Class 12th

108. Kindly Consider the following

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

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(a) Consider circuit in Fig 3.10. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?

(b) Electrons give up energy at the rate of RI² per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = 10²⁹/m3, length of circuit = 10 cm, cross-section = A = (1mm)²

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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

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Kindly Consider the following

107. $\frac{x + 3}{x^{2} - 2 x - 5}$

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$\begin{array}{l} L e t (x + 3) = A \frac{d}{d x} (x^{2} - 2 x - 5) + B \\ \Rightarrow (x + 3) = A (2 x - 2) + B \\ E q u a t i n g t h e c o e f f i c i e n t o f x a n d c o n s t a n t t e r m o n b o t h s i d e, w e g e t \\ W e g e t 2 A = 1 a n d - 2 A + B = 3 \\ A = \frac{1}{2} B = 4 \\ T h e r e f o r e, (x + 3) = \frac{1}{2} (2 x - 2) + 4 - - - - - (1) \\ I = \int \frac{x + 3}{x^{2} - 2 x - 5} d x = \int \frac{\frac{1}{2} (2 x - 2) + 4}{x^{2} - 2 x - 5} d x \\ = \frac{1}{2} \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x + 4 \int \frac{1}{x^{2} - 2 x + 5} d x \\ I = \int \frac{x + 3}{x^{2} - 2 x - 5} d x = \frac{1}{2} I_{1} + 4 I_{2} \\ I_{1} = \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x \\ \begin{array}{l} L e t x^{2} - 2 x - 5 = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow (2 x - 2) d x = d t \end{array} \\ I_{1} = \int \frac{d t}{t} = l o g | t | = l o g | x^{2} - 2 x - 5 | - - - - - (2) \\ I_{2} = \int \frac{1}{x^{2} - 2 x - 5} d x = \int \frac{1}{(x^{2} - 2 x + 1) - 6} d x = \int \frac{1}{{(x - 1)}^{2} - (\sqrt6)^{2}} d x \\ = \frac{1}{2√6} l o g (\frac{x - 1 -\sqrt6}{x - 1 +\sqrt6}) - - - - - (3) \\ S u b s t i t u t i n g (2) a n d (3) i n (1), w e g e t \\ I = \frac{1}{2} l o g | x^{2} - 2 x - 5 | + \frac{4}{2} l o g | \frac{x - 1 -\sqrt6}{x - 1 +\sqrt6} | + C \end{array}$ $\begin{array}{l} \end{array}$

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106. Kindly Consider the following

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

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Kindly go through the solution

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In an experiment with a potentiometer, VB = 10 V. R is adjusted to be 50 Ω (figure). A student wanting to measure voltage E₁of a battery (approx. 8 V) finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the-last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

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

Explanation – let R' be the resistance of potentiometer wire.

Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'

Effective voltage across potentiometer = 10V

The current through main circuit I= $\frac{V}{50 + R}$ = $\frac{10}{50 + R}$

Potential difference across wire of potentiometer

IR'= $\frac{10 R'}{50 + R}$

Since with 50 ohm resistor, null point is not obtained it is possible when

$\frac{10 * R'}{50 + R} < 8$

10R' $< 400 + 8 R'$

2R'<400 or R'<200 ohm

Similarly with 10 ohm resistor, null point is obtained its is only possible when

$\frac{10 * R'}{50 + R} < 8$

2R'>40

R'>40

$\frac{10 * \frac{3}{4} R'}{10 + R'} < 8$

7.5R'<80+8R'

R'>160

160

Any R' between 160 ohm and 200 ohm will achieve.

Since the null point on the last 4^th segment of

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

Explanation – let R' be the resistance of potentiometer wire.

Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'

Effective voltage across potentiometer = 10V

The current through main circuit I= $\frac{V}{50 + R}$ = $\frac{10}{50 + R}$

Potential difference across wire of potentiometer

IR'= $\frac{10 R'}{50 + R}$

Since with 50 ohm resistor, null point is not obtained it is possible when

$\frac{10 * R'}{50 + R} < 8$

10R' $< 400 + 8 R'$

2R'<400 or R'<200 ohm

Similarly with 10 ohm resistor, null point is obtained its is only possible when

$\frac{10 * R'}{50 + R} < 8$

2R'>40

R'>40

$\frac{10 * \frac{3}{4} R'}{10 + R'} < 8$

7.5R'<80+8R'

R'>160

160

Any R' between 160 ohm and 200 ohm will achieve.

Since the null point on the last 4^th segment of the potentiometer, therefore potential drop across 400cm of wire >8V

Potential gradient K $* 400 c m > 8 V$

K*4m>8V

K>2V/m

Similarly potential drop across 300 cm wire <8V

k*3m<8V, K<2*2/3V/m

2*2/3V/m>k>2V/m

less

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105. Kindly Consider the following

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

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Kindly go through the solution

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A room AC runs for 5 a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?

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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= $\frac{ρ l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{ρ l}{A}$ l²= 1.7 $*$ 10^{-8
$* \frac{10}{π * 10^{- 6}} * 81$}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%

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104. Kindly Consider the following

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Class 12th physics ncert exemplar solution class 12th chapter one

Contributor-Level 10

$\begin{array}{l} \frac{6 x + 7}{x^{2} - 9 x + 20} \\ L e t 6 x + 7 = A \frac{d}{d x} (x^{2} - 9 x + 20) + B \\ \Rightarrow 6 x + 7 = A (2 x - 9) + B \\ E q u a t i n g t h e c o e f f i c i e n t o f x a n d c o n s t a n t t e r m, w e h a v e \\ \begin{array}{l} 2 A = 6 a n d - 9 A + B = 7 \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow A = 3 \Rightarrow B = 34 \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} ∴ 6 x + 7 = 3 (2 x - 9) + 34 \end{array} \end{array}$

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Contributor-Level 10

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= $\frac{ρ l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{ρ l}{A}$ l²= 1.7 $*$ 10^{-8
$* \frac{10}{π * 10^{- 6}} * 81$}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%

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Kindly Consider the following

103. $\frac{5 x - 2}{1 + 2 x + 3 x^{2}}$

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