Class 12th

$\begin{array}{l}Here,\\ sinAsinB=\frac{-1}{2}\{cos(A+B)-cos(A-B)\}\\ I={\displaystyle \int sinx}sin2xsin3x\\ ={\displaystyle \int \left[sinx\cdot \frac{1}{2}\{cos(2x-3x)-cos(2x+3x)\}\right]}dx\\ ={\displaystyle \int \{[sin}x\cdot \frac{1}{2}\{cos(-x)-cos5x\}\}dx\\ =\frac{1}{2}{\displaystyle \int sin}xcosx-sinxcos5xdx\\ Now,sin2x=2sinxcosx,\\ =\frac{1}{2}{\displaystyle \int \frac{sin2x}{2}}dx-\frac{1}{2}{\displaystyle \int sin}xcos5xdx\\ =\frac{1}{4}\left[\frac{-cos2x}{2}\right]-\frac{1}{2}{\displaystyle \int \frac{1}{2}}sin(x+5x)+sin(x-5x)dx\\ =\frac{-cos2x}{8}-\frac{1}{4}{\displaystyle \int sin}6x+sin(-4x)dx\\ =\frac{-cos2x}{8}-\frac{1}{4}\left[\frac{-cos6x}{6}+\frac{cos4x}{4}\right]+\; C\\ =\frac{-cos2x}{8}-\frac{1}{8}\left[\frac{-cos6x}{3}+\frac{cos4x}{2}\right]+\; C\\ =\frac{-cos2x}{8}+\frac{cos6x}{24}-\frac{cos4x}{16}+\; C\\ =\frac{1}{4}\left[\frac{1}{6}cos6x-\frac{cos4x}{4}-\frac{cos2x}{2}\right]+\; C.\end{array}$

$\begin{array}{l}I={\displaystyle \int si{n}^{3}x}co{s}^{3}xdx\\ ={\displaystyle \int co{s}^3}xsi{n}^{2}xsinx.dx\\ ={\displaystyle \int co{s}^3}x(1-co{s}^{2}x).sinxdx\\ Putcosx=t\\ \Rightarrow -sin.x.dx=dt\\ I=-{\displaystyle \int t^3}(1-t^2)dt\end{array}$

$\begin{array}{l}=-{\displaystyle \int \left(t^3-t^5\right)}dt=-\left\{\frac{t^4}{4}-\frac{t^6}{6}\right\}+\; C\\ =-\left\{\frac{co{s}^{4}x}{4}-\frac{co{s}^{6}x}{6}\right\}+\; C\\ =\frac{-co{s}^{4}x}{4}+\frac{co{s}^{6}x}{6}+\; C\\ =\frac{1}{6}co{s}^{6}x-\frac{1}{4}co{s}^{4}x+\; C\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- (i)The cesium atoms, are situated at the corners of a cube and Cl atom is situated at the centre of the cube. From the given figure, we can analyse that the chlorine atom is at equal distance from all the eight comers of cube where cesium atoms are placed. Thus, due to symmetry the electric field due to all Cs atoms, on Cl atom will cancel out. Hence net electric field at the centre of cube is zero.

(ii) we know, f=qE  E= Kq/r²= Ke/r²

F= e (E)=e (ke/r²)=ke²/r²

Distance= $\sqrt[]{r^2+r^2+r^2}$ = $\sqrt[]{{0.2}^2+{0.2}^2+{0.2}^2}*$ 10^-9m

F=8.99 $*$ 10⁹ (1.6 $*$ 10^-16)²/ (0.346 $*$ 10^-9)²=1.92 $*$ 10^-9N

$\begin{array}{l}I={\displaystyle \int si{n}^3\left(2x+1\right)dx}\\ ={\displaystyle \int si{n}^2\left(2x+1\right).sin}\left(2x+1\right)dx\\ ={\displaystyle \int \left\{1-co{s}^2\left(2x+1\right)\right\}}sin\left(2x+1\right)dx\\ Puttingcos\left(2x+1\right)=t\\ \Rightarrow -2sin\left(2x+1\right)dx=dt\\ \Rightarrow sin\left(2x+1\right)dx=\frac{-dt}{2}\\ I=\frac{-1}{2}{\displaystyle \int \left(1-t^2\right)}dt\\ =\frac{-1}{2}\left\{t-\frac{t^3}{3}\right\}+\; C\\ =\frac{-1}{2}\left\{cos(2x+1)-\frac{co{s}^3(2x+1){}^3}{3}\right\}+\; C\\ =\frac{-cos(2x+1)}{2}+\frac{co{s}^3(2x+1){}^6}{6}+\; C\\ =\frac{-1}{2}cos(2x+1)+\frac{1}{6}co{s}^3(2x+1)+\; C\end{array}$

