A bottle has an opening of radius a and length b . A cork of length b and radius (a + ∆ a) where (∆ a ≪ a) , is compressed to fit into the opening completely (see figure). If the bulk modulus of cork is B and the

Correct answer: D. (4π μ Bb) ∆ a Pressure = (N)/(A) = (N)/((2π a)b) ⇒ stress = B × strain (N)/((2π a)b) = B 2π a ∆ a × b π a ^ 2 b ⇒ N = B(4π b △ a) f = μ N = μ4π bB Δ a

JEE Main 201610 AprilOnline

Exam: JEE Main
Year: 2016
Date: 10 April
Mode: Online
Subject: Physics
Chapter: Mechanical Properties of Solids

JEE Main 2016 · 10 April · Online · Physics · Mechanical Properties of Solids

A bottle has an opening of radius a and length b . A cork of length...

A bottle has an opening of radius

a

and length

b

. A cork of length

b

and radius \left(a + ∆ a\right)

where \left(∆ a ≪ a\right)

, is compressed to fit into the opening completely (see figure). If the bulk modulus of cork is

B

and the coefficient of friction between the bottle and cork is

\mu

, then the force needed to push the cork into the bottle is

A bottle has an opening of radius a and length b . A cork of length b and radius (a + ∆ a) where (∆ a ≪ a) , is

Options

A $\left(\pi \mu Bb\right)a$
B $\left(2\pi \mu Bb\right) ∆ a$
C $\left(\pi \mu Bb\right) ∆ a$
D $\left(4\pi \mu Bb\right) ∆ a$

Correct answer

D. $\left(4\pi \mu Bb\right) ∆ a$

Step-by-step solution

Pressure = \dfrac{N}{A} = \dfrac{N}{\left(2\pi a\right)b}

<br /><br />

⇒ stress = B \times strain

<br /><br />

\dfrac{N}{\left(2\pi a\right)b} = B\dfrac{2\pi a ∆ a \times b}{{\pi a}^{2}b}

<br /><br />

⇒ N = B\left(4\pi b △ a\right)

<br /><br />

f = \mu N = \mu4\pi bB \Delta a

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