Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, c)

Explanation- Gauss's law states that the total electric flux of an enclosed surface is given by q/? _0, where q is the charge enclosed by the surface.

So total charge inside the surface is = Q-2Q = -Q

Therefore total flux through the surface of the sphere = -Q/? ₀

Now, charge 5Q is lies outside the surface, thus it makes no contribution to electric flux through the given surface. So both option a and c are true.

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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/r³ and it is also conservative.

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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

$\begin{array}{l}\begin{array}{l}Let{x}^3=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}3x^{2}dx=dt\hfill \end{array}\\ I=\int \frac{3x^2}{x^6+1}dx=\int \frac{dt}{t^2+1}\\ \begin{array}{ll}= ta{n}^{–1}t+C\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= ta{n}^{–1}\left(x^3\right) +C\hfill \end{array}\end{array}$

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

$\begin{array}{l}\begin{array}{l}{e}^{2}x+ e^{x}1dx=dt\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}{e}^x\left(x+1\right)dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{e^x(1+x)}{co{s}^2\left(e^{2}x\right)}}dx={\displaystyle \int \frac{dt}{co{s}^{2}t}}\\ \begin{array}{l}={\displaystyle \int se{c}^2}t dt = tan \left(e^x,x\right) + C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\therefore The correct answer is \left(B\right).\hfill \end{array}\end{array}$

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.

$\begin{array}{l}={\displaystyle \int \frac{si{n}^{2}x-co{s}^{2}x}{si{n}^{2}co{s}^{2}x}}dx\\ \begin{array}{l}={\displaystyle \int (se{c}^{2}x-cose{c}^{2}x)}dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= tanx+ cotx+ C.\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Therefore, the correct answer is \left(A\right).\hfill \end{array}\end{array}$

$\begin{array}{l}\frac{1}{cos(x-a)cos(x-b)}=\frac{1}{sin(a-b)}*\left[\frac{sin(a-b)}{cos(x-a)cos(x-b)}\right]\\ =\frac{1}{sin(a-b)}\left[\frac{sin\{(x-b)-(x-a)\}}{cos(x-a)cos(x-b)}\right]\end{array}$

$\begin{array}{l}=\frac{1}{sin(a-b)}[sin(x-b)cos(x-a)-cos(x-b)\cdot sin(x-a)cos(x-a)cos(x-b)\\ =\frac{1}{sin(a-b)}[tan(x-b)-tan(x-a)]\\ =\frac{1}{sin(a-b)}{\displaystyle \int tan}(x-b)-tan\left(x-a\right)]dx\\ =\frac{1}{sin(a-b)}[-log|cos(x-b)|+log|cos(x-a)?]\\ =\frac{1}{sin(a-b)}\left[log\left|\frac{cos(x-a)}{cos(x-b)}\right|\right]+\; C\end{array}$