JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 21)
A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F = $$\sqrt{x}$$N as shown in figure.
_25th_June_Evening_Shift_en_21_1.png)
The value of x = _____________.
Answer
12
Explanation
_25th_June_Evening_Shift_en_21_2.png)
$$F \times {1 \over 2} = 0.2 \times 10 \times {{\sqrt 3 } \over 2}$$
$$ \Rightarrow F = 2\sqrt 3 $$
$$ \Rightarrow F = \sqrt {12} $$ N
$$\therefore$$ $$x = 12$$
Comments (0)
