JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 2)
A bottle has an opening of radius a and length b. A cork of length b and radius (a + $$\Delta $$a) where ($$\Delta $$a < < a) is compressed to fit into the opening completely (See figure). If the bulk modulus of cork is B and frictional coefficient between the bottle and cork is $$\mu $$ then the force needed to push the cork into the bottle is :
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($$\pi $$ $$\mu $$ B b) $$\Delta $$a
(2$$\pi $$ $$\mu $$ B b) $$\Delta $$a
($$\pi $$ $$\mu $$ B b) a
(4$$\pi $$ $$\mu $$ B b) $$\Delta $$a
Explanation
Bulk modulus, B = $${{\Delta P} \over {{{\Delta V} \over V}}}$$
Vi = $$\pi $$ (a + $$\Delta $$a)2b
Vf = $$\pi $$a2b
$$ \therefore $$ $$\Delta $$V = 2$$\pi $$ab$$\Delta $$a
$$ \therefore $$ $${{{\Delta V} \over V}}$$ = $${{2\pi ab\Delta a} \over {\pi {a^2}b}}$$ = $${{2\Delta a} \over a}$$
$$ \therefore $$ $$\Delta $$P = B $$ \times $$ $${{{\Delta V} \over V}}$$
= B $$ \times $$ $${{2\Delta a} \over a}$$
Normal Force (N) = $$\Delta $$P $$ \times $$ A
= B $$ \times $$ $${{2\Delta a} \over a}$$ $$ \times $$ (2$$\pi $$a)b
= 4$$\pi $$B$$\Delta $$ab
$$ \therefore $$ Frictional force = $$\mu $$N = (4$$\pi $$$$\mu $$Bb)$$\Delta $$a
Vi = $$\pi $$ (a + $$\Delta $$a)2b
Vf = $$\pi $$a2b
$$ \therefore $$ $$\Delta $$V = 2$$\pi $$ab$$\Delta $$a
$$ \therefore $$ $${{{\Delta V} \over V}}$$ = $${{2\pi ab\Delta a} \over {\pi {a^2}b}}$$ = $${{2\Delta a} \over a}$$
$$ \therefore $$ $$\Delta $$P = B $$ \times $$ $${{{\Delta V} \over V}}$$
= B $$ \times $$ $${{2\Delta a} \over a}$$
Normal Force (N) = $$\Delta $$P $$ \times $$ A
= B $$ \times $$ $${{2\Delta a} \over a}$$ $$ \times $$ (2$$\pi $$a)b
= 4$$\pi $$B$$\Delta $$ab
$$ \therefore $$ Frictional force = $$\mu $$N = (4$$\pi $$$$\mu $$Bb)$$\Delta $$a
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