Physics

Using conservation of linear momentum, we can write

$mu=M{v}_1\Rightarrow v_1=\frac{mu}{M}........\left(1\right)$           

Using conservation of angular momentum about centre of mass of Rod, we can write

$v_1+\frac{\mathcal{l}\omega }{2}=u\Rightarrow \frac{mu}{M}+\frac{3mu}{M}=u\Rightarrow \frac{m}{M}=\frac{1}{4}$     . (2)

Using definition of e, we can write

$v_1+\frac{\mathcal{l}\omega }{2}=u\Rightarrow \frac{mu}{M}+\frac{3mu}{M}=u\Rightarrow \frac{m}{M}=\frac{1}{4}$  

F = qE Þ Electric force acting on the body

$a=\frac{F}{m}=\frac{q}{m}E=\frac{8*10^{-6}}{10^{-3}}*100=0.80m/s^2$ ->Acceleration of body

$T=2\sqrt[]{\frac{2\mathcal{l}}{a}}=2\sqrt[]{\frac{2*0.1}{0.8}}=1s\Rightarrow$ Time period of Motion

Initially wall will act as observer, so with the help of Doppler's effect, we can write

$f\text{'}=\frac{c}{c-v_0}{f}_0$  

Now wall will act as source of sound of frequency f', so With the help of Doppler's effect, we can write

$f\text{'}\text{'}=\frac{c+v_0}{c}{f}^{\text{'}}=\frac{c+v_0}{c-v_0}{f}_0$              

$\Rightarrow 500=\left(\frac{c+v_0}{c-v_0}\right)400\Rightarrow 5c-5v_0=4c+4v_0\Rightarrow v_0=\frac{c}{9}=\frac{330}{9}m/s=\frac{330}{9}*\frac{18}{5}km/h=132km/h$

Let ball starts its motion with horizontal velocity v₀, so with the help of conservation of mechanical energy, we can write

$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{x}^2\Rightarrow v_0=x\sqrt[]{\frac{k}{m}}=0.05*\sqrt[]{\frac{100}{0.1}}=0.5*\sqrt[]{10}m/s$              

t = Time required to fall the ball =    $\sqrt[]{\frac{2h}{g}}=2*\sqrt[]{\frac{1}{10}}s$

^{$=d=v_0*t=0.5*\sqrt[]{10}*2*\sqrt[]{\frac{1}{10}}=1m$}              

Heat absorbed in cyclic process = Work done = 100? Joule

The equation of wave at any time t will be $y=\frac{1}{1+{\left(x?vt\right)}^2},$ so v * 1 = 2 -> v = 2m/s

In the frame of vehicle, vehicle is in equilibrium under the influence of pseudo force F_P

  $F_P=\frac{m{v}^2}{R}$            

N = mg cos 30° + F_P sin 30°

$\Rightarrow N=mgcos30°+\frac{m{v}^2}{R}sin30°.\left(i\right)$              

, and

$f_S=\frac{m{v}^2}{R}cos30°-mgsin30°$              

By doing (1) * cos 30° - (2) * sin 30°, we have

$\Rightarrow N=\frac{800*10}{0.87-0.2*0.5}=\frac{800*10}{0.77}=10.2*10^{3}kgm/s^2$

Since direction of incident ray (DIR) is from left to write, so considering refraction at point M, we can write

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$              

-> $\frac{1.4}{v}-\frac{1.25}{-40}=\frac{1.4-1.25}{-25}$

$\Rightarrow -\frac{1.4}{v}=\frac{0.15}{25}+\frac{1.25}{40}=\frac{1.2+6.25}{200}=\frac{7.45}{200}$

$\Rightarrow v=-\frac{200*1.4}{7.45}=-37.58cm$

When arm PQ of a rectangular conductor is moving from x = 0 to x = b, the flux $\left(?=ax\right)$ linked with loop increases and while moving from x = b to x = 2b and from x = 2b to x = b, $\left(?=ab\right)$ flux remains constant and then flux $\left(?=ax\right)$ decreases to zero as it moves from x = b to x = 0.

$e_{in}=-\frac{d?}{dt}=-a\frac{dx}{dt}=-av\Rightarrow$ Induced emf

$P=\frac{{\left(e_{in}\right)}^2}{R}\Rightarrow$ Power dissipated

$l_1=\frac{50}{2000}=25mA,andl=\frac{V_i-V_Z}{1000}=\frac{100-50}{1000}=50mA$

$\Rightarrow l_z=l-l_1=25mA$