JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 23)
An inclined plane is bent in such a way that the vertical cross-section is given by $$y = {{{x^2}} \over 4}$$ where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $$\mu$$ = 0.5, the maximum height in cm at which a stationary block will not slip downward is _________ cm.
Answer
25
Explanation
The graph for given equation is shown below
_24th_February_Morning_Shift_en_23_1.png)
At maximum height, the slope of tangent drawn,
$$\tan \theta = {{dy} \over {dx}} = {{2x} \over 4} = {x \over 2}$$ [$$\because$$ $$y = {{{x^2}} \over 4}$$]
$$ \Rightarrow 0.5 = {x \over 2}$$ ($$\because$$ $$\mu$$ = tan$$\theta$$)
$$\Rightarrow$$ x = 1 m
$$\therefore$$ $$y = {{{x^2}} \over 4} = {1 \over 4}$$ = 0.25 m = 25 cm
At maximum height, the slope of tangent drawn,
$$\tan \theta = {{dy} \over {dx}} = {{2x} \over 4} = {x \over 2}$$ [$$\because$$ $$y = {{{x^2}} \over 4}$$]
$$ \Rightarrow 0.5 = {x \over 2}$$ ($$\because$$ $$\mu$$ = tan$$\theta$$)
$$\Rightarrow$$ x = 1 m
$$\therefore$$ $$y = {{{x^2}} \over 4} = {1 \over 4}$$ = 0.25 m = 25 cm
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