Physics

Scalar quantities only have magnitude, which makes sense to combine using ordinary algebra. But vector quantities have both magnitude and direction. Due to this directional aspect, vectors must obey special rules of vector algebra. Vectors have to specifically follow the triangle law or the parallelogram law of addition to be represented in the graph format. These graphical methods account for both magnitude and direction. This makes sure that the resultant vector accurately reflects the combined effect of the individual vectors. If we apply ordinary algebra, we won't be able to know the directional information.

$\omega =2\pi f=2\pi *50=100\pi$

$X_L=wL=100\pi *100*10^{-3}$

$X_C=\frac{1}{wC}=\frac{1}{100\pi }*100*10^{-6}=\frac{100}{\pi }$

$z=10.0086\Omega$

$l=\frac{v}{z}=\frac{220}{10.008}=22A$

volume of wire will remains constant

$\pi r_1^2{\mathcal{l}}_1=\pi r_2^2{\mathcal{l}}_2$

$\Rightarrow {\mathcal{l}}_2=2{\mathcal{l}}_1\left(Given\right)$

$r_2=\frac{r_1}{\sqrt[]{2}}$

$R_2=\frac{\rho {\mathcal{l}}_2}{\pi r_2^2}=\frac{\rho 2{\mathcal{l}}_1}{\pi {\left(\frac{r_1}{\sqrt[]{2}}\right)}^2}$

$=\frac{R_2-R_1}{R_1}*100=\frac{4R_1-R_1}{R_1}*100=\frac{3R_1}{R_1}*100=300\mathrm{\%}$

Volume of 27 identical drops = volume of a bigger drop

$\Rightarrow 27*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$              

R³ = 27r³

R = 3r

Given potential of a small drop = 22v

$V_{bigger}=\frac{k\left(27q\right)}{R}=\frac{27kq}{3r}=9\frac{kq}{r}=9*22=198volt$

For closed vessel

P a T     [v = constant]

->P = kT

$\frac{0.4}{100}=\frac{1}{1}$

T = 250 K

M₁ ->Mass of satellite (1)

M₂ -> Mass of satellite (2)

M_P -> Mass of planet    

Now NLM (2) on S₁

_{$\frac{G{M}_1{M}_P}{R_1^2}=\frac{m_1{v}_1^2}{R_1}$}

Similarly v₂ =   $\sqrt[]{\frac{G{M}_P}{R_2}}$

$\frac{v_1}{v_2}=\sqrt[]{\frac{R_2}{R_1}}=\sqrt[]{\frac{800}{3200}}=\frac{1}{2}=\frac{1}{x}$                

l₁ = M. l of solid sphere about its diameter

$=\frac{2}{5}MR{\text{'}}^2=\frac{2}{5}M{\left(2R\right)}^2=\frac{8}{5}M{R}^2$               

l₂ = M. I of solid cylinder about its axis

$=\frac{M{R}^{12}}{2}=\frac{M{\left(2R\right)}^2}{2}=2M{R}^2$               

I₃ = M. I of solid circular disc about its diameter

$=\frac{M{R}^{12}}{4}=\frac{M{\left(2R\right)}^2}{2}=M{R}^2$               

I₄ = M. I of this circular ring about its diameter

  $=\frac{M{R}^{12}}{2}=\frac{M{\left(2R\right)}^2}{2}=2M{R}^2$              

$\Rightarrow 6M{R}^2+2M{R}^2=\frac{x8M{R}^2}{5}$               

x = 5

For equilibrium along the incline plane is given by,

F cos 60° = mg sin 60°

  $F=\frac{0.2*10*\sqrt[]{3}}{2*1}*2$              

  $=2\sqrt[]{3}N=\sqrt[]{12}N$              

$f=\frac{m{v}^2}{R}$            

$\Rightarrow \frac{m{v}^2}{R}=\mu N$

 

$\frac{m{v}^2}{R}=\mu mg$

$\Rightarrow 30=\sqrt[]{\mu *75*10}$                

  $\mu =\frac{30*30}{750}=1.2$              

              Now, R = 48 m (new Radius of curvature)

    $\mu =1.2$

$V_{max}=\sqrt[]{\mu Rg}$

$=\sqrt[]{1.2*48*10}$            

$=24m/sec$               

Using snell's law –

sin I = $\mu sinr$

sin45° = $\mu sin30°$

$\mu$$=$$\frac{sin45}{sin30}$$=$$\frac{1}{\text{exception 25:}}$$*$$2$$=$√2