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

Case I:

From diagram, ${\mathrm{T}}_1=\mathrm{mg}$ $\therefore v_1=\sqrt{\frac{T_1}{\mu }}=\sqrt{\frac{mg}{\mu }}$     ....(i)

Case II:

From diagram, ${\mathrm{T}}_2\cos \theta =\mathrm{mg},{\mathrm{T}}_2\sin \theta =\mathrm{ma}$
$\therefore {\mathrm{T}}_2^2\left({\sin }^2\theta +{\cos }^2\theta \right)={\mathrm{m}}^2\left({\mathrm{g}}^2+{\mathrm{a}}^2\right)$
${\mathrm{T}}_2=\mathrm{m}{\left({\mathrm{g}}^2+{\mathrm{a}}^2\right)}^{\frac{1}{2}}$
$v_2=\sqrt{\frac{T_2}{\mu }}=\sqrt{\frac{m\sqrt{a^2+g^2}}{\mu }} ....\left(ii\right)$
From (i) and (ii), $\frac{v_1}{v_2}=\sqrt{\frac{g}{{\left(g^2+a^2\right)}^{\frac{1}{2}}}}$
$\Rightarrow \left(\frac{v_2}{v_1}\right)={\left(\frac{g^2+a^2}{g^2}\right)}^{\frac{1}{4}}$
Taking binomial approximation, we get $\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}=1+\frac{1}{4}\frac{{\mathrm{a}}_2}{{\mathrm{g}}_2} \dots .\left[\because (1+\mathrm{x}{)}^{\mathrm{n}}\approx 1+\mathrm{nx}\right]$
$\Rightarrow \frac{51}{50}=1+\frac{1}{4}\frac{a_2}{g_2}$
$\Rightarrow \frac{a_2}{g_2}=\frac{8}{100}$
$\Rightarrow \mathrm{a}=\frac{2\sqrt{2}\mathrm{g}}{10}$
$\Rightarrow \mathrm{a}=2.82 {\mathrm{ms}}^{-2}$