Class 12th

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

Ans: (i) Cell 'B' will act as an electrolytic cell because to less value of E_cell The reactions occurring in the cell are as follows;

At anode: Zn²⁺ + 2e⁻ → Zn

At cathode: Cu (s) → Cu²⁺ + 2e⁻

(ii) Cell 'B' has a higher emf so it acts as a galvanic cell. The reactions are as follows;

At anode: Zn → Zn²⁺ + 2e−

At cathode: Cu²⁺ + 2e− → Cu

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

Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur. However, the average rate cannot be used to predict the rate of reaction at a particular instance as it would be constant for the time interval for which it is calculated. 

So, to express the rate at a particular moment of time we determine the instantaneous rate. Itis obtained when we consider the average rate at the smallest time interval say dt (i.e. when? t approaches zero). 

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

Ans:

As  illustrated in the graph, as the temperature rises, the peak pushes ahead, increasing probable kinetic energy while decreasing the number of molecules utilizing it, resulting in a faster rate of reaction.

 

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

The reaction rate in collision theory is determined by two factors: energy and orientation factor. Certain reactions can be highly exothermic and energetically favoured, meaning the reactants have enough activation energy to collide effectively. However, they do not proceed at a fixed temperature, which is why the reactant molecules are not orientated properly during collisions, resulting in atoms of reactant molecules combining to produce products that do not face each other.

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationis\\ \frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{{\left(1+x^2\right)}^2}\\ \text{since\hspace{0.17em}}itislineardifferentialequationwhereP=\frac{2xy}{1+x^2}andQ=\frac{1}{{\left(1+x^2\right)}^2}\\ \therefore I.F.=e^{{\displaystyle \int P.dx}}=e^{{\displaystyle \int \frac{2xy}{1+x^2}.dx}}=e^{log\left(1+x^2\right)}=\left(1+x^2\right)\\ So,thesolutionis\\ y*I.F.={\displaystyle \int Q*}I.F.dx+c\\ \Rightarrow y\left(1+x^2\right)={\displaystyle \int \frac{1}{{\left(1+x^2\right)}^2}*\left(1+x^2\right)}dx+c\\ \Rightarrow y\left(1+x^2\right)={\displaystyle \int \frac{1}{\left(1+x^2\right)}.}dx+c\\ \Rightarrow y\left(1+x^2\right)=ta{n}^{-1}x+c\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationis\\ \frac{dy}{dx}=e^{x-y}+x^2{e}^{-y}\\ \Rightarrow \frac{dy}{dx}=e^x.e^{-y}+x^2{e}^{-y}\Rightarrow \frac{dy}{dx}=e^{-y}\left(e^x+x^2\right)\\ \Rightarrow \frac{dy}{e^{-y}}=\left(e^x+x^2\right)dx\Rightarrow e^y.dy=\left(e^x+x^2\right)dx\\ Integratingbothsides,wehave\\ \Rightarrow {\displaystyle \int e^y.dy}={\displaystyle \int \left(e^x+x^2\right)}dx\\ \Rightarrow e^y=e^x+\frac{x^3}{3}+c\\ \Rightarrow e^y-e^x=\frac{x^3}{3}+c\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationis\\ \left(e^x+1\right)ydy=\left(y+1\right)e^{x}dx\\ \Rightarrow \frac{y}{y+1}dy=\frac{e^x}{e^x+1}dx\\ Integratingbothsides,wehave\\ \Rightarrow {\displaystyle \int \frac{y}{y+1}dy}={\displaystyle \int \frac{e^x}{e^x+1}}dx\\ \Rightarrow {\displaystyle \int \frac{y+1-1}{y+1}dy}={\displaystyle \int \frac{e^x}{e^x+1}}dx\\ \Rightarrow {\displaystyle \int 1.dy}+{\displaystyle \int \frac{1}{y+1}dy}={\displaystyle \int \frac{e^x}{e^x+1}}dx\\ \Rightarrow y-log\left|y+1\right|=log\left|e^x+1\right|+logk\\ \Rightarrow y=log\left|y+1\right|+log\left|e^x+1\right|+logk\\ \Rightarrow y=log\left|k\left(y+1\right)\left(e^x+1\right)\right|\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationis\frac{dy}{dx}+\frac{y}{x}=sinx\\ \text{\hspace{0.17em}Since}itislineardifferentialequationwhereP=\frac{1}{x}andQ=sinx\\ \therefore I.F.=e^{{\displaystyle \int P.dx}}=e^{{\displaystyle \int \frac{1}{x}.dx}}=e^{logx}=x\\ So,thesolutionis\\ y*I.F.={\displaystyle \int Q*}I.F.dx+c\\ \Rightarrow y*x={\displaystyle \int sinx.x}dx+c\Rightarrow y*x={\displaystyle \int xsinx.}dx+c\\ \Rightarrow yx=x.{\displaystyle \int sinxdx}-{\displaystyle \int \left(D\left(x\right){\displaystyle \int sinxdx}\right)}dx+c\\ \Rightarrow yx=x\left(-cosx\right)-{\displaystyle \int -cosx}dx+c\\ \Rightarrow yx=-xcosx+{\displaystyle \int cosx}dx+c\\ \Rightarrow yx=-xcosx+sinx+c\\ \Rightarrow yx+xcosx=sinx+c\\ \Rightarrow x\left(y+cosx\right)=sinx+c\\ Hence,thecorrectoptionis\left(a\right).\end{array}$