A highly viscous liquid of viscosity coefficient $η$ flows through a fixed horizontal cylindrical tube (fixed from outer surface) of internal radius r, thickness $t (t < < r)$ and length l. Volume of liquid flowing per second is Q and pressure difference across the tube is P. Modulus of rigidity of material of tube is $β$ . Shear strain in the tube will be: $(T a k e, Q = \frac{π p r^{4}}{8 η l})$

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$\frac{π β r^{3}}{16 η Q}$

$\frac{4 η Q}{π β r^{3}}$

$\frac{8 η Q}{π β r^{2}}$

$\frac{β r^{3}}{8 η Q}$

answer is B.

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Detailed Solution

$Q = \frac{π p R^{4}}{8 η l} \Rightarrow \frac{P}{η l} = \frac{8 Q}{π R^{4}} V = \frac{P}{4 η l} {R^{2} - r^{2}} \Rightarrow | \frac{d V}{d r} | = \frac{\Pr}{2 η l} = \frac{4 Q r}{π R^{4}} ∴ S p e a r s t r e s s o n l i q u i d a t r = R τ = η | \frac{d v}{d r} | = \frac{4 Q η}{π R^{3}} S p e a r s t r e s s o n s o l i d = \frac{4 Q η}{π R^{3}} S p e a r s t r e s s o f s o l i d = \frac{4 Q η}{π β R^{3}}$