Class 12th

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- according to the relation $ρ$ = $\frac{m}{n e 2 t}$ also T $\propto \frac{1}{t}$

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Kindly Consider the following

134. $x s i n x$

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

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$\begin{array}{l} \int x s i n x d x = x \int s i n d x - \int \frac{d}{d x} x \int s i n x d x d x \\ = x (- c o s x) - \int (- c o s x) d x \end{array}$

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Consider a simple circuit shown in figure stands for a variable resistance R'. R' can vary from R₀to infinity, r is internal resistance of the battery (r <(a) Potential drop across AB is nearly constant as R' is varied.
(b) Current through R' is nearly a constant as R' is varied.
(c) Current I depends sensitively on R'
(d) I ≥ V/r + R always;

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, d)

Explanation- potential drop across AB is independent to R' also when we decrease the value of R the current will also very large. According to the relation I=e/R+r

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Kindly Consider the following

133. $\frac{1}{x (x^{2} + 1)} e q u a l s$

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

Contributor-Level 10

$\begin{array}{l} \frac{1}{x (x^{2} + 1)} = \frac{x}{x^{2} (x^{2} + 1)} \\ P u t t i n g, x^{2} = t \Rightarrow 2 x d x = d t \end{array}$

$\begin{array}{l} \int \frac{d x}{x (x^{2} + 1)} = \int \frac{x d x}{x^{2} (x^{2} + 1)} = \frac{1}{2} \int \frac{d t}{t (t + 1)} \\ = \frac{1}{2} \int \frac{(t + 1) - t}{t (t + 1)} d t \\ = \frac{1}{2} {\int \frac{d t}{t} - \int \frac{d t}{t + 1}} \\ = \frac{1}{2} [l o g | t | - l o g | t + 1 |] + C \\ = \frac{1}{2} l o g | x^{2} | - \frac{1}{2} l o g | x^{2} + 1 | + C \\ = \frac{2}{2} l o g | x | - \frac{1}{2} l o g | x^{2} + 1 | + c \\ = l o g | x | - \frac{1}{2} l o g | x^{2} + 1 | + c \\ S o, o p t i o n (A) i s c o r r e c t . \end{array}$

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Kirchhoff's junction rule is a reflection of
(a) Conservation of current density vector.
(b) Conservation of charge.
(c) The fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.
(d) The fact that there is no accumulation of charges at a junction.

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Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation-Algebraic sum of the currents flowing towards any point in an electric network is zero. It also tell us about the conservation of charge.

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Kindly Consider the following

132. $\frac{x}{(x - 1) (x - 2)} e q u a l s$

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

Contributor-Level 10

Kindly go through the solution

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Which of the following characteristics of electrons determines the current in a conductor?
(a) Drift velocity alone
(b) Thermal velocity alone
(c) Both drift velocity and thermal velocity
(d) Neither drift nor thermal velocity

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- I=AneV_d , current is directly proportional to drift velocity.

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Kindly Consider the following

131. $\frac{1}{e^{x} - 1}$

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Contributor-Level 10

Putting e^x = t so that e^xdx = dt=>dx = $\frac{d t}{e^{x}} = \frac{d t}{t}$

$\begin{array}{l} ∴ \int \frac{d x}{e^{x} - 1} = \int \frac{d t}{(t - 1) * t} = \int \frac{d t}{t (t - 1)} \\ = \int \frac{t - (t - 1)}{t (t - 1)} d t \\ = \int \frac{d t}{t - 1} - \int \frac{d t}{t} \\ = l o g | t - 1 | - l o g | t | + C \\ = l o g | e^{n} - 1 | - l o g e^{x} +C . \end{array}$

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Kindly Consider the following

130. $\frac{1}{x (x^{4} - 1)} = \frac{x^{3}}{x^{4} (x^{4} - 1)}$

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

Contributor-Level 10

Putting x⁴ = tÞ 4x³dx = dt

$\begin{array}{l} ∴ \int \frac{1}{x (x^{4} - 1)} d x = \frac{1}{4} \int \frac{d t}{t (t - 1)} = \frac{1}{4} \int \frac{t - (t - 1)}{t (t - 1)} d t \\ = \frac{1}{4} {\int \frac{d t}{t - 1} - \int \frac{d t}{t}} \\ = \frac{1}{4} {l o g | t - 1 | - l o g t} + C \end{array}$

$\begin{array}{l} = \frac{1}{4} l o g | \frac{t - 1}{t} | + C \\ = \frac{1}{4} l o g | \frac{x^{4} - 1}{x^{4}} | + C \end{array}$

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A metal rod of length 10 cm and a rectangular cross-section of 1 cm x 1/2 cm is connected to a battery across opposite faces. The resistance will be
(a) Maximum when the battery is connected across 1 cm x 1/2 cm faces
(b) Maximum when the battery is connected across 10 cm x 1 cm faces
(c) Maximum when the battery is connected across 10 cm x 1/2 cm faces
(d) Same irrespective of the three faces

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