The pressure acting on a submarine is 3 × 105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be: (Assume that atmospheric pressure is 1 × 105 Pa density of water is 103 kg m -3, g = 10 m s -2 )

Question

The pressure acting on a submarine is 3 × 105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be: (Assume that atmospheric pressure is 1 × 105 Pa density of water is 103 kg m -3, g = 10 m s -2 ) (A) (200/3) % (B) (200/5) % (C) (5/200) % (D) (3/200) %

Tardigrade Admin · Accepted Answer

P1=ρ g d+P0=3 × 105 P a ∴ ρ g d=2 × 105 P a P2=2 ρ g d+P0 =4 × 105+105=5 × 105 Pa % increase =(P2-P1/P1) × 100 =(5 × 105-3 × 105/3 × 105) × 100=(200/3) %

$\frac{200}{3} %$

$\frac{200}{5} %$

$\frac{5}{200} %$

$\frac{3}{200} %$

Q. The pressure acting on a submarine is 3×105Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be: (Assume that atmospheric pressure is 1×105Pa density of water is 103kgm−3,g=10ms−2 )

3200​%

5200​%

2005​%

2003​%

P1​=ρgd+P0​=3×105Pa ∴ρgd=2×105Pa P2​=2ρgd+P0​ =4×105+105=5×105Pa % increase =P1​P2​−P1​​×100 =3×1055×105−3×105​×100=3200​%

$\frac{200}{3} %$

$\frac{200}{5} %$

$\frac{5}{200} %$

$\frac{3}{200} %$