Class 12th

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, c)

Explanation- In case of meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC = (100 – l) so that Q/P= (100-l)/l. Also P/Q=R/S=>S= (100-l)/l R
When there is no deflection in galvanometer there is no current across the galvanometer, then points B and D are at same potential. That point at which galvanometer shows no deflection is called null point, then potential at B and neutral point D are same. When the jockey contacts a point on the meter wire to

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139. $x^{2} \cdot l o g x$

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$\begin{array}{l} \int x^{2} \cdot l o g x \cdot d x = l o g x \cdot \int x^{2} d x - \int \frac{d}{d x} l o g x \cdot \int x^{2} d x d x \\ = l o g x \cdot \frac{x^{3}}{3} - \int \frac{1}{x} \cdot \frac{x^{3}}{3} d x \\ = \frac{x^{3}}{3} \cdot l o g x - \int \frac{x^{2}}{3} d x \\ = \frac{x^{3}}{3} \cdot l o g x - \frac{1}{3} * \frac{x^{3}}{3} + c \\ = \frac{x^{3}}{3} l o g x - \frac{x^{3}}{9} + c . \end{array}$

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138. $x l o g 2 x$

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$\begin{array}{l} \int x l o g 2 x d x = l o g 2 x \cdot \int x d x - \int \frac{d}{d x} l o g 2 x \int x d x d x \\ = \frac{x^{2}}{2} \cdot l o g 2 x - \int \frac{1}{2 x} * \frac{d (2 x)}{d x} * \frac{x^{2}}{2} d x + C \\ = \frac{x^{2}}{2} \cdot l o g 2 x - \int \frac{1}{2 x} * 2 * \frac{x^{2}}{2} 1^{d x} + C \\ = \frac{x^{2}}{2} l o g 2 x - \frac{1}{2} \int x d x + c \\ = \frac{x^{2}}{2} l o g 2 x - \frac{1}{2} \cdot \frac{x^{2}}{2} + c \\ = \frac{x^{2}}{2} l o g 2 x - \frac{x^{2}}{4} + c \end{array}$

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Kindly Consider the following

137. $x l o g x$

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

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$\begin{array}{l} \int x l o g x d x = l o g x \int x d x - \int \frac{d}{d x} l o g x \cdot \int x d x d x \\ = l o g x * \frac{x^{2}}{2} - \int \frac{1}{x} * \frac{x^{2}}{2} d x \\ = \frac{x^{2}}{2} \cdot l o g x - \frac{1}{2} \int x d x \\ = \frac{x^{2}}{2} l o g x - \frac{1}{2} * \frac{x^{2}}{2} + c \\ = \frac{x^{2}}{2} l o g x - \frac{x^{2}}{4} + c . \end{array}$

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The measurement of an unknown resistance R is to be carried out using Wheatstone bridge as given in the figure. Two students perform an experiment in two ways. The first student takes R₂= 10 Ω and R₁ = 5 Ω. The other student takes R₂ = 1000 Ω, and R₁ = 500 Ω. In the standard arm, both take R₃ = 5 Ω.
Both find R = R₂/R₁, R₃ = 100 Ω within errors.
(a) The errors of measurement of the two students are the same
(b) Errors of measurement do depend on the accuracy with which R₂ and R₁ can be measured
(c) If the student uses large values of R₂ and R₁, the currents through the arms will be feeble

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Answer – (b, c)

Explanation – According to the relation that p/q=r/s if galvanometer shows no deflection. In the above equation $\frac{R 1}{R 2} = \frac{R 3}{R 4}$ the value of R4 depends upon R1 and R2 . if the value of R1 and R2 will be feeble the value of R4 would be affected.

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136. $x^{2} e^{x}$

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$\begin{array}{l} \int x^{2} e^{x} d x = x^{2} \int e^{x} d x - \int \frac{d x^{2}}{d x} \int e^{x} d x d x \\ = x^{2} \cdot e^{x} - \int 2 x e^{x} d x \\ = x^{2} e^{x} - 2 [x \int e^{x} d x - \int \frac{d x}{d x} \int e^{x} d x d x] \\ = x^{2} e^{x} - 2 [x e^{x} - \int e^{x} d x] \\ = x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + c \\ = e^{x} [x^{2} - 2 x + 2] + c \end{array}$

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135. $x s i n 3 x$

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

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$\begin{array}{l} \int x s i n 3 x d x = x \int s i n 3 x d x - \int \frac{d x}{d x} \int s i n 3 x d x d x . \\ = - x \frac{c o s 3 x}{3} + \int \frac{c o s 3 x}{3} \\ = \frac{- x c o s 3 x}{3} + \frac{s i n 3 x}{3 * 3} + c \\ = - \frac{x}{3} c o s 3 x + \frac{1}{9} s i n 3 x + c \end{array}$

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Temperature dependence of resistivity ρ(T) of semiconductors, insulators and metals is significantly based on the following factors:
(a) Number of charge carriers can change with temperature T.
(b) Time interval between two successive collisions can depend on T.
(c) Length of material can be a function of T.
(d) Mass of carriers is a function of T.

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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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134. $x s i n x$

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

Contributor-Level 10

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

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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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