Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (a)

Explanation- field lines are always perpendicular to the surface.

$\begin{array}{l}I={\displaystyle \int si{n^{-}}^1}\left(cosx\right)dx\\ ={\displaystyle \int si{n}^{-1}}\left(sin\left\{\frac{\pi }{2}-x\right\}\right)dx\\ ={\displaystyle \int \{\frac{\pi }{2}-x\}dx}\\ =\frac{\pi }{2}x-\frac{x^2}{2}+\; C\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (a)

Explanation- when a point positive charge brought near an isolated conducting plane, some negative charge develops on the surface of the plane towards the charge and an equal positive charge develops on opposite side of the plane. By process called induction.

$\begin{array}{l}=\frac{cos2x}{co{s}^{2}x+si{n}^{2}x+2sinxcosx}\\ =\frac{cos2x}{1+sin2x}\end{array}$

$\begin{array}{l}={\displaystyle \int \frac{cos2x}{{(cosx+sinx)}^2}}dx\\ ={\displaystyle \int \frac{cos2x}{1+sin2x}}dx\\ \begin{array}{l}Put 1 + sin 2x=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}2 cos 2x dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{cos2x}{{(cosx+sinx)}^2}}dx\\ =\frac{1}{2}{\displaystyle \int \frac{1}{t}}dt=\frac{1}{2}log\left|t\right|+\; C\\ =\frac{1}{2}log\left|1+sin2x\right|+\; C\\ -\frac{1}{2}log\left|{(cosx+sinx)}^2\right|+\; C\\ =log\left|cosx+sinx\right|+C\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (c)

Explanation- As the space is increasing towards right means electric field is decreasing so as force is also decreasing . so force on -p is more than the force on +p 

$\begin{array}{l}\frac{si{n}^{2}x+co{s}^{2}x}{sinxco{s}^{3}x}=\frac{sinx}{co{s}^{3}x}+\frac{1}{sinxcosx}\\ =tanxse{c}^{2}x+\frac{co{s}^{2}x}{\left(\frac{sinxcosx}{co{s}^{2}x}\right)}\\ =tanxse{c}^{2}x+\frac{se{c}^{2}x}{tanx}\\ I={\displaystyle \int \frac{1}{sinxco{s}^{3}x}}dx={\displaystyle \int tan}xse{c}^{2}xdx+{\displaystyle \int \frac{se{c}^{2}x}{tanx}}dx\\ \begin{array}{l}Put tanx=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Se{c}^{2}x dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{1}{sinxco{s}^{3}x}}dx={\displaystyle \int tanxse{c}^{2}xdx+{\displaystyle \int \frac{se{c}^{2}x}{tanx}}dx}\\ ={\displaystyle \int t}dt+{\displaystyle \int \frac{1dt}{t}}\\ =\frac{t^2}{2}+log\left|t\right|+\; C\\ =\frac{1}{2}ta{n}^{2}x+log\left|tanx\right|+\; C\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (b)

Explanation-The charges may be located anywhere inside the surface, if the surface      is so chosen that there are some charges inside and some outside, the electric field on the left side of equation is due to all charges, both inside and outside S.

$\begin{array}{l}=\frac{cos2x+(1-cos2x)}{co{s}^{2}x}\\ =\frac{1}{co{s}^{2}x}=se{c}^{2}x\\ ={\displaystyle \int \frac{cos2x+2si{n}^{2}x}{co{s}^{2}x}}dx\\ \begin{array}{l}={\displaystyle \int se{c}^2}xd x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= tanx+ C\hfill \end{array}\end{array}$

$\begin{array}{l}\frac{si{n}^{3}x+co{s}^{3}x}{si{n}^{2}x\cdot co{s}^{2}x}=\frac{si{n}^{3}x}{si{n}^{2}x\cdot co{s}^{2}x}+\frac{co{s}^{3}x}{si{n}^{2}x.co{s}^{2}x}\\ =\frac{sinx}{co{s}^{2}x}+\frac{cosx}{si{n}^{2}x}\\ =tanxsecx+cotxcosecx.\\ ={\displaystyle \int \frac{si{n}^{3}x+co{s}^{3}x}{si{n}^{2}x\cdot co{s}^{2}x}}dx\\ \begin{array}{l}={\displaystyle \int \left(tanxsecx+cotxcosecx\right)dx}\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= secx-cosecx+ C\hfill \end{array}\end{array}$

$ta{n}^{4}x=ta{n}^{2}xta{n}^{2}x$

$=(se{c}^{2}x-1)ta{n}^{2}x$

$=se{c}^{2}x.ta{n}^{2}x-ta{n}^{2}x$

$=se{c}^{2}x.ta{n}^{2}x-(se{c}^{2}x-1)$

I $\begin{array}{l}\begin{array}{}\end{array}=se{c}^{2}x.ta{n}^{2}x-se{c}^{2}x+1\end{array}$

$\begin{array}{l} I={\displaystyle \int ta{n}^4}xdx={\displaystyle \int se{c}^2}x.ta{n}^{2}xdx-{\displaystyle \int se{c}^2}xdx+{\displaystyle \int 1}.dx\end{array}$

$\begin{array}{l}={\displaystyle \int se{c}^2}x.ta{n}^{2}xdx-tan\pi +x+C-----\left(i\right)\end{array}$

$\begin{array}{l}LetI={\displaystyle \int se{c}^2}x.ta{n}^{2}xdx\end{array}$

$\begin{array}{l}Puttanx=t\end{array}$

$\begin{array}{l}se{c}^{2}xdx=dt\end{array}$

$\begin{array}{l}{I}_1={\displaystyle \int se{c}^2}x.ta{n}^{2}xdx\end{array}$

$\begin{array}{l}={\displaystyle \int t^2}dt\end{array}$

$=\frac{t^3}{3}=\frac{ta{n}^{3}x}{}$

$\frac{3}{}I={\displaystyle \int ta{n}^4}xdx$

$=\frac{1}{3}ta{n}^{3}x-tanx+x+\; C$