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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2020
  • Hard
  • Numerical

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

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(B) \(75\)

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\(\mathrm{F}=\mu \mathrm{m} \mathrm{g}\dots (1)\)

\(\mathrm{F}\left(\mathrm{b}+\frac{\mathrm{a}}{2}\right)=\mathrm{mg} \frac{\mathrm{a}}{2}\dots (2)\)

\(\mu m g\left(b+\frac{a}{2}\right)=m g \times \frac{a}{2}\)

\(\left(b+\frac{a}{2}\right) \mu=\frac{a}{2}\)

\(0.4=\mu=\frac{a}{2 b+a}\)

\(0.8 b+0.4 a=a\)

\(0.8 \mathrm{b}=0.6 \mathrm{a}\)

\(\frac{b}{a}=\frac{3}{4}\)

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