Question

The pressure at half the depth of a lake is equal to two-third pressure at the bottom of the lake. So the depth 'h' of the lake is (\(ρ\) = density of water in the lake, \(g\) = acceleration due to gravity, \(P_0\) = atmospheric pressure).

• A. \(\frac{2P_0}{ρg}\)
• B. \(\frac{P_0}{ρg}\)
• C. \(\frac{2P_0}{3ρg}\)
• D. \(\frac{P_0}{2ρg}\)

Hint:

Write absolute pressure at half depth and at the bottom.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The absolute pressure at depth \(d\) below the surface of a liquid is \(P = P_0 + \rho g d\).

Step 2: Key Formula or Approach:
At half depth: \(P_0 + \rho g\frac h2\). At the bottom: \(P_0 + \rho g h\). The given condition is \(P_{half} = \frac23 P_{bottom}\).

Step 3: Detailed Explanation:
\(P_0 + \frac{\rho g h}{2} = \frac23\left(P_0 + \rho g h\right)\).
Multiply by \(6\): \(6P_0 + 3\rho g h = 4P_0 + 4\rho g h\).
So \(2P_0 = \rho g h\).
\[ h = \frac{2P_0}{\rho g} \]

Final Answer:
The depth is \(h = \frac{2P_0}{\rho g}\), option (A). \[ \boxed{\frac{2P_0}{\rho g}} \]