Physics

The law of gravitation holds good for any pair of bodies in the universe.

This is correct statement.

The weight of any person becomes zero when the person is at the centre of the Earth.

Weight = mg

= m * 0 [g = 0 at the centre of the Earth]

= 0

$\Rightarrow I_{g}G=\left(l-l_g\right)8\Rightarrow l_g*72=\left(l-l_g\right)8\Rightarrow 9l_g=l-l_g$

$10l_g=l{l}_g=\frac{1}{10}l$

% of I which passes through galvanometer

$$ $l_g=10\mathrm{\%}$  of I

To keep,   $\Delta {\alpha }_{cm}=0$

      $M_1\Delta x_1+M_2\Delta X_2=0$          

$\Rightarrow 10*6+30\Delta X_2=0$      

$\Delta X_2=\frac{-10*6}{30}$

$\Delta X_2=-2cm$        

i.e. 2 cm towards the 10 kg block.

Assume both the block is moving with common acceleration 'a' then

$a=\frac{F}{m+M}.......(i)$

N = mg

N = 2g

$f=ma=\frac{mF}{m+M}$

For no slipping

$\Rightarrow \frac{2F}{10}\le g$

$F\le 5g$

$\begin{array}{l}F\le 5*9.8\\ F\le 49.0N\end{array}$

Given, m = 0.6 $\pm$ $$ 0.006

$\Rightarrow p=\frac{m}{v}=\frac{m}{\pi r^2\mathcal{l}}$

$\frac{\Delta \rho }{\rho }*100=\frac{\Delta m}{m}*100+2\frac{\Delta r}{r}*100+\frac{\Delta \mathcal{l}}{\mathcal{l}}*100$

= 1 + 2 * 1 + 1

From the physics perspective, in uniform circular motion, speed and angular velocity are constant. That makes the angular acceleration and tangential acceleration are zero. Non-uniform motion, on the other hand, has varying speed and angular velocity, where angular acceleration is non-zero. 

Let $V_{R{G}_r}$ ${}_{{}_{}}$ (velocity of Rain w.r.t. ground or w.r.t to standing Girl)

${\overrightarrow{V}}_{R{G}_r}=V_x\widehat{i}+V_y\widehat{j}$

$tan\left(270+45°\right)=-cot45°=\frac{V_y}{V_x}=-1$

Now, velocity of Rain w.r.t. moving Girl

${\overrightarrow{V}}_{R{G}_m}=\left(V_x-15\sqrt[]{2}\right)\widehat{i}+V_y\widehat{j}=\left(15\sqrt[]{2}-15\sqrt[]{2}\right)\widehat{i}+V_y\widehat{j}=V_y\widehat{j}$

${\overrightarrow{V}}_{R{G}_m}=-15\sqrt[]{2}\widehat{j}\Rightarrow {\overrightarrow{V}}_{R{G}_m}=\frac{-30}{\sqrt[]{2}}\widehat{j}\Rightarrow \left|{\overrightarrow{V}}_{R{G}_m}\right|=\frac{30}{\sqrt[]{2}}$

We know that horizontal speed will remain constant

20 cos = v cos ……… (i)

Along y-axis

$U_y=20sin\alpha ,V_y=vsin\beta$

$V_y=V_y+a_{y}t$

$$ $Vsin\beta =20sin\alpha -10*10$  ……. (ii)

$\Rightarrow (ii)÷(i)$

$\Rightarrow tan\beta =tan\alpha -5sec\alpha$

In forward bias diode act as a short circuit wire. Hence, the equivalent circuit is now.

So,  ${}_{}\frac{}{}$ $V_{ab}=\frac{30}{5+10}*5=10\mathrm{V}$

$\mathrm{}50.00$

Net force on motor will be

$F_{\text{min}}=[920+68(10\left)\right]g+6000$

$=22000\mathrm{N}$

So, required power for motor

$\begin{array}{rr}{P}_{\text{min}}& ={\stackrel{?}{F}}_{\text{min}}\cdot \stackrel{?}{v}\\ & =22000*3\\ & =66000\text{watt}\end{array}$