QJEEM24Phe91e: JEE Main 2024 Physics - Class 11 JEEM24Phe91e Solution: A block of mass is placed on a surface having vertical…

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:

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

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

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

Simplify:

$\mu \cos\theta \geq \sin\theta$

Divide through by $\cos\theta$ :

$\mu \geq \tan\theta$

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4. Maximum Slope Without Slipping:

From the above condition:

$\tan\theta \leq \mu$

Substitute $\mu = 0.5$ :

$\tan\theta \leq 0.5$

Thus:

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

$x \leq 1$

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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.