Class 12th

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

Mid point of BC is $\frac{1}{2}\left(5\widehat{i}+\left(\alpha -2\right)\widehat{j}+9\widehat{k}\right)$  

$\overline{AB}=\widehat{i}+\left(\alpha -4\right)\widehat{j}+\widehat{k}$  

$\overline{AC}=\widehat{i}+\left(-2-\alpha \right)\widehat{j}+\widehat{k}$

For a = 1,   $\overline{AB}$ and $\overline{AC}$  will be collinear. So for non collinearity

a = 2

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2\left(1+4+2-16\right)}{1+2^2+2^2}$  

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$\left|\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z+1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=0$

->2x – z = 1

option (B) satisfies.

According to question, we can write

  $d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2*125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

Let AB $\equiv x-2y+1=0$  

AC  $\equiv 2x-y+1=0$  

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through

orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_P=\frac{{\mu }_{0}I}{2\pi r}*2$

$\Rightarrow 300*10^{-6}=\frac{4\pi *10^{-7}}{2\pi *4*10^{-2}}*2$                                                                         

=> I = 30 A

Line $\perp$  to the normal

->3p + 2q – 1 = 0

$\left(2,-1,-3\right)$ lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin =  $\left|\frac{5}{\sqrt[]{15^2+{\left(-22\right)}^2+1^2}}\right|=\sqrt[]{\frac{5}{142}}$  

Process-AB $\Rightarrow$ Isobaric,                                                           

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

According to Law of discharging of capacitor, we can write

$Q=Q_0{e}^{-\frac{t}{\tau }}\Rightarrow \frac{Q_2}{8}=Q_0{e}^{-\frac{t_2}{\tau }}\Rightarrow t_{2}3\tau \mathcal{l}n\left(2\right),and$

^{$U=\frac{Q^2}{2C}=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow \frac{1}{2}\left(\frac{Q_0^2}{2C}\right)=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow t_1=\frac{\tau }{2}\mathcal{l}n\left(2\right)$}

^{$\Rightarrow \frac{t_1}{t_2}=\frac{1}{6}$}

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$  

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

-> $C=+\frac{\pi }{2}$

now at x =  $\mathcal{l}n\sqrt[]{3}$  

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$