MHT CET · Physics · Mechanical Properties of Fluids
Water rises to a height \(3 \mathrm{~cm}\) in a capillary tube. If cross-sectional area of capillary tube is reduced to \(\frac{1}{0}\) th initial area then water will rise to a height of
- A \(9 \mathrm{~cm}\)
- B \(6 \mathrm{~cm}\)
- C \(7 \mathrm{~cm}\)
- D \(8 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(9 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{h}_{1} \mathrm{r}_{1}=\mathrm{h}_{2} \mathrm{r}_{2}\)
\(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\)
\(\mathrm{~A} \propto \mathrm{r}^{2}\)
\(\sqrt{\mathrm{A}} \propto \mathrm{r}\)
\(\therefore \quad \mathrm{h}_{2}=\mathrm{h}_{1} \sqrt{\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}}=3 \sqrt{9}=3 \times 3=9 \mathrm{~cm}\)
\(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\)
\(\mathrm{~A} \propto \mathrm{r}^{2}\)
\(\sqrt{\mathrm{A}} \propto \mathrm{r}\)
\(\therefore \quad \mathrm{h}_{2}=\mathrm{h}_{1} \sqrt{\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}}=3 \sqrt{9}=3 \times 3=9 \mathrm{~cm}\)
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