Physics

Zener break down occurs in p-n junction having p and n both : Heavily doped and have narrow depletion layer.

$T=2\pi \sqrt[]{\frac{L}{g}}or{T}^2=4{\pi }^2\left(\frac{L}{g}\right)$

$\Rightarrow g=4{\pi }^2\left(\frac{L}{T^2}\right)$

$\Rightarrow \frac{\Delta g}{g}\mathrm{\%}=\left(\frac{\Delta L}{L}+\frac{2\Delta T}{T}\right)\mathrm{\%}=\left[\frac{1mm}{1m}+\frac{2*0.01}{1.95}\right]*100\mathrm{\%}=\left(0.001+0.0102\right)*100=1.13\mathrm{\%}$

R = $2\Omega$

L = 2 mH

E = 9V

$i=\frac{\epsilon }{2R}=\frac{9v}{4\Omega }=2.25A$

Just after the switch 'S' is closed, the inductor acts as open circuit.

According to KTG, the gas exerts pressure because its molecule :

Suffer change in momentum when impinge on the walls of container.

$F_{1}cos45°+F_{2}cos45°+F_3=\frac{m{v}^2}{r}$

$\Rightarrow \frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(2r\right)}^2}=\frac{m{v}^2}{r}$

$\Rightarrow v=\sqrt[]{\frac{Gm}{4r}\left(2\sqrt[]{2}+1\right)}$

Putting m = 1 kg and r = 1 m,

$v=\frac{1}{2}\sqrt[]{G\left(1+2\sqrt[]{2}\right)}$

Velocity of block in equilibrium, in first case,

$v=A\omega =A.\sqrt[]{\frac{k}{M}}$

Velocity of block in equilibrium, is second case,

$v\text{'}=A\text{'}\omega \text{'}=A\text{'}\sqrt[]{\frac{k}{M+m}}$

From conservation of momentum,

Mv = (M + m) v'

$\Rightarrow MA\sqrt[]{\frac{k}{M}}=\left(M+m\right)A\text{'}\sqrt[]{\frac{k}{M+m}}\Rightarrow A\text{'}=A\sqrt[]{\frac{M}{M+m}}$

Since, $\frac{x^2}{\alpha kT}$ $\frac{{}^{}}{}$ should be dimensionless.

So, dimension of $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$ $\frac{{}^{}}{{}^{}{}^{}}{}^{}{}^{}$

Dimension of $\alpha {\beta }^2$ ${}^{}$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$ ${}^{}{}^{}{}^{}$

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$

$\frac{dq}{dt}=t=20t+8t^2$

$\Rightarrow {\displaystyle \underset{0}{\overset{q}{\int }}dq={\displaystyle \underset{0}{\overset{15}{\int }}\left(20t+8t^2\right)dt}}$

$\Rightarrow q=20*\frac{15^2}{2}+8.\frac{15^3}{3}=11250C$

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

  $=\left(l_L+l_C\right)$  

Therefore current through R circuit at resonance will be zero