Question

The enclosed air inside a closed box of rigid walls has pressure \(P_1\). The density of air inside this box is constant. On heating the gas, the pressure of the enclosed air is increased from \(P_1\) to \(P_2\). Now it is observed that the sound travels \(1.4\) times faster than at pressure \(P_1\). The ratio \(P_2\) by \(P_1\) is

• A. \(2.25\)
• B. \(1.96\)
• C. \(1.5\)
• D. \(1.4\)

Hint:

Speed of sound is \(v=\sqrt{\gamma P/\rho}\) and the density is fixed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a gas, \(v=\sqrt{\dfrac{\gamma P}{\rho}}\). In a rigid closed box the density stays constant.

Step 2: Key Formula or Approach
So \(v\propto\sqrt P\) and \(\dfrac{v_2}{v_1}=\sqrt{\dfrac{P_2}{P_1}}\).

Step 3: Detailed Explanation
\[ 1.4=\sqrt{\frac{P_2}{P_1}} \Rightarrow \frac{P_2}{P_1}=1.4^2=1.96 \]

Final Answer:
The ratio \(P_2/P_1\) is 1.96, option (B). \[ \boxed{1.96\ \text{(B)}} \]