$\begin{array}{l}Here,cosAcosB=\frac{1}{2}\{cos(A+B)+cos(A-B)\}\\ I={\displaystyle \int cos2x\left(cos4xcos6x\right)dx}\\ ={\displaystyle \int cos}2x\left[\frac{1}{2}cos(4x+6x)+cos(4x-6x)\right]dx\\ ={\displaystyle \int cos2x\left[\frac{1}{2}(cos10x+cos(-2x))\right]dx}\\ =\frac{1}{2}{\displaystyle \int cos}2xcos10x+cos2xcos(-2x)dx\\ =\frac{1}{2}{\displaystyle \int cos}2xcos10x+cos2xdx\left[\therefore cos\left(-x\right)=cosx\right]\\ =\frac{1}{2}\left[\frac{1}{2}\{cos2x+10x+cos2x-10\}+\left\{\frac{1+cos4x}{2}\right\}\right]dx\\ =\frac{1}{4}{\displaystyle \int cos}12x+cos8x+1+cos4x\cdot dx\\ =\frac{1}{4}[\frac{sin12x}{12}+\frac{sin8x}{8}+x+\frac{cos4x}{4x}]+\; C\end{array}$

Explanation- (i) F= $\frac{Kq2}{r2}$ = 9 $*$ 10^{9 $*$} (34.8 $*$ 10³)²/ (10^-2)²=1.09 $*$ 10²³N

                      (ii) F= $\frac{Kq2}{r2}$ = 9 $*$ 10^{9 $*$} (34.8 $*$ 10³)²/ (100)²=1.09 $*$ 10¹⁵N

                      (iii) F= $\frac{Kq2}{r2}$ = 9 $*$ 10⁹ $*$ (34.8 $*$ 10³)²/ (10⁶)²=1.09 $*$ 10⁷N

$\begin{array}{l}Here,\\ sinAcosB=\frac{1}{2}\left\{sin(A+B)+sin(A-B)\right\}\\ sin3xcos4x=\frac{1}{2}(sin(3x+4x)+sin(3x-4x))\\ Then,\\ {\displaystyle \int sin}3xcos4xdx={\displaystyle \int \frac{1}{2}}[sin(3x+4x)+sin\left(3x-4x\right)dx\\ =\frac{1}{2}{\displaystyle \int sin}(3x+4x)+sin\left(3x-4x\right)dx\\ =\frac{1}{2}{\displaystyle \int sin}7x+sin(-x)dx\\ =\frac{1}{2}\left[-\frac{cos7x}{7}+cosx\right]+c\\ =\frac{-cos7x}{14}+\frac{cosx}{2}+c\end{array}$

$\begin{array}{l}Here,\\ si{n}^2\left(2x+5\right)=\frac{1-cos2(2x+5)}{2}\\ =\frac{1-cos(4x+10)}{2}\\ Then,\\ ={\displaystyle \int si{n}^2}(2x+5)dx={\displaystyle \int \frac{1-cos(4x+10)}{2}}dx\\ =\frac{1}{2}{\displaystyle \int 1}\cdot dx-\frac{1}{2}{\displaystyle \int cos}(4x+10)dx\\ =\frac{1}{2}\cdot x-\frac{1}{2}\frac{sin(4x+10)}{4}+\; C\\ =\frac{x}{2}-\frac{1}{8}sin(4x+10)+\; C\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- 1 molar mass of Al has N_A (Avogadro number)= 6.023 $*$ 10²³

M= $\frac{NA}{M}*$ m= 6.023 $*$ 10²³/27 $*$ 0.75 = 1.6 $*$ 10²²

Magnitude of positive and negative charges in one paisa coin = Nze

As aluminium has 13 electron and 13 protons so

=  1.6 $*$ 10²² $*$ 13 $*$ 1.6 $*$ 10^-19=33.28 $*$ 10³C.