Water rises to a height of 2 cm in a capillary tube. If crosssectional area of the tube is reduced to 16 1 th of initial area, then water will rise to a height of

h = r ρ g 2 T cos θ ∴ h 1 h 2 = r 2 r 1 Area A = π r 2 ∴ A ∝ r 2 ∴ A 2 A 1 = r 2 2 r 1 2 or r 2 r 1 = A 2 A 1 = 16 r 2 r 1 = 4 ∴ h 1 h 2 = 4 or h 2 = 4 h 1 = 4 × 2 = 8 cm

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PhysicsFluidSurface TensionMedium2 minPYQ_2021

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Water rises to a height of $2 cm$ in a capillary tube. If crosssectional area of the tube is reduced to $\frac{1}{16}^{th}$ of initial area, then water will rise to a height of

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$h = \frac{2 T cos θ}{r ρ g} ∴ \frac{h _{2}}{h _{1}} = \frac{r _{1}}{r _{2}} Area A = π r^{2} ∴ A \propto r^{2} ∴ \frac{A _{1}}{A _{2}} = \frac{r _{1}^{2}}{r _{2}^{2}} or \frac{r _{1}}{r _{2}} = \frac{A _{1}}{A _{2}} = 16 \frac{r _{1}}{r _{2}} = 4 ∴ \frac{h _{2}}{h _{1}} = 4 or h_{2} = 4 h_{1} = 4 \times 2 = 8 cm$

Stream:MHT_CETSubject:PhysicsTopic:FluidSubtopic:Surface Tension

⏱ 2mℹ️ Source: PYQ_2021

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