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QJEEM22Ph9a2f: JEE Main 2022 Physics - Class 11 JEEM22Ph9a2f Solution: A block of mass 200 g is kept stationary on a smooth inclined…

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

JEE Main 2022 (Online) 25th June Evening Shift Physics - Laws of Motion Question 57 English

The value of x = _____________.

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JEE Main 2022 (Online) 25th June Evening Shift Physics - Laws of Motion Question 57 English Explanation

F×12=0.2×10×32F \times {1 \over 2} = 0.2 \times 10 \times {{\sqrt 3 } \over 2}

⇒F=23\Rightarrow F = 2\sqrt 3

⇒F=12\Rightarrow F = \sqrt {12} N

∴\therefore x=12x = 12

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