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Question: A circular tube of radius ‘R’ and cross-sectional radius ‘r’ (r \<\< R) is filled completely with ir...

A circular tube of radius ‘R’ and cross-sectional radius ‘r’ (r << R) is filled completely with iron balls of radius ‘r’. Iron balls are just fitting into the tubes. The tension in the tube when it is rotated about its axis perpendicular to its plane with angular velocity ‘w’ –

A

43\frac{4}{3}prw2r3R

B

43πρω2r2R2\frac{4}{3}\pi\rho\omega^{2}r^{2}R^{2}

C

23πρω2r3R\frac{2}{3}\pi\rho\omega^{2}r^{3}R

D

23πρω2r2R2\frac{2}{3}\pi\rho\omega^{2}r^{2}R^{2}

Answer

23πρω2r2R2\frac{2}{3}\pi\rho\omega^{2}r^{2}R^{2}

Explanation

Solution

Q r < < R, mass/length can be given by

l = m2r\frac{m}{2r}= 43πr3ρ2r\frac{\frac{4}{3}\pi r^{3}\rho}{2r}.23πr2ρ\frac{2}{3}\pi r^{2}\rho

Now,

Tension in tube is given by

T = (l . w2R)R = 23πr2ρ\frac{2}{3}\pi r^{2}\rho. w2.R.R

= 23πρω2r2R2\frac{2}{3}\pi\rho\omega^{2}r^{2}R^{2} Q