Properties Of Matter Ques 48
- A submarine experience a pressure of $5.05 \times 10^{6} Pa$ at a depth of $d _1$ in a sea. When it goes further to a depth of $d _2$, it experiences a pressure of $8.08 \times 10^{6} Pa$, then $d _2-d _1$ is approximately (density of water $=10^{3} kg / m^{3}$ and acceleration due to gravity $=10 ms^{-2}$ ) (2019 Main, $10 Apr$ II)
(a) $500 m$
(b) $400 m$
(c) $600 m$
(d) $300 m$
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Answer:
Correct Answer: 48.(d)
Solution:
Formula:
- Pressure inside a fluid volume open to atmosphere is
$$ p=p _0+h \rho g $$
where, $p=$ pressure at depth $h, h=\operatorname{depth}$,
$$ \rho=\text { density of fluid } $$
and $g=$ acceleration due to gravity.
In problem given,
when $h=d _1$, pressure $p _1=5.05 \times 10^{6} Pa$ and when $h=d _2$, pressure $p _2=8.08 \times 10^{6} Pa$
So, we have
$$ \begin{aligned} & p _1=p _0+d _1 \rho g=5.05 \times 10^{6} \\ & \text { and } \quad p _2=p _0+d _2 \rho g=8.08 \times 10^{6} \\ & \Rightarrow \quad p _2-p _1=\left(d _2-d _1\right) \rho g=3.03 \times 10^{6} \\ & \Rightarrow \quad d _2-d _1=\frac{3.03 \times 10^{6}}{\rho g} \end{aligned} $$
Given, $\rho=10^{3} kgm^{-3}$
and $\quad g=10 ms^{-2}$
$\therefore d _2-d _1=\frac{3.03 \times 10^{6}}{10^{3} \times 10}$
$$ =303 m \approx 300 m $$
BETA
