At a depth of 1000 m in an ocean, what is the force acting on the window area 20 cm × 20 cm of a submarine, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is .1.03 × 103 kgm -3, g=10 ms -2)

Question

At a depth of 1000 m in an ocean, what is the force acting on the window area 20 cm × 20 cm of a submarine, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is .1.03 × 103 kgm -3, g=10 ms -2) (A) 2 × 105 N (B) 3 × 105 N (C) 4 × 105 N (D) 5 × 105 N

Tardigrade Admin · Accepted Answer

The pressure outside the submarine is P=Pa+ ρ g h and the pressure inside it is Pa. Hence, the net pressure acting on the window is the garage pressure Pg=ρ g h. Since the area of the window is A=0.04 m 2, the force acting on it is F=Pg ⋅ A=ρ g h . A =(1.03 × 103 kg m -3) .(10 m s -2)(1000 m ) ×(0.04 m 2) =4.12 × 105 N =4 × 105 N

Q. At a depth of 1000 m in an ocean, what is the force acting on the window area $20 c m \times 20 c m$ of a submarine, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is $1.03 \times 1 0^{3} k g m^{- 3}, g = 10 m s^{- 2})$

$2 \times 1 0^{5} N$

$3 \times 1 0^{5} N$

$4 \times 1 0^{5} N$

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

Q. At a depth of 1000 m in an ocean, what is the force acting on the window area 20cm×20cm of a submarine, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is 1.03×103kgm−3,g=10ms−2)

2×105N

3×105N

4×105N

5×105N

Q. At a depth of 1000 m in an ocean, what is the force acting on the window area $20 c m \times 20 c m$ of a submarine, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is $1.03 \times 1 0^{3} k g m^{- 3}, g = 10 m s^{- 2})$

$2 \times 1 0^{5} N$

$3 \times 1 0^{5} N$

$4 \times 1 0^{5} N$

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