Physics

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

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

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

$h\nu =3.4-1.51=1.89eV$

As we know that radius of circular path in magnetic field is given as

$r=\frac{\sqrt[]{2mK}}{qB}\Rightarrow K=\frac{r^2{q}^2{B}^2}{2m}$              

$\Rightarrow K=\frac{{\left(7*10^{-3}\right)}^2*\left(1.6*10^{-19}\right)*{\left(5*10^{-4}\right)}^2}{2*9.1*10^{-31}}eV=107.7*10^{-2}eV$  = 1.08 eV

$\Rightarrow ?=h\nu -K=1.89-1.08=0.81eV$              

Method-I : Using Kirchhoff's Law for loop $\overrightarrow{BDABB},$  we can write

$-140+20l+6l_1=0\Rightarrow 10l+3l_1=70.......\left(i\right)$              

Using Kirchhoff's Law for loop $\overrightarrow{ACBBA},$  

we can write

$5\left(l-l_1\right)+90-6l_1=0$              

->-5l + 11l₁ = 90 .(ii)

Adding equation (1) with twice the equation (2), we have

$25l_1=250\Rightarrow l_1=\frac{250}{25}=10A$              

Method-II : Using concept of equivalent cell, we can write

$V_{AB}=\frac{\frac{140}{20}+\frac{0}{6}+\frac{90}{5}}{\frac{1}{20}+\frac{1}{6}+\frac{1}{5}}=\frac{2100+0+5400}{15+50+60}$

$=\frac{7500}{125}=60Volt$

$\Rightarrow l_1=\frac{60}{6}=10A$              

As we know that equivalent half life of radioactive material decays by simultaneous emissions, is given as

$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{700}+\frac{1}{1400}=\frac{3}{1400}\Rightarrow T=\frac{1400}{3}years$

$\Rightarrow t=\frac{\mathcal{l}n\left(3\right)}{\lambda }=\frac{\mathcal{l}n\left(3\right)}{\mathcal{l}n\left(2\right)}T=\frac{1.1}{0.693}*\frac{1400}{3}\approx 740.74years$

Using conservation linear momentum, we can write

$Mv=\frac{h}{\lambda }=\frac{h\nu }{c}\Rightarrow v=\frac{h\nu }{Mc}$                                                           

Loss of internal energy = Gain in Kinetic energy + energy of gamma ray

= $\frac{1}{2}M{v}^2+h\nu =\frac{1}{2}M{\left(\frac{h\nu }{Mc}\right)}^2+h\nu =h\nu \left[1+\frac{h\nu }{2M{c}^2}\right]$