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MCQMediumJEE 2024Friction (Static, Kinetic, Rolling)

JEE Physics 2024 Question with Solution

In the given arrangement of a doubly inclined plane, two blocks of masses MM and mm are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.250.25. The value of mm, for which M=10 kgM = 10 \, \text{kg} will move down with an acceleration of 2 m/s22 \, \text{m/s}^2, is:

  • A

    9 kg9 \, \text{kg}

  • B

    4.5 kg4.5 \, \text{kg}

  • C

    6.5 kg6.5 \, \text{kg}

  • D

    2.25 kg2.25 \, \text{kg}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two blocks on a doubly inclined plane are connected by a light string over an ideal pulley. M=10 kgM = 10 \, \text{kg}, μ=0.25\mu = 0.25, and block MM moves down with acceleration a=2 m/s2a = 2 \, \text{m/s}^2.

Find: The value of mm.

For block MM, along the plane:

Mgsin⁡53∘−μMgcos⁡53∘−T=MaMg \sin 53^\circ - \mu Mg \cos 53^\circ - T = Ma

Substituting the given values:

10×10×sin⁡53∘−0.25×10×10×cos⁡53∘−T=10×210 \times 10 \times \sin 53^\circ - 0.25 \times 10 \times 10 \times \cos 53^\circ - T = 10 \times 2 100×45−0.25×100×35−T=20100 \times \frac{4}{5} - 0.25 \times 100 \times \frac{3}{5} - T = 20 80−15−T=2080 - 15 - T = 20 T=45 NT = 45 \, \text{N}

For block mm, along the plane:

T−mgsin⁡37∘−μmgcos⁡37∘=maT - mg \sin 37^\circ - \mu mg \cos 37^\circ = ma

Substituting T=45 NT = 45 \, \text{N}:

45−m×10×35−0.25×m×10×45=m×245 - m \times 10 \times \frac{3}{5} - 0.25 \times m \times 10 \times \frac{4}{5} = m \times 2 45−6m−2m=2m45 - 6m - 2m = 2m 45=10m45 = 10m m=4.5 kgm = 4.5 \, \text{kg}

Therefore, the required mass is 4.5 kg4.5 \, \text{kg}. The correct option is B.

Using direct substitution after finding tension

Given: M=10 kgM = 10 \, \text{kg}, μ=0.25\mu = 0.25, a=2 m/s2a = 2 \, \text{m/s}^2.

Find: The value of mm.

First write the equation for block MM and substitute the standard values sin⁡53∘=45\sin 53^\circ = \frac{4}{5} and cos⁡53∘=35\cos 53^\circ = \frac{3}{5}:

10gsin⁡53∘−μ(10g)cos⁡53∘−T=10×210g \sin 53^\circ - \mu (10g) \cos 53^\circ - T = 10 \times 2

This gives:

T=80−15−20=45 NT = 80 - 15 - 20 = 45 \, \text{N}

Now use the equation for block mm:

T−mgsin⁡37∘−μmgcos⁡37∘=2mT - mg \sin 37^\circ - \mu mg \cos 37^\circ = 2m

Substituting T=45 NT = 45 \, \text{N}, sin⁡37∘=35\sin 37^\circ = \frac{3}{5}, and cos⁡37∘=45\cos 37^\circ = \frac{4}{5}:

45=10m45 = 10m m=4.5 kgm = 4.5 \, \text{kg}

Therefore, the correct option is B.

Common mistakes

  • A common mistake is taking friction on block mm in the wrong direction. Since block MM moves down, block mm moves up its plane, so friction on mm acts down the plane. Use friction opposite to the actual direction of motion.

  • Students often use incorrect trigonometric values for the incline components. Here, sin⁡53∘=45\sin 53^\circ = \frac{4}{5}, cos⁡53∘=35\cos 53^\circ = \frac{3}{5}, sin⁡37∘=35\sin 37^\circ = \frac{3}{5}, and cos⁡37∘=45\cos 37^\circ = \frac{4}{5}. Interchanging these gives the wrong tension and wrong mass.

  • Another mistake is forgetting that the acceleration magnitude is the same for both blocks because the string is light and inextensible. Do not assign different accelerations to MM and mm.

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