Class 12th

88. Kindly Consider the following

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

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Kindly go through the solution

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Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region,
(a) The electric field is necessarily zero
(b) The electric field is due to the dipole moment of the charge distribution only
(c) The dominant electric field is ∝ 1/r³, for large r, where r is the distance from an origin in this regions r
(d) The work done to move a charged particle along a closed path, away from the region, will be zero

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (c), (d)

Explanation- electric field is not necessarily zero may it become zero by their algeabric sum. For dipole the electric field is always ∝ 1/r³and it is also conservative.

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87. Kindly Consider the following

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

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Kindly go through the solution

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The electric field at a point is
(a) Always continuous
(b) Continuous if there is no charge at that point
(c) Discontinuous only if there is a negative charge at that point (d) discontinuous if there is a charge at that point

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (b), (d)

Explanation- if we place a charge then we must experience some forces but if there would be no charge so field is continuous

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Kindly Consider the following

86. $\frac{3 x^{2}}{x^{6} + 1}$

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$\begin{array}{l} \begin{array}{l} L e t x^{3} = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} 3 x^{2} d x = d t \end{array} \\ I = \int \frac{3 x^{2}}{x^{6} + 1} d x = \int \frac{d t}{t^{2} + 1} \\ \begin{array}{l} = t a n^{- 1} t + C \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = t a n^{- 1} (x^{3}) + C \end{array} \end{array}$

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Kindly Consider the following

85. $\int \frac{e^{2} (1 + x)}{c o s^{2} (e^{2} x)}$

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

Contributor-Level 10

$\begin{array}{l} \begin{array}{l} L e t e^{2} x = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} e^{2} x + e^{x} 1 d x = d t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} e^{x} (x + 1) d x = d t \end{array} \\ = \int \frac{e^{x} (1 + x)}{c o s^{2} (e^{2} x)} d x = \int \frac{d t}{c o s^{2} t} \\ \begin{array}{l} = \int s e c^{2} t d t = t a n (e^{x}, x) + C \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} ∴ T h e c o r r e c t a n s w e r i s (B) . \end{array} \end{array}$

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Multiple Choice Questions

(One or More than One Correct Answer Type)

If Ф_s E . dS = 0 over a surface, then

(a) The electric field inside the surface and on it is zero
(b) The electric field inside the surface is necessarily uniform
(c) The number of flux lines entering the surface must be equal tothe number of flux lines leaving it
(d) All charges must necessarily be outside the surface

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (c), (d)

Explanation- It is only possible when charges must be outside the surface or field line entering or leaving the surface are equal.

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Kindly Consider the following

84. $\int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} c o s^{2} x} d x is equal to$

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$\begin{array}{l} = \int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} c o s^{2} x} d x \\ \begin{array}{l} = \int (s e c^{2} x - c o s e c^{2} x) d x \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} = t a n x + c o t x + C . \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} T h e r e f o r e, t h e c o r r e c t a n s w e r i s (A) . \end{array} \end{array}$

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Kindly Consider the following

83. $\frac{1}{c o s (x - a) c o s (x - b)}$

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

Contributor-Level 10

$\begin{array}{l} \frac{1}{c o s (x - a) c o s (x - b)} = \frac{1}{s i n (a - b)} * [\frac{s i n (a - b)}{c o s (x - a) c o s (x - b)}] \\ = \frac{1}{s i n (a - b)} [\frac{s i n {(x - b) - (x - a)}}{c o s (x - a) c o s (x - b)}] \end{array}$

$\begin{array}{l} = \frac{1}{s i n (a - b)} [s i n (x - b) c o s (x - a) - c o s (x - b) \cdot s i n (x - a) c o s (x - a) c o s (x - b) \\ = \frac{1}{s i n (a - b)} [t a n (x - b) - t a n (x - a)] \\ = \frac{1}{s i n (a - b)} \int t a n (x - b) - t a n (x - a)] d x \\ = \frac{1}{s i n (a - b)} [- l o g | c o s (x - b) | + l o g | c o s (x - a) ?] \\ = \frac{1}{s i n (a - b)} [l o g | \frac{c o s (x - a)}{c o s (x - b)} |] + C \end{array}$

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A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed .

(a) Perpendicular to the diameter

(b) Parallel to the diameter

(c) At an angle tilted towards the diameter

(d) At an angle tilted away from the diameter

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