(3)

Here, $\overrightarrow{F}=3\hat{j}\text{N},$     $\overrightarrow{r}=2\hat{k}\text{m}$

Torque, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=2\hat{k}\times 3\hat{j}=-6\hat{i}\text{Nm}$

(4)

[IMAGE 55]

Moment of the force is,

$\overrightarrow{\tau }=(\overrightarrow{r}-{\overrightarrow{r}}_0)\times \overrightarrow{F}$

Here, ${\overrightarrow{r}}_0=2\hat{i}-2\hat{j}-2\hat{k}$

and $\overrightarrow{r}=2\hat{i}+0\hat{j}-3\hat{k}$

$\therefore$    $\overrightarrow{r}-{\overrightarrow{r}}_0=(2\hat{i}+0\hat{j}-3\hat{k})-(2\hat{i}-2\hat{j}-2\hat{k})=0\hat{i}+2\hat{j}-\hat{k}$

$\therefore$    $\overrightarrow{\tau }=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& -1\\ 4& 5& -6\end{array}\right|$$=-7\hat{i}-4\hat{j}-8\hat{k}$

(3)

For the conservation of angular momentum about origin, the torque $\overrightarrow{\tau }$ acting on the particle will be zero.

By definition, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Here, $\overrightarrow{r}=2\hat{i}-6\hat{j}-12\hat{k}$ and $\overrightarrow{F}=\alpha \hat{i}+3\hat{j}+6\hat{k}$

$\therefore$   $\overrightarrow{\tau }=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -6& -12\\ \alpha & 3& 6\end{array}\right|$

           $=\hat{i}(-36+36)-\hat{j}(12+12\alpha )+\hat{k}(6+6\alpha )$

            $=-\hat{j}(12+12\alpha )+\hat{k}(6+6\alpha )$

But  $\overrightarrow{\tau }=0$

$\therefore$      $12+12\alpha =0 \text{or} \alpha =-1$ and  $6+6\alpha =0 \text{or} \alpha =-1$

(1)

According to law of conservation of angular momentum

        $mvr=mv\text{'}r\text{'}$

        $v_0{R}_0=v\left(\frac{R_0}{2}\right);$ $v=2v_0$                              ...(i)

$\therefore$    $\frac{K_0}{K}=\frac{\frac{1}{2}m{v}_0^2}{\frac{1}{2}m{v}^2}={\left(\frac{{\displaystyle v_0}}{{\displaystyle v}}\right)}^2$

or    $\frac{K}{K_0}={\left(\frac{{\displaystyle v}}{{\displaystyle v_0}}\right)}^2={\left(2\right)}^2 \text{(Using (i))}$

          $K=4K_0=2m{v}_0^2$