JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 1)
The pressure acting on a submarine is 3 $$\times$$ 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 $$\times$$ 105 Pa density of water is 103 kg m$$-$$3, g = 10 ms$$-$$2)
(Assume that atmospheric pressure is 1 $$\times$$ 105 Pa density of water is 103 kg m$$-$$3, g = 10 ms$$-$$2)
$${{200} \over 5}$$%
$${{200} \over 3}$$%
$${{3} \over 200}$$%
$${{5} \over 200}$$%
Explanation
P = P0 + h$$\rho$$g = 3 $$\times$$ 105 Pa
$$ \Rightarrow $$ h$$\rho$$g = 3 $$\times$$ 105 $$-$$ 1 $$\times$$ 105
$$ \Rightarrow $$ h$$\rho$$g = 2 $$\times$$ 105
$$ \therefore $$ 2h$$\rho$$g = 4 $$\times$$ 105
$$ \therefore $$ P' = P0 + 4 $$\times$$ 105
$$ \therefore $$ P' = 5 $$\times$$ 105 Pa
$$ \therefore $$ % increase in pressure = $${{P' - P} \over P} \times 100$$
$$ = {{(5 - 3) \times {{10}^5}} \over {3 \times {{10}^5}}} \times 100$$
$$ = {{200} \over 3}$$%
$$ \Rightarrow $$ h$$\rho$$g = 3 $$\times$$ 105 $$-$$ 1 $$\times$$ 105
$$ \Rightarrow $$ h$$\rho$$g = 2 $$\times$$ 105
$$ \therefore $$ 2h$$\rho$$g = 4 $$\times$$ 105
$$ \therefore $$ P' = P0 + 4 $$\times$$ 105
$$ \therefore $$ P' = 5 $$\times$$ 105 Pa
$$ \therefore $$ % increase in pressure = $${{P' - P} \over P} \times 100$$
$$ = {{(5 - 3) \times {{10}^5}} \over {3 \times {{10}^5}}} \times 100$$
$$ = {{200} \over 3}$$%
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