\rho_{air}×g}$

$\implies{\sf (75-60)×\rho_{Hg}×g=H×\rho_{air}×g}$

$\implies{\sf 15×\rho_{Hg}=H×\rho_{air}}$

${\sf \pink{15cm=15×10^{-2}m}}$

$\implies{\sf 15×10^{-2}×\rho_{Hg}=H×\rho_{air}}$

$\implies{\sf H= \dfrac{15×10^{-2}×\rho_{Hg}}{\rho_{air}}}$

$\implies{\sf H = 15×10^{-2}×10^4}$

$\implies{\sf H = 15×10^{2} }$

$\implies{\sf H = 1500m}$

$\implies{\bf \red{H=1.5Km}}$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Option (a) 1.5 Km

Height of hill will be ${\bf \red{1.5Km}}$

Answer 2

Answer:

answer is 1.5 cm. 2.5 cm also