Physics 3 Marks Question

1. Equivalent capacitance $(\text{C}_{n}) = \frac{\text{C}}{3} + \text{C}$

$ = \frac{4\text{C}}{3} = \frac{40}{3}\mu\text{F}$

2. $\text{Charge on } \text{C}_{4}, \text{q}_{4} = \text{C}_{4}\times\text{V}=10\times500 \mu\text{C}$

$ = 5\times10^{-3}\text{C} = 5 \text{mC}$

$\text{Charge on } \text{C}_ {1},\text{C}_{2},\text{C}_{3}\text{ is same and is equal to } \frac{\text{C}}{3}\times\text{V}$

$ = \frac{5}{3}\times10^{-3}\text{C}$

$ = 1.67\text{mC}$.

1. Electric flux through a Gaussian surface,

$\varphi = \frac{\text{total enclosed charge }}{\in_{0}}$

Net charge enclosed inside the shell q = 0

$\therefore$ Electric flux through the shell $\frac{\text{q}}{\in_{o}} = 0 $

even when the student writes - Electric flux through the shell is zero as electric field inside the shell is zero.

2. Gauss Law - Electric flux through a Gaussian surface is ¹/∈₀ times the net charge enclosed within it.

Alternate Answer

$\oint\overrightarrow{\text{E}}.\overrightarrow{\text{d}S} = \frac{\text{q}}{\text{E}_{o}}$

3. Force on the charge at the centre i.e. Charges ^𝑄/2 =0

$\text{F}_{A} = \frac{1}{4\pi\text{E}_{o}}\frac{2\text{Q}(\text{Q} + \text{Q} / 2)}{\text{x}^{2}}$

$ = \frac{1}{4\pi\text{E}_{o}}\frac{3\text{Q}^{2}}{\text{x}^{2}}$.

1. $\beta = \frac{\lambda\text{D}}{\text{d}}$

$ = \frac{500\times10^{-9}\times1}{10^{-3}}$

= 0.5 mm or 5 x 10^-4 m

2. $\beta_{0} = \frac{2\lambda\text{D}}{\text{a}} = 10 \beta$

$\text{a} = \frac{2\times500\times10^{-9}\times1}{10\times5\times10^{-4}}$

𝑎 =2 ×10⁻⁴ m or 0.2 mm.

1. Current in the circuit $\text{I} = \frac{\text{V}}{\sqrt{\text{R}^{2} + \bigg(\frac{1}{\text{C}\omega}\bigg)^{2}}}$

$\text{I} = \frac{220}{\sqrt{100^{2} + \bigg(\frac{1}{\frac{100}{\pi}\times10^{-6}\times2\pi\times50}}\bigg)^{2}}$

$ = \frac{2.2}{\sqrt{2}}\text{A} = 1.55\text{A}$

2. Voltage across the resistor$ = 100\times1.55 \text{V}$

$ = 155 \text{ volt}$

Voltage across the capacitor $ = 100\times1.55 \text{V}$

$ = 155 \text{ volt}$

Yes,

The sum of the two voltages is greater than 220 V but the voltage across the resistor and the capacitor are not in phase.

1. $\text{P}_{av} = \text{I}_{av}\text{x }\text{ e}_{av}\cos\phi$

For an ideal inductor, $\phi = \frac{\pi}{2}$

$\therefore\text{P}_{av} = \text{l}_{av} \text{x } \text{ e}_{av}\cos\frac{\pi}{2}$

$\text{P}_{av} = 0 .$

2. Brightness decreases:

Because as iron rod is inserted inductance increases. Thus, current decreases and brightness decreases.

