MHT CET · Physics · Mechanical Properties of Fluids
Water rises to height \(2 \cdot 2 \mathrm{~cm}\) in glass capillary tube. The height to which same water rises in another capillary having \(\frac{1}{4}\) area of cross-section is
- A \(16 \cdot 4 \mathrm{~cm}\)
- B \(4 \cdot 4 \mathrm{~cm}\)
- C \(8 \cdot 4 \mathrm{~cm}\)
- D \(2 \cdot 2 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(B) \(4 \cdot 4 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{h}_{1}=2.2 \mathrm{~cm}, \mathrm{~h}_{2}=?\)
\(\mathrm{~h}_{1} \mathrm{r}_{1}=\mathrm{h}_{2} \mathrm{r}_{2}\)
\(\mathrm{~h}_{1} \sqrt{\mathrm{A}_{1}}=\mathrm{h}_{2} \sqrt{\mathrm{A}_{2}}\)
\(\mathrm{~h}_{2}=\frac{\mathrm{h}_{1} \sqrt{\mathrm{A}_{1}}}{\sqrt{\mathrm{A}_{2}}}=2.2 \times 2=4.4 \mathrm{~cm}\)
\(\mathrm{~h}_{1} \mathrm{r}_{1}=\mathrm{h}_{2} \mathrm{r}_{2}\)
\(\mathrm{~h}_{1} \sqrt{\mathrm{A}_{1}}=\mathrm{h}_{2} \sqrt{\mathrm{A}_{2}}\)
\(\mathrm{~h}_{2}=\frac{\mathrm{h}_{1} \sqrt{\mathrm{A}_{1}}}{\sqrt{\mathrm{A}_{2}}}=2.2 \times 2=4.4 \mathrm{~cm}\)
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