MHT CET · Physics · Mechanical Properties of Fluids
A drum of radius ' \(R\) ' full of liquid of density ' \(d\) ' is rotated at angular velocity ' \(\omega\) ' rad/s. The increase in pressure at the centre of the drum will be
- A \(\frac{\omega^2 R^2 d}{2}\)
- B \(\frac{\omega^2 \mathrm{Rd}}{2}\)
- C \(\frac{\omega R d^2}{2}\)
- D \(\frac{\omega^2 \mathrm{R}^2 \mathrm{~d}^2}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\omega^2 R^2 d}{2}\)
Step-by-step Solution
Detailed explanation
\(P_1+\frac{1}{2} \rho v^2+\rho g h=\text { constant } \quad...(i) \text { and } v=R \omega \)
\( \text {At centre } R=0 \Rightarrow v_1=0 \quad...(ii) \therefore \text { From (i), } \)
\( P_1+\frac{1}{2} \rho v_1^2+\rho g h=P_2+\frac{1}{2} \rho v_1^2+\rho g h \)
\( P_1+0=P_2+\frac{1}{2} \rho v_1^2 \)
\( \therefore P_1=P_2+\frac{1}{2} \rho(R \omega)^2 \)
\( \therefore P_1-P_2=\frac{\omega^2 R^2 d}{2} \ldots(\because r o m(i i)\)
\( \text {At centre } R=0 \Rightarrow v_1=0 \quad...(ii) \therefore \text { From (i), } \)
\( P_1+\frac{1}{2} \rho v_1^2+\rho g h=P_2+\frac{1}{2} \rho v_1^2+\rho g h \)
\( P_1+0=P_2+\frac{1}{2} \rho v_1^2 \)
\( \therefore P_1=P_2+\frac{1}{2} \rho(R \omega)^2 \)
\( \therefore P_1-P_2=\frac{\omega^2 R^2 d}{2} \ldots(\because r o m(i i)\)
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