Question

In an open tube manometer, the volume of the gas enclosed in the bulb is \(250\,\mathrm{cm^3}\) and the difference in the heights of the liquid levels in the manometer is \(125\,\mathrm{cm}\). The density of the liquid in the manometer is \[ \left( g=10\,\mathrm{m\,s^{-2}},\; P_{\text{atm}}=100\,\mathrm{kPa} \right) \]

• A. \(12.5\,\mathrm{kg\,m^{-3}}\)
• B. \(0.16\,\mathrm{kg\,m^{-3}}\)
• C. \(4.8\,\mathrm{kg\,m^{-3}}\)
• D. \(8\times10^{3}\,\mathrm{kg\,m^{-3}}\)

Hint:

For an open tube manometer, \[ \boxed{ \Delta P=\rho gh } \] where \(h\) is the difference in the liquid levels.

Answer 1

The Correct Option is D

Step 1: Use the pressure relation for an open tube manometer. Pressure difference is \[ \Delta P=\rho gh. \] Here, \[ \Delta P=P_{\text{atm}} =100\,\text{kPa} =10^5\,\text{Pa}, \] and \[ h=125\,\text{cm}=1.25\,\text{m}. \]

Step 2: Calculate the density. \[ \rho = \frac{\Delta P}{gh} = \frac{10^5}{10\times1.25} = 8000\,\mathrm{kg\,m^{-3}}. \] Thus, \[ \boxed{\rho=8\times10^{3}\,\mathrm{kg\,m^{-3}}.} \] Hence, \[ \boxed{(D)} \] is the correct answer.