A block of mass m is placed on a surface having vertical cross section given by y=x^24. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

Correct answer: (A) 1/4 m. At equilibrium, tanθ=μ. Now, tanθ is slope of the curve. Therefore, tanθ=dy/dx=x/2 Now, x/2=0.5 ⇒x=1 Required height y=x^24=1/4 m.

Q65 / 90 · JEE Main 2024 (30 Jan Shift 2) · Physics

A block of mass $m$ is placed on a surface having vertical cross section given by $y = \frac{x^{2}}{4}$ . If coefficient of friction is $0.5$ , the maximum height above the ground at which block can be placed without slipping is:

Correct answer
A
$\frac{1}{4} m$
B
$\frac{1}{2} m$
C
$\frac{1}{6} m$
D
$\frac{1}{3} m$

Solution

At equilibrium, $\tan θ = μ$ .
Now, $\tan θ$ is slope of the curve. Therefore,
$\tan θ = \frac{d y}{d x} = \frac{x}{2}$
Now,
$\frac{x}{2} = 0.5 \Rightarrow x = 1$
Required height
$y = \frac{x^{2}}{4} = \frac{1}{4} m$ .

Hence, the correct answer is option: 'A'.

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A block of mass m is placed on a surface having vertical cross section given by y=x24. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

Solution

Relevant Questions

A block of mass $m$ is placed on a surface having vertical cross section given by $y = \frac{x^{2}}{4}$ . If coefficient of friction is $0.5$ , the maximum height above the ground at which block can be placed without slipping is: