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Question

Physics Question on mechanical properties of solids

A bottle has an opening of radius aa and length bb. A cork of length bb and radius (a+Δa)(a+ \Delta a) where (Δa<<a)(\Delta a < < a) is compressed to fit into the opening completely (See figure). If the bulk modulus of cork is BB and frictional coefficient between the bottle and cork is μ\mu then the force needed to push the cork into the bottle is :

A

(2πμBb)Δa(2 \pi \mu B\, b) \Delta\,a

B

(πμBb)Δa(\pi \mu B\, b) \Delta a

C

(πμBb)a(\pi \mu \,B\, b ) a

D

(4πμBb)Δa(4 \pi \mu\, B \,b ) \Delta \,a

Answer

(4πμBb)Δa(4 \pi \mu\, B \,b ) \Delta \,a

Explanation

Solution

βΔVV=ΔP\beta \frac{\Delta V }{ V }=-\Delta P
Vi=π(a+Δa)2bV_{i}=\pi(a+\Delta a)^{2} b
Vf=πa2bV_{f}=\pi a^{2} b
ΔV2πabΔa\Delta V \simeq-2 \pi \,ab \Delta a
ΔVV=2πabΔaπa2b=2Δaa\frac{\Delta V }{ V }=\frac{-2 \pi \,a b \Delta a }{\pi a ^{2} b }=\frac{-2 \Delta a }{ a }
ΔP=2βΔaa\Rightarrow \Delta P =\frac{2 \beta \Delta a }{ a }
Normal force =2βΔaa2παb=\frac{2 \beta \Delta a }{ a } 2 \pi \alpha b
=4πβbΔa=4 \pi \beta b \Delta a
friction =μN=\mu N
=4πμβbΔa=4 \pi \,\mu \beta\, b\, \Delta\, a