Class 12th

Kindly go through the solution

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

Explanation- according to ohms law

I= $\frac{V_{eff}}{R_{eff}+R}$

If voltage and resistance increase

V'= nVeff, R'= nReff

I'= $\frac{\mathrm{n}\mathrm{V}\mathrm{e}\mathrm{f}\mathrm{f}}{\mathrm{n}\mathrm{R}\mathrm{e}\mathrm{f}\mathrm{f}+\mathrm{n}\mathrm{R}}=$ $\frac{V_{eff}}{R_{eff}+R}$ =

Kindly go through the solution

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

Explanation- R= $\rho l$ /A

Resistance of first conductor, RA= $\frac{\rho l}{\pi {(10}^{-3}*0.5)}$ 2

Resistance of second conductor, RB= $\frac{\rho l}{\pi \left. {{(10}^{-6}\right)-(0.5*0.5*{10}^{-6})\}}$

Now ratio $\frac{R_A}{R_B}=\frac{\frac{\rho l}{\pi {(10}^{-3}*0.5)}2}{\frac{\rho l}{\pi \left. {{(10}^{-6}\right)-(0.5*0.5*{10}^{-6})\}}}$ = 3:1

$\frac{x}{{(x-1)}^2(x-2)}=\frac{A}{(x-1)}+\frac{B}{{(x-1)}^2}+\frac{C}{x+2}$

= A(x- 1)(x + 2) + B(x + 2) + C(x- 1)²

= A(x² + 2x-x- 2) + B(x + 2) + C(x² + 1 - 2x)

= A(x² + x- 2) + B(x + 2) + C(x²- 2x + 1)

Comparing the co-efficient we get, A + C = 0 ¾ (1)

A + B - 2C = 1 ¾ (2)

- 2A + 2B + C = 0 ¾ (3)

EQⁿ (3) - 2 ´ eⁿQ. (2),

- 2A + 2B + C (2A + 2B - 4C) = 0 - 2 ´ 1.

=> - 4A + 5C = - 2 ¾ (4)

Eqⁿ (4) + 4 ´ Eqⁿ (1) we get,

- 4A + 5C + 4A + 4C = - 2 + 4 ´ 0

=> 9C = - 2

$\begin{array}{l}\Rightarrow C=-\frac{2}{9}\\ PuttingC=-\frac{2}{9}, in----- \left(1\right)\\ Putting value of A and C in \left(3\right) we get,\\ -2*\frac{2}{9}+2B-\frac{2}{9}=0\\ \Rightarrow 2B=\frac{2}{9}+\frac{4}{9}\\ \Rightarrow 2B=\frac{6}{9}:\frac{2}{3}\\ \Rightarrow B=\frac{1}{3}.\end{array}$

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

Explanation- according to ohms law

I= $\frac{E+E}{R+r1+r2}$

The potential difference across terminal is

V=e-Ir= E- $\frac{2E}{R+r1+r2}$ r1=0

E= $\frac{2Er1}{R+r1+r2}$

1= $\frac{2r1}{R+r1+r2}$

R+r1+r2 = 2r1

R= r1-r2

$\frac{x}{\left(x^2+1\right)(x-1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}$

=> x = (Ax + B)(x- 1) + C(x² + 1)

= Ax²- Ax + Bx- B + Cx² + C.

Comparing the co-efficients we get,

A + C = 0 ---- (1)

- A + B = 1 ---- (2)

- B + C = 0 ---- (3)

Adding (1) and (2) we get,

A + C - A + B = 0 + 1

=> B + C = 1 --- (4)

Adding (4) + (3) we get,

- B + C + B + C = 0 + 1

=> 2C = 1

And C = B from (3)