1. New Electric field energy values are: = $\frac{1}{2}\text{k}(\text{c}_{1}\text{v}^{2}_{1}) \text{ and}\frac{1}{2}\text{k}(\text{c}_{2}\text{v}^{2}_{1}).$
2. New charges are: =$\frac{1}{2}\text{KC}_{1}\text{V}_{1}\text{ and }\frac{1}{2}\text{KC}_{2}\text{V}_{2}.$
3. New P.D values are: V₁ and V₂

(The battery remains connected to the capacitors)

Alternate Answer

1. New Electric field energy values are: $\frac{1}{2}\frac{\text{Q}^{2}}{\text{KC}_{1}} \text{and } \frac{1}{2}\frac{\text{Q}^{2}}{\text{KC}_{2}}.$
2. New charges are: Q and Q as before.
3. New P.D values are: $\frac{\text{Q}}{\text{KC}_{1}} \text{and }\frac{\text{Q}}{\text{KC}_{2}}.$

1. No power is dissipated when R = 0 $(\text{or }\phi = 90^{o} ) $
2. Maximum power is dissipated when X_L= X_C

Alternate Answer

1. Zero.
2. We have $\text{C}_{series} = \frac{3\mu\text{F}}{3} =1 \mu\text{F}$

Also, $\text{C}_{parallel} =( 3 + 3 +3 ) =9 \mu\text{F}$

Energy stored $ = \frac{1}{2}\text{CV}^{2}$

$\therefore$Energy in series combination $ =\frac{1}{2}1 \times10^{-6}\times\text{V}^{2}$

Energy in parallel combination $ =\frac{1}{2}9 \times10^{-6}\times\text{V}^{2}$

$\therefore$ Ratio = 1:9.

1. Low coercivity and high permeability.
2. The net magnetic flux through any closed surface is zero/

$\oint\text{B}.\text{ds} = 0 $

$\oint\text{E}.\text{ds} = \frac{\text{q}}{\in_{o}}$/The net electric flux through any closed surface is $\frac{1}{\in_{o}}$ times the net charge.

which indicates magnetic monopoles do not exist/ magnetic poles always exists in pairs.

Alternate Answer

1. Curvature of the earth.
2. Insufficient height of the receiving antenna.

1. Let there be $\text{N}_{0}$ radioactive nuclei at t =0.

If N is the number of nuclei left over at t=t, we have

$\frac{-\text{dN}}{\text{dt}}\propto\text{N}$

Or $\frac{-\text{dN}}{\text{dt}} = \lambda\text{N}(\lambda =\text{decay constant })$

$\therefore\frac{\text{dN}}{\text{N}} = -\lambda\text{dt}$

Or $l \text{nN} = - \lambda\text{t} + constant $

$\therefore \text{ At } \text{t} = 0 , $ we have

$l\text{nN}_{0} = constant $

$l \text{nN} = - \lambda\text{t} + l\text{nN}_{0}$

Or $l\text{n}(\frac{\text{N}}{\text{N}_{0}}) = -\lambda\text{t}$

$\therefore \text{N} = \text{N}_{0}\text{e}^{-\lambda\text{t}}$

2. Mean life $ = \frac{1}{\text{decay constant }}$

Alternate Answer

$\tau = \frac{1}{\lambda}$.

Alternate Answer

1. When light with energy $\text{hv} > \text{(𝑒𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑝)} \text{𝐸}_{𝑔}$ falls on photodiode, electron-hole pairs are generated.
2. Due to electric field at the junction, electrons and holes are separated before they combine.
3. Electrons are collected on n-side and holes are collected on p-side giving rise to an emf and current flows in external load.

3. It is easier to observe the change in the current, with change in the light intensity, when reverse bias is applied.

Alternate Answer

1. From graph of i² – t

Average power consumed in resistor R

$\text{P}_{av} = \frac{1}{\int_{0}^{T}\text{dt}}.\int_{0}^{T}\text{i}^{2}\text{R dt}$

$ = \frac{\text{i}_{m}^{2}\text{R}}{\text{T}}\int_{0}^{T}\sin^{2}\omega\text{t }\text{ dt} - - - - - (1)$

$ = \frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}\int_{0}^{T}(1 -\cos2\omega\text{t})\text{dt}$

$ = \frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}\bigg[\int_{0}^{T}\text{dt} - \int_{0}^{T}\cos2\omega\text{t }\text{ dt}\bigg] - - - - - (2)$

$ =\frac{\text{i}_{m}^{2}\text{R}}{2\text{T}}[\text{T} - 0 ]$
$ = \frac{\text{i}_{m}^{2}\text{R}}{2}$

2. In case of ac

$\text{P}_{av} =\frac{\text{V}_{rms}^{2}}{\text{R}} = \frac{\text{V}_{eff}^{2}}{\text{R}}$

