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In a u-tube as shown in a figure, water and oil are in the left side and right side of the tube respectively. The heights from the bottom for water and oil columns are $15 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. The density of the oil is $\left[\right.$ take $\left.\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^3\right]$
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The correct answer is:
$750 \mathrm{~kg} / \mathrm{m}^3$
According to Pascal's law "Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel."
In the given situation as shown in the figure below
Pressure due to water column of height $15 \mathrm{~cm}=$ Pressure due to oil column of height $20 \mathrm{~cm}$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{h}_{\mathrm{w}} \rho_{\mathrm{w}} \mathrm{g} & =\mathrm{h}_0 \rho_0 \mathrm{~g} \\
15 \rho_w & =20 \rho_0 \Rightarrow \rho_0=\frac{15}{20} \rho_\omega \\
\rho_0 & =\frac{15}{20} \times 1000\left(\because \text { given, } \rho_w=1000 \mathrm{~kg} \mathrm{~m}^{-3}\right) \\
& =750 \mathrm{kgm}^{-3}
\end{aligned}
$$
In the given situation as shown in the figure below
Pressure due to water column of height $15 \mathrm{~cm}=$ Pressure due to oil column of height $20 \mathrm{~cm}$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{h}_{\mathrm{w}} \rho_{\mathrm{w}} \mathrm{g} & =\mathrm{h}_0 \rho_0 \mathrm{~g} \\
15 \rho_w & =20 \rho_0 \Rightarrow \rho_0=\frac{15}{20} \rho_\omega \\
\rho_0 & =\frac{15}{20} \times 1000\left(\because \text { given, } \rho_w=1000 \mathrm{~kg} \mathrm{~m}^{-3}\right) \\
& =750 \mathrm{kgm}^{-3}
\end{aligned}
$$
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