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Physics Question 46 – JEE-MAIN 2026

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A block takes t time to slide down a plane inclined at 45∘ to the horizontal. If the surface is made smooth (frictionless), the block takes time t2 to slide down the plane. The coefficient of friction between the block and the inclined plane is (α100). The value of α is _______.

Relate the distance, time, and acceleration using kinematic equations, and then determine the acceleration in terms of forces using Newton's second law for both cases (with and without friction).

Step 1: Determine Acceleration in Both Cases✦ Active

Let L be the length of the inclined plane. For motion starting from rest, the distance covered is L=12at2, which implies a=2Lt2.

For the frictionless case, the time taken is t2=t2. The acceleration is a2=gsin⁡θ. So, we have L=12(gsin⁡θ)(t2)2=18gsin⁡θt2 (Equation 1).

For the case with friction, the time taken is t1=t. The acceleration is a1=g(sin⁡θ−μcos⁡θ). So, we have L=12g(sin⁡θ−μcos⁡θ)t2 (Equation 2).

Step 2: Solve for the Coefficient of Friction (μ)○ Expand

Equating the two expressions for L from Equation 1 and Equation 2:

18gsin⁡θt2=12g(sin⁡θ−μcos⁡θ)t2

Cancel gt2 from both sides and simplify:

18sin⁡θ=12(sin⁡θ−μcos⁡θ) sin⁡θ=4(sin⁡θ−μcos⁡θ) sin⁡θ=4sin⁡θ−4μcos⁡θ 4μcos⁡θ=3sin⁡θ μ=34sin⁡θcos⁡θ=34tan⁡θ
💡 Teacher's Secret Hint

Ensure careful algebraic manipulation to avoid errors when equating the expressions for length.

Step 3: Calculate the Value of α○ Expand

Given the angle of inclination θ=45∘. We know that tan⁡45∘=1.

Substitute this value into the expression for μ:

μ=34×1=34

The coefficient of friction is given as (α100). Therefore:

α100=34 α=34×100=3×25=75
💡 Teacher's Secret Hint

Remember the trigonometric values for common angles like 45∘.

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