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Properties of Matter question

2016 · 10 Apr · Shift 1 · Q53
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Properties of Matter question

2016 · 10 Apr · Shift 1 · Q53

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A bottle has an opening of radius a and length b. A cork of length b and radius (a + Δ\DeltaΔ a) where (Δ\DeltaΔ 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 μ\muμ then the force needed to push the cork into the bottle is : JEE Main 2016 (Online) 10th April Morning Slot Physics - Properties of Matter Question 242 English
  1. A
    (πμ\pi \muπμ B b) Δ\DeltaΔ a
  2. B
    (2 πμ\pi \muπμ B b) Δ\DeltaΔ a
  3. C
    (πμ\pi \muπμ B b) a
  4. D
    (4 πμ\pi \muπμ B b) Δ\DeltaΔ a
View written solutionFree

Correct answer: D

The question text appears truncated, but from the options and standard JEE Physics results, it is the usual friction/viscous resistance while inserting a slightly oversized cork into a bottle neck problem.

We interpret it as:

  • bottle opening radius =a= a=a
  • neck length =b= b=b
  • cork length =b= b=b
  • cork radius =a+Δa= a+\Delta a=a+Δa, where Δa≪a\Delta a \ll aΔa≪a
  • bulk modulus of cork material =B= B=B
  • coefficient of friction between cork and bottle =μ= \mu=μ

We need the force required to push the cork in.

1. Compression of the cork

The cork radius must reduce from (a+Δa)(a+\Delta a)(a+Δa) to aaa.

So the radial strain is

Δaa.\frac{\Delta a}{a}.aΔa​.

Since the cork is compressed laterally, the corresponding stress is of order

stress=BΔaa.\text{stress} = B\frac{\Delta a}{a}.stress=BaΔa​.

Thus the normal pressure between cork and bottle wall is

p=BΔaa.p = B\frac{\Delta a}{a}.p=BaΔa​.

2. Contact area

The cork is in contact with the cylindrical inner wall of the neck over length bbb.

Hence contact area is the curved surface area:

A=2πab.A = 2\pi a b.A=2πab.

3. Total normal force

Normal force exerted by the wall on the cork is

N=pA=(BΔaa)(2πab)=2πBb Δa.N = pA = \left(B\frac{\Delta a}{a}\right)(2\pi ab) = 2\pi B b\,\Delta a.N=pA=(BaΔa​)(2πab)=2πBbΔa.

4. Friction force required to push the cork

The resisting friction is

F=μN=μ(2πBb Δa).F = \mu N = \mu(2\pi B b\,\Delta a).F=μN=μ(2πBbΔa).

Therefore,

F=2πμBb Δa.F = 2\pi \mu B b\,\Delta a.F=2πμBbΔa.

5. Match with the options

This corresponds to:

Option B:

(2πμBb) Δa(2\pi \mu B b)\,\Delta a(2πμBb)Δa

6. Comparison with stored answer

Stored correct answer is D:

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

But from the standard derivation, the force comes out to be

2πμBb Δa.2\pi \mu B b\,\Delta a.2πμBbΔa.

So I do not agree with the stored answer.

A factor of 222 extra would arise only if one incorrectly doubles the normal reaction, but for cylindrical wall contact the normal force is simply pressure times curved surface area, giving the result above.

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