JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(\mu=0.4,\) the maximum possible value of \(100 \times \frac{b}{a}\) for a box not to topple before moving is
- A \(80\)
- B \(75\)
- C \(85\)
- D \(82\)
Answer & Solution
Correct Answer
(B) \(75\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=\mu \mathrm{m} \mathrm{g}\dots (1)\) \(\mathrm{F}\left(\mathrm{b}+\frac{\mathrm{a}}{2}\right)=\mathrm{mg} \frac{\mathrm{a}}{2}\dots (2)\) \(\mu m g\left(b+\frac{a}{2}\right)=m g \times \frac{a}{2}\) \(\left(b+\frac{a}{2}\right) \mu=\frac{a}{2}\)…
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