Q188 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
The atmospheric pressure on the surface of the fluid is $P_0$ . The depth $h$ at which pressure is $n$ times the atmospheric pressure is
- A$\dfrac{nP_0}{\rho g}$
- B$\dfrac{(n+1)P_0}{\rho g}$
- Correct answerC$\dfrac{(n-1)P_0}{\rho g}$
- D$\dfrac{n^2 P_0}{\rho g}$
Solution
The pressure $P$ at a depth $h$ below the surface of a fluid of density $\rho$ is given by
$P = P_0 + \rho g h$
Given that the pressure at depth $h$ is $n$ times the atmospheric pressure, we have
$P = nP_0$
Substituting this into the pressure equation gives
$nP_0 = P_0 + \rho g h$
$\rho g h = nP_0 - P_0$
$\rho g h = (n - 1)P_0$
$h = \dfrac{(n - 1)P_0}{\rho g}$
Answer: $\dfrac{(n-1)P_0}{\rho g}$
$P = P_0 + \rho g h$
Given that the pressure at depth $h$ is $n$ times the atmospheric pressure, we have
$P = nP_0$
Substituting this into the pressure equation gives
$nP_0 = P_0 + \rho g h$
$\rho g h = nP_0 - P_0$
$\rho g h = (n - 1)P_0$
$h = \dfrac{(n - 1)P_0}{\rho g}$
Answer: $\dfrac{(n-1)P_0}{\rho g}$
Hence, the correct answer is option: 'C'.