A drum of radius ' R ' full of liquid of density ' d ' is rotated at angular velocity ' ω ' rad/s. The increase in pressure at the centre of the drum will be (a) (ω² R² d)/(2) (b) (ω² R d)/(2) (c) (ω R d²)/(2) (d) (ω² R² d²)/(2)

Answer: (a). In a rotating liquid, dP/dr = d ω² r. Integrating from centre to rim: Δ P = ∫₀R d ω² r dr = (d ω² R²)/(2) So the pressure difference between rim and centre is (ω² R² d)/(2); the answer is (a).

A drum of radius ' R ' full of liquid of density ' d ' is rotated at… | MHT CET 2024 (2 May, Shift 2)

Q11MHT CET 2024 · 2 MAY · SHIFT 2Difficult

Pressure in a Fluid

A drum of radius ' $R$ ' full of liquid of density ' $d$ ' is rotated at angular velocity ' $\omega$ ' $\text{rad/s}$ . The increase in pressure at the centre of the drum will be

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Answer: (a)

In a rotating liquid, $\dfrac{dP}{dr} = d\,\omega^2 r$ .
Integrating from centre to rim: $\Delta P = \int_0^R d\,\omega^2 r\,dr = \dfrac{d\,\omega^2 R^2}{2}$
So the pressure difference between rim and centre is $\dfrac{\omega^2 R^2 d}{2}$ ; the answer is (a).

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