Class 12th

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

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

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

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

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

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

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

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

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

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

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (d) 

Explanation- Total flux inside a conducting sphere is depend upon the algeabric sum of total charge inside that surface, which is same for all fig. As it does not depend upon shape of a body

$\begin{array}{l}\begin{array}{l}ta{n}^{3}2xsec 2x= ta{n}^{2}2xtan 2xsec 2x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l} \{se{c}^2(2x)-1\} tan 2xsec 2x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= se{c}^{2}2xtan 2xsec 2x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}-tan 2xsec 2x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}I ={\displaystyle \int ta{n}^3}2x.sec 2x dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int ta{n}^2}2xtan 2xsec 2x dx-{\displaystyle \int tan}2x.sec 2x dx\hfill \end{array}\\ ={\displaystyle \int ta{n}^2}2xtan2xsec2xdx-\frac{sec2x}{2}+\\ \begin{array}{l}Put sec 2x=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow 2sec 2xtan 2x dx=dt\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}I={{\displaystyle \int tan}}^{3}2x.sec 2x dx\hfill \end{array}\\ =\frac{1}{2}{\displaystyle \int t^2}dt-\frac{sec2x}{2}+\; C\\ =\frac{t^3}{6}-\frac{sec2x}{2}+\; C\\ =\frac{{(sec2x)}^3}{6}-\frac{sec2x}{2}+\; C\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- when a positive point charge is brought near an isolated conducting sphere without touching the sphere, then the free electron in the sphere are attracted towards the positive charge. This leaves an excess positive charge on right side of sphere . And also by induction the negative charge is setup across left side. So field lines first goes from positive charge (+q) to negative charge of sphere then from positive charge of sphere to negative charge anywhere in the universe.

$\begin{array}{l}\frac{cosx-sinx}{1+sin2x}=\frac{cosx-sinx}{\left(si{n}^{2}x+co{s}^{2}x\right)+2sinxcosx}\\ =\frac{cosx-sinx}{{(sinx+cosx)}^2}\\ \left[?si{n}^2+co{s}^2=1\&sin2x=2sinxcosx\right]\\ I={\displaystyle \int \frac{cosx-sinx}{1+sin2x}}dx={\displaystyle \int \frac{cosx-sinx}{{(sinx+cosx)}^2}dx}\\ \begin{array}{l}Put sinx+ cosx=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}(cosx-sinx)dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{dt}{t^2}}={\displaystyle \int t^{-2}}dt\\ =-t^{-1}+=-\frac{1}{t}+\; c\\ =\frac{1}{-sinx+cosx}+\; c\end{array}$