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#### A $10 \Omega, 20\, \mathrm{mH}$ coil carrying constant current is connected to a battery of $20 \mathrm{~V}$ through a switch. Now after switch is opened current becomes zero in $100 \mu \mathrm{s}$. The average e.m.f. induced in the coil is____________ $\mathrm{V}$.Option: 1 400Option: 2 -Option: 3 -Option: 4 -

$\mathrm{R= 10\, \Omega }$
$\mathrm{L= 20 \times 10^{-3}H}$

Current in the steady state ,
$\mathrm{I_{0}= \frac{v}{R}= 2A}$

$\mathrm{Avg.Emf\: induced= e_{avg}= L\left |\frac{\Delta I}{\Delta t} \right |}$
$\mathrm{= 200\times10^{-3}\left | \frac{2}{100\times 10^{-6}}\right |}$
$\mathrm{\Rightarrow e_{avg}= 400}$

#### Two capacitors $\mathrm{A}$  and $\mathrm{B}$ having capacitors  $\mathrm{10 \mu \mathrm{F}}$  and  $\mathrm{20 \mu \mathrm{F}}$  are connected in series with a $\mathrm{12 \mathrm{~V}}$ battery. The ratio of the charges on.Option: 1 $0.5: 1 \\$Option: 2 $1: 1 \\$Option: 3 $2: 1 \\$Option: 4 $2: 4$

when capacitors are connected in series they carry same charge. the ratio is  $1: 1$

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#### Equivalent capacitor between A & B  is. Option: 1 $\mathrm{ \frac{4}{5} \mathrm{~F} \\ }$Option: 2 $\mathrm{3 / 5 \mathrm{~F} }$Option: 3 $\mathrm{ 2 \mathrm{~F} \\ }$Option: 4 $\mathrm{ 4 \mathrm{~F} }$

Equivalent circuit-

$\mathrm{C_{\text {eq. }} =\frac{|\times|}{1+1}+1 }$
$=\frac{1}{2}+1=3 / 2 \mathrm{~F}$
$\mathrm{c_{e q}=\frac{3}{2} F}$ and F are connected in series.

$\mathrm{C_{e q}=\frac{\frac{3}{2} \times 1}{\frac{3}{2}+1}=\frac{3}{2} \times \frac{2}{5}=\frac{3}{5} F }$

#### In the region of space the electric field is given by  $\mathrm{\vec{E}=8 \hat{i}+4 \hat{j}+3 \hat{k}}$ . The electric flux through a surface of area 100 units in the xy plane is -Option: 1 800 UnitsOption: 2 300 unitsOption: 3 400 unitsOption: 4 1500 units.

we know that,

$\mathrm{ \phi=\vec{E} \cdot \vec{A} }$
$\mathrm{ \phi=(8 \hat{\imath}+4 \hat{\jmath}+3 \hat{k}) \cdot 100 \hat{i} }$
$\mathrm{ \phi=300}$

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#### Two point charges $\mathrm{+16 q }$ and  $\mathrm{-4 q }$  located at $\mathrm{ x=0}$ and $\mathrm{ x=L}$ . The location of a point on the x-axes from $\mathrm{ x=0}$, at which the net electric field due to there two charges is zero is -Option: 1 $\mathrm{ l }$Option: 2 $\mathrm{ 2 l}$Option: 3 $\mathrm{ L / 2}$Option: 4 $\mathrm{L / 6}$

$\mathrm{\frac{k \cdot 16 q}{(l+x)^2} =\frac{4 k q}{x^2}}$
$\mathrm{\frac{x}{l+x} =\frac{1}{2} }$
$\mathrm{2 x =l+x }$

$\mathrm{\\ l =x }$

#### A point charge of  $\mathrm{0.009 \mu c}$  is placed at origin. The intensity of electric field due to this charge at point $\mathrm{(\sqrt{2}, \sqrt{7}, 0)}$ is -Option: 1 $\mathrm{(3 \sqrt{2} \hat{i}+3 \sqrt{7} \hat{j}) \mathrm{N} / \mathrm{C} }$Option: 2 $\mathrm{(2 \sqrt{3} \hat{i}+3 \sqrt{7} \hat{j}) \mathrm{N} / \mathrm{C} }$Option: 3 $\mathrm{(3 \sqrt{2} \hat{i}+7 \sqrt{3} \hat{j}) N / C }$Option: 4 $\mathrm{(\sqrt{2} \hat{i}+7 \sqrt{3} \hat{j}) N / C}$

key points.-

$\mathrm{\vec{E}=\frac{k q}{\gamma^2} \hat{\gamma} }$
$\mathrm{\vec{E}=\frac{k q}{\gamma^3} \vec{\gamma} }$
Now

$\mathrm{\vec{\gamma}=x \hat{\imath}+y \hat{\jmath} }$

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#### The inward and outward electric flee for a closed surface in units of  $\mathrm{N}-\mathrm{m}^2 / \mathrm{c}$  are respectively $8 \times 10^3$ and  $\mathrm{4 \times 10^3}$. Then the total charge inside the surface is -Option: 1 $4 \times 10^3 \mathrm{C}$Option: 2 $-4 \times 10^3 \mathrm{c}$Option: 3 $\frac{-4 \times 10^3}{\epsilon} c$Option: 4 $-4 \times 10^3 \epsilon_0$

$\mathrm{ \therefore \quad \phi=\frac{Q}{\epsilon_0} }$
$\mathrm{ Q=\phi \epsilon }$..........(1)
$\mathrm{ \phi=\epsilon_0\left[-8 \times 10^3+4 \times 10^3\right] }$
$\mathrm{ \phi=-4 \times 10^3 \epsilon_0 }$

#### A charge $\mathrm{q}$ is placed at the centre of a cube. Then the flux passing through one face of cube will be -Option: 1 $\mathrm{q / t_0 }$Option: 2 $\mathrm{9 / 2t_0}$Option: 3 $\mathrm{q / 3 \epsilon_0 }$Option: 4

Total flux  $\mathrm{=\frac{1}{\epsilon_0} \times q }$
flex through one face $\mathrm{=\frac{1}{6}}$

$\therefore \quad$ flux    $\mathrm{=\frac{1}{\epsilon_0} \times 2 \times \frac{1}{6} }$
$\mathrm{ =2 / 6 \epsilon_0}$

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#### Current drawn by the battery just after switch 's' closed is.Option: 1 $\mathrm{ 2 V / 3 R \\ }$Option: 2 $\mathrm{ 3 V / 2 R \\ }$Option: 3 $\mathrm{ 2 V / R \\ }$Option: 4 $\mathrm{2 / R }$

when switch  S is closed this circuit will become

current
$\mathrm{i=\frac{V}{R_{\text {eq }}} }$    $\mathrm{=\frac{V}{3 R / 2}}$
$\mathrm{=\frac{2 V}{3 R}}$

#### What is the time constant of the circuit ?Option: 1 4 sec.Option: 2 6 sec.Option: 3 9 sec.Option: 4 5sec.

$\mathrm{C_{\text {net }}}$ $=0.5+0.5=1 \mathrm{f} \\$

$\mathrm{R_{\text {net }}=\frac{10 \times 10}{10+10}=5 \Omega }$
$\text { time constant }$

$\tau_{\text {net }} =R_{\text {net }} C_{\text {net }} \\$

$=5 \times 1=5 \mathrm{sec}$