Class 12th

According to Young's double slit experiment, we can write

$\beta =\frac{\lambda D}{d}\Rightarrow \Delta \beta ={\beta }_2-{\beta }_1=\frac{\lambda }{d}\Delta D$

$\Rightarrow \lambda =\frac{d\Delta \beta }{\Delta D}=\frac{1*10^{-3}*3*10^{-5}}{5*10^{-2}}=60*10^{-8}m=600nm$

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$         

Time required = 4 * $t_{\frac{1}{2}}=20yrs$  

$\frac{dE}{dT}$ is the temperature Co-efficient of cell. The cell having less variation of EMF, with respect to temperature have high efficiency.

According to definition of displacement current, we can write

$I_d={\epsilon }_0\frac{d?}{dt}={\epsilon }_0\frac{d\left(ES\right)}{dt}={\epsilon }_0\frac{d\left(\frac{V}{\mathcal{l}}S\right)}{dt}=\frac{{\epsilon }_{0}S}{\mathcal{l}}\left(\frac{dV}{dt}\right)$

$\Rightarrow \mathcal{l}=\frac{8.85*10^{-12}*40*10^{-4}*10^6}{4.425*10^{-6}}8*10^{-3}m$

According to Work energy theorem, we can write

$K_f-K_i=W_{ElectricForce}\Rightarrow \frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2=-eV\Rightarrow v^2=v_0^2-\frac{2eV}{m}$

$\Rightarrow v^2={\left(6.0*10^5\right)}^2-\frac{2*1.6*10^{-19}}{9*10^{-31}}=\frac{324-128}{9}*10^{10}=\frac{196}{9}*10^{10}\Rightarrow V=\frac{14}{3}*10^{5}m/s$

According to Kirchhoff's Law, we can write

-20 + 2000I + 600 * 5I = 0 $\Rightarrow I=\frac{20}{5000}A$  

Reading of voltmeter = 2000I = 2000 * $\frac{20}{5000}=8volt$  

According to Concept of resonance tube, we can write

$\frac{\lambda }{4}+e={\mathcal{l}}_{1}and\frac{3\lambda }{4}+e={\mathcal{l}}_2\Rightarrow \frac{\lambda }{2}={\mathcal{l}}_2-{\mathcal{l}}_1\Rightarrow \frac{V}{2\nu }={\mathcal{l}}_2-{\mathcal{l}}_1$

$\Rightarrow {\mathcal{l}}_2=\frac{v}{2\nu }+{\mathcal{l}}_1=\frac{336}{400}+0.20=0.84+0.20=1.04m=104cm$

$Molality=\frac{molesofsolute*1000}{wtofsolvent\left(gm\right)}$  

$Molality=\frac{35*1000*1.46}{36.5*100}$ = 14.0 M

According to question, we can write

$\frac{2V}{2+2r}=\frac{V}{2+\frac{r}{2}}$

$\Rightarrow 2+2r=4+r\Rightarrow r=2\Omega$

According to relation between field and potential, we can write

  ^{$\overrightarrow{E}=-\frac{dV}{dx}\left(\widehat{i}\right)=-\frac{d\left(3x^2\right)}{dx}\left(\widehat{i}\right)=-6x\left(\widehat{i}\right)$}

^{${\overrightarrow{E}}_{\left(1,0,3\right)}=-6\left(\widehat{i}\right)N/C$}