Class 12th

Kindly Consider the following

128. $\frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)}$

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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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Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.4.
(a) The battery that runs the potentiometer should have voltage of 8 V.
(b) The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.

(c) The first portion of 50 cm of wire itself should have a potential drop of 10 V.
(d) Potentiometer is usually used for comparing resistances and not voltages.

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation-The potential drop along the wires of potentiometer should be greater than emfs of cells.
In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than 10 V.

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Kindly Consider the following

127. $\frac{c o s x}{(1 - s i n x) (2 - s i n x)}$

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Contributor-Level 10

Putting sin x = t cos xdx = dt

$\begin{array}{l} \int \frac{d t}{(1 - t) (2 - t)} = \int \frac{(2 - t) - (1 - t)}{(1 - t) (2 - t)} d t . \\ = \int \frac{d t}{1 - t} - \int \frac{d t}{2 - t} \\ = \frac{l o g | 1 - t |}{(- 1)} - \frac{l o g | 2 - t |}{(- 1)} + c \\ = l o g | \frac{2 - t}{1 - t} | + c \\ = l o g | \frac{2 - s i n x}{1 - s i n x} | + c \end{array}$

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New answer posted

a year ago

Kindly Consider the following

126. $\frac{1}{x (x^{n} + 1)}$

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

Contributor-Level 10

$\frac{1}{x (x^{n} + 1)} = \frac{x^{n - 1}}{x^{n - 1} \cdot x (x^{n} + 1)} = \frac{x^{n - 1}}{x^{n} (x^{n} + 1)}$

Putting xⁿ = t Þ n xⁿ^-¹dx = dt

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

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A resistance R is to be measured using a meter bridge, a student chooses the standard resistance S to be 100 Ω He finds the null point at l₁= 2.9 cm. He is told to attempt to improve the accuracy. Which of the following is a useful way?
(a) He should measure l₁, more accurately
(b) He should change S to 1000 ? and repeat the experiment
(c) He should change S to 3 ? and repeat the experiment
(d) He should have given up hope of a more accurate measurement with a meter bridge.

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Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (c)

Explanation – R/S= (l₁/100-l₁)= 100 (2.9/100-2.9)= 100/97.1=2.98ohm

So he should change S to almost 3 ohm and repeat the experiment.

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New answer posted

a year ago

Kindly Consider the following

125. $\frac{1}{x^{4} - 1}$

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Contributor-Level 10

Kindly go through the solution

$\begin{array}{l} \RightarrowA = \frac{1}{2} \\ A n d f r o m (1) w e g e t, - \frac{- 1}{2} \\ ∴ \int \frac{d x}{y^{2} - 1} = \frac{1}{2} \int \frac{d x}{y - 1} - \frac{1}{2} \int \frac{d x}{y + 1} \\ = \frac{1}{2} \int \frac{d x}{x^{2} - 1} - \frac{1}{2} \int \frac{d x}{x^{2} - 1} {y = x^{2}} \\ = \frac{1}{2} * \frac{1}{2 \cdot 1} l o g | \frac{x - 1}{x + 1} | - \frac{1}{2} t a n^{- 1} x + C \\ = \frac{1}{4} l o g | \frac{x - 1}{x + 1} | - \frac{1}{2} t a n^{- 1} x + C \end{array}$

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New answer posted

a year ago

Kindly Consider the following

124. $\frac{3 x - 1}{{(x + 2)}^{2}}$

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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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New answer posted

a year ago

Two batteries of emf e1 and e2 (e2>e1) and internal resistance r1 and r2 respectively are connected in parallel as shown in fig

(a) Two equivalent emf e_eq of the cells is between e1 and e2 i.e. e1eq

(b) The equivalent emf e_eq is samller than e₁

(c) The e_eq is given by e_eq=e₁+e₂ always

(d) E_eq is independent of internal resistance r₁ and r₂

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Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- eeq= $\frac{e_{2} r_{1 +} e_{2} r_{1}}{r_{1} + r_{2}}$

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New answer posted

a year ago

Kindly Consider the following

123. $\frac{2}{(1 - x) (1 + x^{2})}$

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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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New answer posted

a year ago

Consider a current carrying wire (current 7) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is
(a) Source of emf
(b) Electric field produced by charges accumulated on the surface of wire
(c) The charges just behind a given segment of wire which push them just the right way by repulsion
(d) The charges ahead.

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