Class 12th

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

Explanation- applying kirchhoff's junction rule I₁ = I+I₂

Applying kirchhoff's rule in outer loop containing 10V cell

10= IR+10I₁……………. (1)

Applying kirchhoff's rule in outer loop containing 2V cell

2= 5I₂-RI= 5 (I₁-I)-RI

4= 10I₁-10I-RI………… (2)

From 1 and 2

6=3RI+10I

2=I (R+10/3)

V= I (R+R_eff)

After comparing V=2V, R_eff= 10/3 ohm

Since effective internal resistance R_eff of two cells 10/3 ohm, being the parallel combination 5 ohm and 10 ohm . the equivalent circuit is

$\begin{array}{l}I=\int \frac{dx}{x^2+2x+2}=\int \frac{dx}{(x^2+2x+1)+1}\\ =\int \frac{dx}{{(x+1)}^2+1}dx\\ \begin{array}{l}= \left[ta{n}^{–1}\left(x+1\right)\right] +C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Hence, the correct Answer is \left(B\right).\hfill \end{array}\end{array}$

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

Explanation- according to ohm's law V= IR

I= 6/6 = 1A

I= AneV_d  or V_d= i/neA

$\begin{array}{l}Let\left(x+3\right)=A\frac{d}{dx}\left(x^2–2x–5\right)+B\\ \Rightarrow \left(x+3\right)=A\left(2x–2\right)+B \\ Equatingthecoefficientofxandconstanttermonbothside,weget\\ Weget 2A=1 and –2A+B=3 \\ A= \frac{1}{2} B=4 \\ Therefore,\left(x+3\right)=\frac{1}{2}(2x-2)+4-----(1)\\ I=\int \frac{x+3}{x^2-2x-5}dx=\int \frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5}dx\\ =\frac{1}{2}\int \frac{2x-2}{x^2-2x-5}dx+4\int \frac{1}{x^2-2x+5}dx\\ I=\int \frac{x+3}{x^2-2x-5}dx=\frac{1}{2}{I}_1+4I_2\\ I_1=\int \frac{2x-2}{x^2-2x-5}dx\\ \begin{array}{ll}Let{x}^2– 2x– 5 =t\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow \left(2x–2\right)dx=dt\hfill \end{array}\\ I_1=\int \frac{dt}{t}=log\left|t\right|=log\left|x^2-2x-5\right|-----(2)\\ I_2=\int \frac{1}{x^2-2x-5}dx=\int \frac{1}{\left(x^2-2x+1\right)-6}dx=\int \frac{1}{{(x-1)}^2-(\surd 6){}^2}dx\\ =\frac{1}{\mathrm{2\surd 6}}log\left(\frac{x-1-\surd 6}{x-1+\surd 6}\right)-----(3)\\ Substituting\left(2\right)and\left(3\right)in\left(1\right),weget\\ I=\frac{1}{2}log\left|x^2-2x-5\right|+\frac{4}{2}log\left|\frac{x-1-\surd 6}{x-1+\surd 6}\right|+C\end{array}$$\begin{array}{l}\end{array}$

Kindly go through the solution

Kindly go through the solution

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

Explanation – power consumption in a day i.e in 5  = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $*$ I

2000W= 220V $*$ l or l= 9A approx.

R= $\frac{\rho l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{\rho l}{A}$ l²= 1.7 $*$ 10^{-8 $*\frac{10}{\pi *{10}^{-6}}*81$} j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000}*$ 100=0.2%

Power loss in Al wire =1.6 $*$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000}*$ 100= 0.32%