A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force $\mathrm{F}=\sqrt{\mathrm{x}} \mathrm{N}$ as shown in figure. The value of $x=$ _____.

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Laws of Motion

A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force

F = x N

as shown in figure. The value of

x =

_____.

Answer: 12

Solution

For a block on a smooth inclined plane at

60°

held by horizontal force

F

, resolving forces parallel and perpendicular to the incline:

F cos 60° = m g sin 60°

. With

m = 0.2

kg,

m g = 2

N. Substituting:

F \times \frac{1}{2} = 2 \times \frac{3}{2}

, so

F = 23

N. Given

F = x

, we have

x = 23

, thus

x = 12

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Integer answer

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