Laws of Motion

JEE Main level, numerical answer, medium

96 of 210

Question 96 of 210

A block of mass 200g200g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F=xNF=\sqrt{x}N as shown in figure.

The value of
x=_____.

Show answer and solution

Answer: 12


Balancing the forces in horizontal direction
Nsin⁡θ=FN\sin \theta =F
And balancing the forces in vertical direction
Ncos⁡θ=mgN\cos \theta =mg
⇒tan⁡θ=Fmg\Rightarrow \tan \theta =\dfrac{F}{mg}⇒F=mgtan⁡θ\Rightarrow F=mg\tan \theta
F=0.2×10×3=12NF=0.2\times 10\times \sqrt{3}=\sqrt{12}N. Therefore, x=12x=12

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