MHT CET · Physics · Mechanical Properties of Fluids
Water rises to a height of \(15 \mathrm{~mm}\) in a capillary tube having cross-sectional area 'A'.
If cross-sectional area of the tube is made ' \(\frac{A^{\prime}}{3}\) then the water will rise to a height
of
- A \(15 \sqrt{3} \times 10^{-3} \mathrm{~m}\)
- B \(20 \sqrt{3} \times 10^{-3} \mathrm{~m}\)
- C \(5 \sqrt{3} \times 10^{-3} \mathrm{~m}\)
- D \(10 \sqrt{3} \times 10^{-3} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(15 \sqrt{3} \times 10^{-3} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{h} \propto \frac{1}{\mathrm{r}}\) since \(\mathrm{A}=\pi \mathrm{r}^{2}, \quad \sqrt{\mathrm{A}} \propto \mathrm{r}\)
\(\therefore \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{A}}}\)
\(\therefore \frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\sqrt{\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}}=\sqrt{3} \quad \because \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=3\)
\(\therefore \mathrm{h}_{2}=\sqrt{3} \mathrm{~h}_{1}=\sqrt{3} \times 15 \times 10^{-3} \mathrm{~m} \quad=15 \sqrt{3} \times 10^{-3} \mathrm{~m}\)
\(\therefore \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{A}}}\)
\(\therefore \frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\sqrt{\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}}=\sqrt{3} \quad \because \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=3\)
\(\therefore \mathrm{h}_{2}=\sqrt{3} \mathrm{~h}_{1}=\sqrt{3} \times 15 \times 10^{-3} \mathrm{~m} \quad=15 \sqrt{3} \times 10^{-3} \mathrm{~m}\)
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