So, from Eqⁿ (1),

A = - C

$\begin{array}{l}A=-\frac{1}{2}.\\ \therefore {\displaystyle \int \frac{x}{\left(x^2+1\right)(x-1)}}dx={\displaystyle \int \frac{\left(-\frac{1}{2}x+\frac{1}{2}\right)}{x^2+1}}dx+{\displaystyle \int \frac{1}{2}}dx.\\ =\frac{1}{2}{\displaystyle \int \frac{1-x}{x^2+1}}dx+\frac{1}{2}{\displaystyle \int \frac{1}{x-1}}dx.\\ =\frac{1}{2}{\displaystyle \int \frac{1}{x^2+1}}dx-\frac{1}{4}{\displaystyle \int \frac{2xdx}{x^2+1}}+\frac{1}{2}log|x-1|+c\\ =\frac{1}{2}ta{n}^{-1}x-\frac{1}{4}log\left(x^2+1\right)+\frac{1}{2}log|x-1|+c.\end{array}$

$\frac{1-x^2}{x(1-2x)}$ which is not a proper fraction

So, division is missing

$\begin{array}{l}So,\frac{1-x^2}{x(1-2x)}=\frac{1}{2}+\frac{(\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1)}{x(1-2x)}\\ =\frac{1}{2}+\frac{1}{2}\frac{(2-x)}{x(1-2x)}\\ Let\frac{2-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{(1-2x)}\end{array}$

=> 2 -x = A(1 - 2x) + B x.

   So, - 2A + B = - 1

   A = 2.

   So, B = - 1 + 2A = - 1 + 2 ´ 2 = - 1 + 4 = 3

$\begin{array}{l}\therefore {\displaystyle \int \frac{1-x^{2}dx}{x(1-2x)}}={\displaystyle \int \frac{1}{2}}dx+{\displaystyle \int \frac{1}{2}}\frac{(2-x)}{x(1-2x)}dx\\ =\frac{x}{2}+\frac{1}{2}\left\{{\displaystyle \int \frac{A}{x}}dx+{\displaystyle \int \frac{B}{1-2x}}dx\right\}\\ =\frac{x}{2}+\frac{1}{2}\{\left\{\frac{2}{x}dx+{\displaystyle \int \frac{3}{(1-2x)}}dx\right\}\\ =\frac{x}{2}+log\left|x\right|+\frac{3}{2}log\frac{|1-2x|}{(-2)}+c\\ =\frac{x}{2}+log\left|x\right|-\frac{3}{4}log|1-2x|+c\end{array}$

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

Explanation- when all resistance are in parallel

$\frac{1}{R_p}=\frac{1}{R_1}+\dots \dots \dots \dots ..+\frac{1}{R_n}$ by multiplying this equation by R_min we have

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\dots \dots \dots \dots ..+\frac{R_{min}}{R_n}$

But there exist a term in RHS $\frac{R_{min}}{R_{min}}=1$ and other terms are positive so we have

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\dots \dots \dots \dots ..+\frac{R_{min}}{R_n}$ >1

This shows that equivalent resistance is less than maximum resistance available.

But when all resistance are in series

R_s =R₁+R₂………R_n

here must be R_max value in RHS

R_s= R₁+……R_max+….R_n

And R_s> R_max

So equivalent resistance is less than R_max

$\frac{2x}{x^2+3x+2}=\frac{2x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$

2x = A(x + 2) + B(x + 1).

= (A + B) x + (2A + B).

Comparing the coefficients we get,

A + B = 2 ---(1)

2A + B = 0 ---- (2).

So, Eqⁿ (2) - (1),

2A + B - (A + B) = 0 - 2

Þ A = - 2

And B = 2 - A = 2 - (-2) = 2 + 2 = 4

B = 4

$\begin{array}{l}So,\frac{2x}{(x+1)(x+2)}=\frac{-2}{x+1}+\frac{4}{x+2}.\\ \therefore {\displaystyle \int \frac{2xdx}{(x+1)(x+2)}}=-2{\displaystyle \int \frac{dx}{x+1}}+4{\displaystyle \int \frac{dx}{x+2}}\\ =-2log|x+1|+4log|x+2|+c\\ =4log|x+2|-2log|x+1|+c\end{array}$