In the following sequence of reactions,what is $D$?
$Toluene$ $\xrightarrow{[O]} A$ $\xrightarrow{SOCl_2} B$ $\xrightarrow{NaN_3} C$ $\xrightarrow{\text{Heat}} D$

$A$ body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{mv^2}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle?

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:

Given that,$a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution,then $a, b$ and $c$ lie in

$A$ particle is performing $U.C.M.$ along the circumference of a circle of diameter $50 \ cm$ with frequency $2 \ Hz$. The acceleration of the particle in $m/s^2$ is

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod perpendicular to its length. The acceleration of the centre of mass of the rod is: