The driver of a car records a period of $\frac{\pi }{3}$ seconds for a pendulum of 1m hung from the roof. The acceleration of the car is $\left(\mathrm{g}=10{\mathrm{ms}}^{-2}\right)$

Given, $T=\frac{\pi }{3}\text{ second, }l=1m$ . As the car accelerates it will given a pseudoforce on bob and the bob will tilt itself from the vertical by an angle $\theta ={\tan }^{-1}\left(\frac{a}{g}\right)$ and the time period of oscillation is $T=2\pi \sqrt{\frac{l}{\sqrt{g^2+a^2}}}$ squaring both sides with T and L ${\left(\frac{\pi }{3}\right)}^2=4{\pi }^2\frac{1}{\sqrt{g^2+a^2}}\Rightarrow \sqrt{g^2+a^2}=36$ $a^2={36}^2-{10}^2;a=\sqrt{{36}^2-{10}^2}=34.5{\mathrm{ms}}^{-2}$  so, choice (4) is correct.