Physics

$V_{rms}\alpha \sqrt[]{T}|T_i=127°C=400k.$

$V_{rm{s}^1}=2V_{rms},m=0.056kg=56g$   

T_f = 4T_i = 4 * 400 = 1600 k

$Q=n{c}_v\Delta T=\left(\frac{m}{M}\right)C_v\Delta T$        

=  $\left(\frac{56}{28}\right)\frac{5}{2}R\Delta T$

=  $2*\frac{5}{2}*2*1200$

Q = 12 * 10³ cal

= 12 k cal

At steady state condition -> Heat conducted through slab AB = Heat conducted though slab BC

$\frac{k_{1}A}{AB}\left(100-80\right)=\frac{kA}{B{C}_R}\left(80-0\right)$

-> $\frac{k_1}{16}*20=\frac{k}{8}*80$ $\frac{{}_{}}{}\frac{}{}$

k1 = 8 k

  $B_C=\frac{{\mu }_{0}l}{2r}=B$ ${}_{}\frac{{}_{}}{}$ -(1)

$\begin{array}{l}{B}_P=\frac{{\mu }_{0}l{r}^2}{2{\left(r^2+\frac{r^2}{4}\right)}^{3/2}}\\ B_P=\frac{{\mu }_{0}l{r}^2}{2*r^3{\left(\frac{5}{4}\right)}^{3/2}}\end{array}$

$B_P=\frac{{\mu }_{0}l}{2r{\left(\frac{5}{4}\right)}^{3/2}}$

$\Rightarrow B_P=\frac{B{4}^{3/2}}{5^{3/2}}$

$B_P=B\frac{2^3}{{\left(\sqrt[]{5}\right)}^3}$

$B_P={\left(\frac{2}{\sqrt[]{5}}\right)}^{3}B$

A = 30p cm² = 30p * 10^-4 m²

d = 1mm            = 10^-3 m

E = (dielectric strength for breakdown)

= 3.6 * 10⁷ v/m

Q = 7 * 10^-6 C

$\frac{\sigma }{{\in }_0}=E$

        $\Rightarrow \frac{Q}{kA{\in }_0}=E$

$k=\frac{Q}{A{\in }_{0}E}$        

$=\frac{7*10^{-6}*4\pi *9*10^9}{30\pi *10^{-4}*1*3.6*10^7}$

$=\frac{7*4\pi *9}{30\pi *3.6}$

$=\frac{7*4*9*10}{30*36}$

$=\frac{70}{30}$

$k=\frac{7}{3}=2.33$

0.25 = 1 - $\frac{T_2}{T_1}$ $\frac{{}_{}}{{}_{}}$

0.25 = 1 - $\frac{\left(27+273\right)}{T_1}\Rightarrow \frac{300}{T_1}=1-0.25$ $\frac{}{{}_{}}\frac{}{{}_{}}$

$\Rightarrow \frac{300}{T_1}=0.75$

$T_1=\frac{300*100}{75}$

T1 = 400 k

Now, efficiency increases by 100%

${\eta }_2=100\mathrm{\%}{\eta }_1+{\eta }_1$

$=2{\eta }_1$

= 0.25 * 2

= 0.50

0.50 = 1 - $\frac{T_2\text{'}}{T_1}$ $\frac{{}_{}}{{}_{}}$

$\Rightarrow \frac{T_2\text{'}}{T_1}=0.50$

$T_1\text{'}=\frac{300}{0.5}$

$T_1\text{'}=600k$

$$ $T_1\text{'}-T_1=$  (600 – 400) = 200 k or 200° C

Baseband Signal frequency 3.5 MHz

& Carrier signal frequency = 3.5 GHz

${\upsilon }_C=3.5GHz=3.5*10^9$

$\lambda =\frac{C}{{\upsilon }_c}=\frac{3*10^8}{3.5*10^9}=\frac{3}{3.5}*10$

$=\frac{30}{35}*10$

= $\frac{60}{7}$ $\frac{}{}$

Size of antenna = $\frac{\lambda }{4}=\frac{60}{7*4}$ $\frac{}{}\frac{}{}$

= 21.4 mm

$A{\to }^{105}B{+}^{115}C$

$\Rightarrow Q=\left[105*6.4+115*6.4\right]-\left[220*5.6\right]Mev$

Q = 176 MeV

According to Rutherford, e- revolves around in nucleus in circular orbit. Thus e- is always accelerating (centripetal acceleration). An accelerating change emits EM radiation and thus e- should loose energy and finally should collapse in the nucleus.

Velocity of wave in a medium = $\frac{1}{\sqrt[]{{\mu }_m{\in }_m}}$ $\frac{}{\text{exception 25:}}$

$\begin{array}{l}v=\frac{1}{\sqrt[]{{\mu }_0{\mu }_r{\in }_r{\in }_0}}\\ v=\frac{3*10^8}{\sqrt[]{1.61*6.44}}=0.931*10^{8}m/sec\end{array}$

$\begin{array}{l}B={\mu }_{m}H\\ ={\mu }_r{\mu }_{0}H\\ =1.61*4\pi *10^{-7}*4.5*10^{-2}\\ =1.61*4\pi *4.5*10^{-9}\end{array}$

$\frac{E}{B}=v$

E = vB   = 0.931 * 10⁸ * 1.61 * 4p * 4.5 * 10^-9

= [0.931 * 1.61 * 4p * 4.5] * 10^-1

= 8.476

» 8.476 v m^-1

$y_1=5sin2\pi \left(x-vt\right)cm$

$y_2=3sin2\pi \left(x-vt+1.5\right)cm$

$?=\left(phaseAngle\right)=2\pi *1.5=3\pi$

$y_R=\sqrt[]{y_1^2+y_2^2+2y_1{y}_{2}cos?}$

$\begin{array}{l}=\sqrt[]{5^{2}T{3}^2+2*5*3cos3\pi }\\ =\sqrt[]{25+9+\left(30*-1\right)}\end{array}$

$\begin{array}{l}=\sqrt[]{34-30}\\ =\sqrt[]{4}\\ =2cm\end{array}$