Class 12th

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 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

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

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

Let PT perpendicular to QR

$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}=\lambda \Rightarrow T\left(2\lambda -1,3\lambda -2,2\lambda +1\right)$ therefore

$2\left(2\lambda -5\right)+3\left(3\lambda -4\right)+2\left(2\lambda -6\right)=0\Rightarrow \lambda =2$

$T\left(3,4,5\right)\therefore PT=\sqrt[]{1+4+4}=3\therefore QT=\sqrt[]{26-9}=\sqrt[]{17}$

$\therefore \Delta PQR=\frac{1}{2}*2\sqrt[]{17}*3=3\sqrt[]{17}$  

Therefore square of  $ar\left(\Delta PQR\right)$ = 153.

According to Lorentz's Force, we can write

$\overrightarrow{F}={\overrightarrow{F}}_{Electric}+{\overrightarrow{F}}_{Magnetic}=q\overrightarrow{E}+q\left(\overrightarrow{v}*\overrightarrow{B}\right),so$  

Statement I is correct but Statement II is incorrect

$f\text{'}\left(x\right)=\frac{4x^2-1}{x}$ so f (x) is decreasing in $\left(0,\frac{1}{2}\right)and\left(\frac{1}{2},\infty \right)$ $\Rightarrow a=\frac{1}{2}$  

Tangent at y² = 2x is y = mx + $\frac{1}{2m}$ it is passing through (4, 3) therefore we get m = $\frac{1}{2}or\frac{1}{4}$  

So tangent may be  $y=\frac{1}{2}x+1ory=\frac{1}{4}x+2buty=\frac{1}{2}x+1$  passes through (-2, 0) so rejected.

Equation of normal  $\frac{x}{9}+\frac{y}{36}=1$  

A salt bridge is a U-shaped tube that connects the two halves (oxidation and reduction) of the cell with an electrochemical cell, such as a galvanic cell. It allows the flow of ions in both halves and maintains the electricity to stay neutral.

${\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}={\left(\frac{f\left(x\right)}{x}\right)}^3\Rightarrow x^3{\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}=f^3\left(x\right),$ differentiating w.r.to x

$x^{3}f\left(x\right)+3x^2\frac{f^3\left(x\right)}{x^3}=3f^2\left(x\right)f\text{'}\left(x\right)\Rightarrow 3y^2\frac{dy}{dx}=x^{3}y=\frac{3y^3}{x}\Rightarrow 3xy\frac{dy}{dx}=x^4+3y^2$  

After solving we get  $y^2=\frac{x^4}{3}+c{x}^2$  also curve passes through (3, 3) Þ c = -2

$\therefore y^2=\frac{x^4}{3}-2x^2$ which passes through $\left(\alpha ,6\sqrt[]{10}\right)$ $\therefore \frac{{\alpha }^4-6{\alpha }^2}{3}=360\Rightarrow \alpha =6$