Home
Class 11
PHYSICS
The height of a mercury barometer is 75 ...

The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ration of density of mercury to that of air is `10^(4)`. The height of the hill is

A

250 m

B

2.5 km

C

1.25 km

D

750 m

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will use the information provided about the heights of the mercury barometer at sea level and at the top of the hill, as well as the ratio of the densities of mercury and air. ### Step-by-Step Solution: 1. **Identify Given Values:** - Height of mercury barometer at sea level (h1) = 75 cm - Height of mercury barometer at the top of the hill (h2) = 50 cm - Ratio of density of mercury (ρ_m) to density of air (ρ_a) = 10^4 2. **Convert Heights to Meters:** - h1 = 75 cm = 0.75 m - h2 = 50 cm = 0.50 m 3. **Calculate the Pressure Difference:** - The pressure at sea level (P1) can be expressed as: \[ P_1 = \rho_m \cdot g \cdot h_1 \] - The pressure at the top of the hill (P2) can be expressed as: \[ P_2 = \rho_m \cdot g \cdot h_2 \] - The pressure difference (ΔP) is given by: \[ \Delta P = P_1 - P_2 = \rho_m \cdot g \cdot (h_1 - h_2) \] 4. **Express the Pressure Difference in Terms of Air Density:** - The pressure difference due to the height of the hill (h) can be expressed as: \[ \Delta P = \rho_a \cdot g \cdot h \] 5. **Equate the Two Expressions for Pressure Difference:** - Setting the two expressions for ΔP equal gives: \[ \rho_m \cdot g \cdot (h_1 - h_2) = \rho_a \cdot g \cdot h \] - We can cancel out g from both sides: \[ \rho_m \cdot (h_1 - h_2) = \rho_a \cdot h \] 6. **Substitute the Density Ratio:** - From the ratio of densities, we have: \[ \frac{\rho_m}{\rho_a} = 10^4 \implies \rho_m = 10^4 \cdot \rho_a \] - Substitute this into the equation: \[ 10^4 \cdot \rho_a \cdot (h_1 - h_2) = \rho_a \cdot h \] - Cancel ρ_a (assuming it is not zero): \[ 10^4 \cdot (h_1 - h_2) = h \] 7. **Calculate the Height of the Hill:** - Substitute the values of h1 and h2: \[ h = 10^4 \cdot (0.75 - 0.50) = 10^4 \cdot 0.25 = 2500 \text{ m} \] - Convert meters to kilometers: \[ h = 2.5 \text{ km} \] ### Final Answer: The height of the hill is **2.5 kilometers**.
Promotional Banner

Topper's Solved these Questions

  • PROPERTIES OF MATTER

    DC PANDEY|Exercise JEE Advanced|57 Videos
  • PROPERTIES OF MATTER

    DC PANDEY|Exercise More than one option is correct|21 Videos
  • PROJECTILE MOTION

    DC PANDEY|Exercise Level - 2 Subjective|10 Videos
  • RAY OPTICS

    DC PANDEY|Exercise Integer type q.|15 Videos

Similar Questions

Explore conceptually related problems

the height of a mercury barometer is 76 cm at sea level and 50 cm at the top of a hill. Determine the height of the hill. (Densities of air and mercury are 1.36 kg m^(-3) and 13.6 xx 10^(3) kg m^(-3)) hint : Difference of pressure =76 cm - 50 cm = 26 cm of mercury 0.26 xx 9.8 xx 13.6 xx 10^(3) Nm^(-2) pressure due to h metre of air = h xx 9.8 xx 1.36 ]

The height of mercury which exerts the same pressure as 20 cm of water column, is

What will be the height of mercury column in a barometer it it is taken to the moon?

The column of mercury in a barometer is 76 cm Hg. Calculate the atmosphere pressure I the density of mercury =13600kgm^(-3) (Take g=10ms^(-2) )