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QJEEM20Ph135a: JEE Main 2020 Physics - Class 11 JEEM20Ph135a Solution: Consider a uniform cubical box of side a on a rough floor that…

JEE Main 2020 (Online) 7th January Evening Slot Physics - Rotational Motion Question 141 English Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is μ\mu = 0.4, the maximum possible value of 100 × ba{b \over a} for box not to topple before moving is .......

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JEE Main 2020 (Online) 7th January Evening Slot Physics - Rotational Motion Question 141 English Explanation
Torque about Center Of Mass

Fb + f a2{a \over 2} = N a2{a \over 2} ...(1)

For not sliding

F = f = μ\mu N ....(2)

N = Mg ...(3)

∴\therefore μ\mu N (b+a2)\left( {b + {a \over 2}} \right) = Mg a2{{a \over 2}}

⇒\Rightarrow 4b = 3 aa

⇒\Rightarrow b = 0.75 aa

which is not possible as as b can maximum be equal to 0.5 aa

∴\therefore ba{b \over a} = 0.5

So 100 × ba{b \over a} = 50

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