The force acting on a window of area 50 cm × 50 cm of a submarine at a depth of 2000 m in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water = 103 kg m-3, g = 10 m s-2)

Question

The force acting on a window of area 50 cm × 50 cm of a submarine at a depth of 2000 m in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water = 103 kg m-3, g = 10 m s-2) (A) 5 × 105 N (B) 25 × 105 N (C) 5 × 106 N (D) 25 × 106 N

Tardigrade Admin · Accepted Answer

Here, h - 2000 m, ρ = 103 kg m-3, g = 10 m s-2 The pressure outside the submarine is P = Pa + ρ gh where Pa is the atmospheric pressure Pressure inside the submarine is Pa Hence, net pressure acting on the window is gauge pressure Gauge pressure, Pg=P-Pa= ρ gh =103 Kg m-3 × 10 m s-2 × 2000 m =2 × 107 pa Area of a window is A = 50 cm × 50 cm = 2500 × 10-4 m2 Force acting on the window is F = Pg A =2 × 107 pa × 2500 × 10-4 m2 =5 × 106 N

Q. The force acting on a window of area $50 c m \times 50 c m$ of a submarine at a depth of $2000 m$ in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water $= 1 0^{3} k g m^{- 3}$ , $g = 10 m s^{- 2}$ )

$5 \times 1 0^{5} N$

$25 \times 1 0^{5} N$

$5 \times 1 0^{6} N$

$25 \times 1 0^{6} N$

10⁶N

Q. The force acting on a window of area 50cm×50cm of a submarine at a depth of 2000m in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water =103kgm−3, g=10ms−2)

5×105N

25×105N

5×106N

25×106N