Q151 / 180 · NEET 2026 (Cancelled) · Physics

A submarine is designed to withstand an absolute pressure of $100$ atm. How deep can it go below the water surface ?
(Consider the density of water $= 1000$ kg m$^{-3}$, $1$ atm $= 1 \times 10^{5}$ Pa and gravitational acceleration $g = 10$ m/s$^2$)

  • Correct answer
    A
    $990$ m
  • B
    $9900$ m
  • C
    $99$ m
  • D
    $9000$ m

Solution

The absolute pressure at a depth $h$ below the water surface is given by $P = P_0 + \rho g h$, where $P_0$ is the atmospheric pressure.

Given:
$P = 100$ atm $= 100 \times 10^5$ Pa
$P_0 = 1$ atm $= 1 \times 10^5$ Pa
$\rho = 1000$ kg m$^{-3}$
$g = 10$ m/s$^2$

Substituting the values into the equation:

$100 \times 10^5 = 1 \times 10^5 + 1000 \times 10 \times h$

$99 \times 10^5 = 10^4 \times h$

$h = \dfrac{99 \times 10^5}{10^4}$

$h = 990$ m

Answer: $990$ m

Hence, the correct answer is option: 'A'.

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