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JEE Physics 2025 Question with Solution

Two blocks of masses mm and MM, (M>mM > m), are placed on a frictionless table as shown in figure. A massless spring with spring constant kk is attached with the lower block. If the system is slightly displaced and released then (μ\mu = coefficient of friction between the two blocks)

A horizontal spring fixed to a wall is attached to block M on a frictionless table, with smaller block m placed on top of M and coefficient of friction mu marked between the blocks.
  • A

    A, B, D Only

  • B

    B, C, D Only

  • C

    C, D, E Only

  • D

    A, B, C Only

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Two blocks of masses mm and MM move on a frictionless table, with the upper block in contact through static friction. The spring constant is kk.

Find: Which statements among A, B, C, D, E are correct, and hence the correct option.

When the two blocks move together without slipping, they behave like a single body of mass M+mM+m. The restoring force is provided by the spring.

Using Newton's second law for the combined system,

(M+m)a=−kx(M+m)a=-kx

So,

a=−kxM+ma=-\frac{kx}{M+m}

This is the equation of SHM, hence

ω2=kM+m\omega^2=\frac{k}{M+m}

Therefore,

T=2πM+mkT=2\pi\sqrt{\frac{M+m}{k}}

So statement A is correct.

From the same equation,

∣a∣=kxM+m|a|=\frac{kx}{M+m}

for displacement magnitude xx from the mean position. So statement B is correct.

For the upper block, the only horizontal force is static friction. Therefore,

f=maf=ma

Substituting the acceleration,

f=m(k∣x∣M+m)=mk∣x∣M+mf=m\left(\frac{k|x|}{M+m}\right)=\frac{mk|x|}{M+m}

Hence the friction depends on kk, not directly on μ\mu. So statement C is incorrect.

For no slipping, the required friction must not exceed the maximum static friction.

frequired=mω2Af_{\text{required}}=m\omega^2A

At the limiting condition,

mω2Amax⁡=μmgm\omega^2A_{\max}=\mu mg

Thus,

Amax⁡=μgω2A_{\max}=\frac{\mu g}{\omega^2}

Using

ω2=kM+m\omega^2=\frac{k}{M+m}

we get

Amax⁡=μ(M+m)gkA_{\max}=\frac{\mu (M+m)g}{k}

So statement D is correct.

The maximum friction between the two blocks is determined by the normal reaction between them, which is

N=mgN=mg

Therefore,

fmax⁡=μmgf_{\max}=\mu mg

not μ(M+m)g\mu (M+m)g. So statement E is incorrect.

Hence the correct statements are A, B, D only. Therefore, the correct option is A.

Statement-wise Check

Given: The lower block attached to the spring and the upper block move together unless slipping occurs.

Find: Evaluate each statement separately.

  1. For the full system,
F=−kxF=-kx

and total mass is M+mM+m. Hence,

(M+m)a=−kx(M+m)a=-kx

which gives SHM with time period

T=2πM+mkT=2\pi\sqrt{\frac{M+m}{k}}

So A is true.

  1. From the same result,
a=−kxM+ma=-\frac{kx}{M+m}

Therefore the magnitude is

kxM+m\frac{kx}{M+m}

So B is true.

  1. For the top block,
f=ma=m⋅k∣x∣M+mf=ma=m\cdot \frac{k|x|}{M+m}

Hence the expression in C is wrong because it contains μ\mu instead of kk. So C is false.

  1. At extreme position, required friction is maximum:
f=mω2Af=m\omega^2A

No slipping requires

mω2A≤μmgm\omega^2A\le \mu mg

At the limiting value,

Amax⁡=μ(M+m)gkA_{\max}=\frac{\mu (M+m)g}{k}

So D is true.

  1. Maximum static friction acts between the two blocks only, so
fmax⁡=μN=μmgf_{\max}=\mu N=\mu mg

Thus E is false.

Therefore the true statements are A, B, D, which matches option A. The solution also notes a discrepancy in labeling: the page's statement-combination option (1) corresponds to output label A here.

Common mistakes

  • Using only mass MM in the time period formula is incorrect because both blocks move together when there is no slipping. Treat the oscillating mass as M+mM+m.

  • Taking friction as μN\mu N at every instant is wrong. Static friction is self-adjusting and equals the force needed to accelerate the upper block, up to the maximum value μmg\mu mg.

  • Using the normal force as (M+m)g(M+m)g for friction between the blocks is incorrect. The contact normal between the upper and lower blocks is only mgmg.

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