JEE Main 2024 — Laws Of Motion Question with Solution

Given:

The equation of the surface: $$y=\frac{x^2}{4}$$.

Coefficient of friction: $$\mu =0.5$$.

Gravitational acceleration: $$g$$ (assumed constant).

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1. Slope of the Surface:

The slope of the surface at any point is given by:

$$\tan \theta =\frac{dy}{dx}$$

From the equation of the surface:

$$y=\frac{x^2}{4}$$

Differentiating with respect to $x$:

$$\frac{dy}{dx}=\frac{x}{2}$$

Thus, the slope at any point is:

$$\tan \theta =\frac{x}{2}$$

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2. Forces Acting on the Block:

At the point where the block is placed:

Weight of the block acts vertically downward: $mg$.

Normal force acts perpendicular to the surface.

Frictional force acts parallel to the surface, opposing the component of the weight that causes slipping.

3. Condition for No Slipping:

The block will not slip if the frictional force is sufficient to counteract the component of the gravitational force parallel to the slope. The frictional force is:

$$f=\mu N$$

The normal force $N$ is given by:

$$N=mg\cos \theta$$

The component of weight parallel to the slope is:

$$F_{\text{parallel}}=mg\sin \theta$$

For the block to not slip:

$$f\ge F_{\text{parallel}}$$

Substitute $f=\mu N$ and $N=mg\cos \theta$:

$$\mu (mg\cos \theta )\ge mg\sin \theta$$

Simplify:

$$\mu \cos \theta \ge \sin \theta$$

Divide through by $\cos \theta$:

$$\mu \ge \tan \theta$$

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4. Maximum Slope Without Slipping:

From the above condition:

$$\tan \theta \le \mu$$

Substitute $\mu =0.5$:

$$\tan \theta \le 0.5$$

Thus:

$$\frac{x}{2}\le 0.5$$

$$x\le 1$$

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5. Maximum Height:

The height $y$ of the block at $x=1$ is obtained from the surface equation:

$$y=\frac{x^2}{4}$$

Substitute $x=1$:

$$y=\frac{1^2}{4}=\frac{1}{4}\text{m}$$

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Final Answer:

The maximum height above the ground at which the block can be placed without slipping is:

$$\boxed{\frac{1}{4}\text{m}}$$

Thus, the correct option is Option D.