A heavy ball of mass M is suspended from the ceiling of a car by a light string of mass m   m ≪ M…

1. $\frac{g}{10}$
2. $\frac{g}{20}$
3. $\frac{g}{5}$
4. $\frac{g}{30}$

Solution

For standing car, $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{Mg}{\mu }}$



For accelerating car,

$\upsilon \prime =\sqrt{\frac{\sqrt{{\left(Mg\right)}^2+{\left(Ma\right)}^2}}{\mu }}$

$\Rightarrow \frac{60.5}{60}=\sqrt{\frac{\sqrt{{\left(Mg\right)}^2+{\left(Ma\right)}^2}}{Mg}}$

$\Rightarrow {\left(1+\frac{1}{120}\right)}^2=\frac{\sqrt{{\left(Mg\right)}^2+{\left(Ma\right)}^2}}{Mg}$

$\Rightarrow 1+\frac{2}{120}=\sqrt{1+\frac{a^2}{g^2}}$

$\approx \frac{2}{120}=\frac{1}{2}\frac{a^2}{g^2}$

$\therefore a=g\sqrt{\frac{4}{120}}=\frac{g}{\sqrt{30}}\approx \frac{g}{5}$

Asked in: JEE Main 2019 (09 Jan Shift 1)

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