Class 12th

Explanation- A= 10^-4m²

So d= 10^-3 and i= 100^-4A

I= 100W/m²

$\mathrm{?}=600\mathrm{n}\mathrm{m}=600*{10}^{-9}$

${\mathrm{?}}_{\mathrm{N}\mathrm{a}}=0.97\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}=\mathrm{A}*\mathrm{d}$ 10^-4(10^-3)=10^-7m³

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}23\mathrm{k}\mathrm{g}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{m}$

So volume Na atoms=23/0.97m³

Volume occupied by one Na atom= $\frac{23}{0.97*6*{10}^{26}}=3.95*{10}^{-26}{\mathrm{m}}^3$

Number of Na atoms in target $\frac{{10}^{-7}}{3.95*{10}^{-26}}=2.53*{10}^{18}$

So energy falling per sec= $\frac{\mathrm{n}\mathrm{h}\mathrm{c}}{\mathrm{?}}=\mathrm{I}\mathrm{A}$

So n= $\frac{\mathrm{I}\mathrm{A}\mathrm{?}}{\mathrm{h}\mathrm{c}}$ = $\frac{100*{10}^{-4}*660*{10}^{-9}}{6.62*{10}^{-34}*3*{10}^8}=3.3*{10}^{16}$

N=P $*\mathrm{n}*\mathrm{N}\mathrm{a}=\mathrm{P}*3.3*{10}^{16}*2.53*{10}^{18}$

I = 100 $*{10}^{-6}={10}^{-4}$ A

I=Ne= $\mathrm{P}*3.3*{10}^{16}*2.53*{10}^{18}$ ( ${10}^{-4}$ A)

P= 7.48 $*{10}^{-21}$ it is less than 1.

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol: 

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthaty=mx+cwilltouchtheellipse\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1if{c}^2=a^2{m}^2+b^2\\ Hereequationofstraightlineisxcos\alpha +ysin\alpha =pandthatofellipseis\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ xcos\alpha +ysin\alpha =p\\ \Rightarrow ysin\alpha =-xcos\alpha +p\\ \Rightarrow y=-x\frac{cos\alpha }{sin\alpha }+\frac{p}{sin\alpha }\Rightarrow y=-xcot\alpha +\frac{p}{sin\alpha }\\ Compaingwithy=mx+c,weget\\ m=-cot\alpha andc=\frac{p}{sin\alpha }\\ So,accordingtothecondition,weget{c}^2=a^2{m}^2+b^2\\ \frac{p^2}{si{n}^2\alpha }=a^2{\left(-cot\alpha \right)}^2+b^2\\ \Rightarrow \frac{p^2}{si{n}^2\alpha }=a^2\frac{co{s}^2\alpha }{si{n}^2\alpha }+b^2\Rightarrow p^2=a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha \\ Hence,a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha =p^{2}Henceproved.\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetusconsiderthatthecompanyincreasestheannualsubscriptionbyRs.x\\ So,xisthenumberofsubscriberswhodiscontinuetheservices.\\ \therefore Totalrevenue,R\left(x\right)=\left(500-x\right)\left(300+x\right)\\ =150000+500x-300x-x^2\\ =-x^2+200x+150000\\ Differentiatingbothsidesw.r.t.x,weget{R}^{\text{'}}\left(x\right)=-2x+200\\ Forlocalmaximaandlocalminima,R^{\text{'}}\left(x\right)=0\\ -2x+200=0\Rightarrow x=100\\ R^{\text{'}\text{'}}\left(x\right)=-2<0Maxima\\ So,R\left(x\right)ismaximumatx=100\\ Hence,inordertogetmaximumprofit,thecompanyshouldincreaseitsannualsubscription\\ byRs.100.\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=x^5-5x^4+5x^3-1\\ f^{\text{'}}\left(x\right)=5x^4-20x^3+15x^2\\ Forlocalmaximaandlocalminima,f^{\text{'}}\left(x\right)=0\\ \therefore 5x^4-20x^3+15x^2=0\Rightarrow 5x^2\left(x^2-4x+3\right)=0\\ \Rightarrow 5x^2\left(x^2-3x-x+3\right)=0\Rightarrow x^2\left(x-3\right)\left(x-1\right)=0\\ \therefore x=0,x=1andx=3\\ Now{f}^{\text{'}\text{'}}\left(x\right)=20x^3-60x^2+30x\\ \Rightarrow f^{\text{'}\text{'}}{\left(x\right)}_{atx=0}=20{\left(0\right)}^3-60{\left(0\right)}^2+30\left(0\right)=0whichisneitherMaximanorMinima.\\ \therefore f\left(x\right)hasthepointofinflextionatx=0\\ \Rightarrow f^{\text{'}\text{'}}{\left(x\right)}_{atx=1}=20{\left(1\right)}^3-60{\left(1\right)}^2+30\left(1\right)\\ =20-60+30=-10<0Maxima\\ \Rightarrow f^{\text{'}\text{'}}{\left(x\right)}_{atx=3}=20{\left(3\right)}^3-60{\left(3\right)}^2+30\left(3\right)\\ =540-540+90=90>0Minima\\ Themaximumvalueofthefunctionatx=1\\ f\left(x\right)={\left(1\right)}^5-5{\left(1\right)}^4+5{\left(1\right)}^3-1\\ =1-5+5-1=0\\ Theminimumvalueofthefunctionatx=3\\ f\left(x\right)={\left(3\right)}^5-5{\left(3\right)}^4+5{\left(3\right)}^3-1\\ \text{\&thins}\end{array}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

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

Sol.

$\begin{array}{l}Letxbethelengthofthecube\\ \therefore VolumeofthecubeV=x^3\dots \left(i\right)\\ Giventhat\frac{dV}{dt}=K\\ Differentiatingeq\left(i\right)w.r.t.t,weget\\ \frac{dV}{dt}=3x^2.\frac{dx}{dt}=K\left(constant\right)\\ \therefore \frac{dx}{dt}=\frac{K}{3x^2}\\ Nowsurface\text{?}\text{?}areaofthecube,S=6x^2\\ Differentiatingbothsidesw.r.t.t,weget\\ \frac{dS}{dt}=6.2.x.\frac{dx}{dt}=12x.\frac{K}{3x^2}\\ \Rightarrow \frac{dS}{dt}=\frac{4K}{x}\Rightarrow \frac{dS}{dt}\propto \frac{1}{x}\left(4K=constant\right)\\ Hence,surface\text{?}\text{?}areaofthecubevariesinverselyasthelengthoftheside.\end{array}$

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

Sol.

$\begin{array}{l}GiventhatL=200{\left(10-t\right)}^2\\ whereLrepresentsthenumberoflitresofwaterinthepool.\\ Differentiatingbothsidesw.r.t.,t,weget\\ \frac{dL}{dt}=200*2\left(10-t\right)\left(-1\right)=-400\left(10-t\right)\\ Buttherateatwhichthewaterisrunningout\\ =-\frac{dL}{dt}=400\left(10-t\right)\dots \left(i\right)\\ Rateatwhichthewaterisrunningafter5seconds\\ =400\left(10-5\right)=2000L/s\left(finalrate\right)\\ Forinitialrateputt=0\\ =400\left(10-0\right)=4000L/s\\ Theaveragerateatwhichthewaterisrunningout\\ =\frac{Initialrate+Finalrate}{2}=\frac{4000+2000}{2}=\frac{6000}{2}=3000L/s\\ Hence,therequiredrate=3000L/s.\end{array}$