MHT CET · Physics · Mechanical Properties of Fluids
Water rises up to a height \(h\) in a capillary tube immersed vertically in water. When this whole arrangement is taken to a depth \(d\) in a mine, the water level rises up to a height \(h^{\prime}\). If \(R\) is the radius of the earth, then the ratio \(\frac{h}{h^{\prime}}\) is
- A \(1+\frac{d}{R}\)
- B \(1-\frac{d}{R}\)
- C \(\frac{R+d}{R-d}\)
- D \(\frac{R-d}{R+d}\)
Answer & Solution
Correct Answer
(B) \(1-\frac{d}{R}\)
Step-by-step Solution
Detailed explanation
On the surface of the earth \(T=\frac{r h \rho g}{2}\) for water
\(\therefore h=\frac{2 T}{r \rho g}\)
At a depth \(d\), the value of acceleration due to gravity changes to
\(g^{\prime}=g\left(1-\frac{d}{R}\right)\), therefore
\(g^{\prime}=g\left(1-\frac{d}{R}\right)\), therefore
\(
\begin{aligned}
& h^{\prime}=\frac{2 T}{r \rho g^{\prime} d}=\frac{2 T}{r \rho g\left(1-\frac{d}{R}\right)} \\
& \therefore \frac{h}{h^{\prime}}=\frac{2 T}{r \rho g} \times \frac{r \rho g\left(1-\frac{d}{R}\right)}{2 T}=1-\frac{d}{R}
\end{aligned}
\)
\(\therefore h=\frac{2 T}{r \rho g}\)
At a depth \(d\), the value of acceleration due to gravity changes to
\(g^{\prime}=g\left(1-\frac{d}{R}\right)\), therefore
\(g^{\prime}=g\left(1-\frac{d}{R}\right)\), therefore
\(
\begin{aligned}
& h^{\prime}=\frac{2 T}{r \rho g^{\prime} d}=\frac{2 T}{r \rho g\left(1-\frac{d}{R}\right)} \\
& \therefore \frac{h}{h^{\prime}}=\frac{2 T}{r \rho g} \times \frac{r \rho g\left(1-\frac{d}{R}\right)}{2 T}=1-\frac{d}{R}
\end{aligned}
\)
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