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JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF FLUIDS

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Application +4 | -1
Calculate the diameter of a capillary tube in which mercury is depressed by 1.21 cm1.21 \mathrm{~cm}. Given surface tension for mercury is 540×10−3 Nm−1540 \times 10^{-3} \mathrm{~Nm}^{-1}, the angle of contact with glass is 140∘140^{\circ} and density of mercury is 13.6×103 kg m−313.6 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}.
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Correct option: A
1.026×10−3 m1.026 \times 10^{-3} \mathrm{~m}
Option A (1.026×10−3 m1.026 \times 10^{-3} \mathrm{~m}) is correct because it is the diameter calculated using the formula 2r=4σcos⁡θhρg2r = \frac{4 \sigma \cos \theta}{h \rho g}.

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