$\Rightarrow\text{R} = \frac{\text{V}_{rms}^{2}}{\text{P}_{av}}$

$ = \frac{220\times220}{100}$

$ = 484\Omega.$

Resultant potential across LCR

$\text{V}^{2} = \text{V}^{2}_{\text{R}} + (\text{V}_{L} - \text{V}_{c} ) ^{2}$

$\text{V}_{R} = \text{iR},\text{V}_{L} = \text{i}\text{X}_{L},\text{V}_\text{c} = \text{i X}_{c}$

On solving

$\text{Z} = \sqrt{\text{R}^{2} + (\text{X}_{L} - \text{X}_{C})^{2}}$

With increase in ω, current first increases (up to ω_o ) and then decreases.

Method – I

$\text{Z} =\text{E}/\text{I}$

Calculation of $\text{Z} =\frac{12}{0.5}= 24\Omega$

$\cos\phi = \frac{\text{R}}{\text{Z}}$

Calculation of R = 12Ω.

Method – II

$\text{Z} = \text{E}/\text{I}$

Z = 24Ω

$\tan \phi = \frac{\text{X}_{L}}{\text{R}}$

R = 12Ω.

Method I

$\text{E} = \text{B} $ 1 v 

$\text{I} =\frac{\text{B1v}}{\text{R}} = \text{I} = 0.5\text{A}$

$\text{F} = \text{B}$ I 1 

Calculation of $\text{F} = 3.75\times10^{-3}\text{N}$.

Method II

$\text{F} = \frac{\text{B}^{2}1^{2}\text{v}}{\text{R}}$

Correct substitution and calculation of $\text{F} = 3.75\times10^{-3}\text{N}$.

1. $X_L=\omega_L=(1000\times100\times10^{-3})\Omega=100\Omega$

$X_C=\frac{1}{\omega C}=\bigg(\frac{1}{1000\times2\times10^{-6}}\bigg)\Omega=500\Omega$

Phase angle

$\tan\Phi=\frac{X_L-X_C}{R}$

$\tan\Phi=-\frac{\pi}{4}$

As  $X_C>X_L$ (phase angle is negative), hence current leads voltage

2. To make power factor unity

$X_{C'}>X_L$

$\frac{1}{WC'}=100$

$C'=10\mu\text{F}$

$C'=C+C_1$

$10=2+C_1$

$C_1=8\mu\text{F}$

1. Increases

$\text{X}_{L} = \omega_{L}$

As number of turns decreases, L decreases, hence current through bulb increases. /voltage across bulb increases.

2. Decreases 

Iron rod increases the inductance, which increases X_L, hence½ current through the bulb decreases/voltage across bulb decreases. 

3. Increases

condition (X_C = X_L) the current through the bulb will become maximum/increase.

1. $\omega = \frac{1}{\sqrt{\text{LC}}}$

$ = \frac{1}{\sqrt{5\times80\times10^{-6}}} = 50\text{radian / s }$

2. Current at resonance

$\text{I}_{rms} = \frac{\text{V}_{rms}}{\text{R}} = \frac{240}{40}\text{A} = 6\text{A}$

3. V_rms across capacitor

$\text{V}_{rms} = \text{I}_{rms}\text{X}_{c}$

$ = 6\times\frac{1}{50\times80\times10^{-6}}\text{V}$

$ = \frac{6\times10^{6}}{4\times10^{3}}\text{V} = \frac{6000}{4}\text{V} = 1500\text{V}.$

1. $\omega = \frac{1}{\sqrt{\text{LC}}}$

$ = \frac{1}{\sqrt{200\times10^{-3}\times500\times10^{-6}}} = 100 \text{ radian s }^{-1}$

 $\big(\text{or } \text{ v} = \frac{\omega}{2\pi} = \frac{50}{\pi} \text{Hz}\approx51.9\text{Hz})$

2. $\text{I} = \frac{\text{V}}{\text{R}} = \frac{100}{10} = 10 \text{A}$
3. Q-factor $\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}$ $\big(\text{Q-factor } = \frac{\omega_{\circ}\text{L}}{\text{R}}\big)$

 = 2.

1. Now, voltage drope accurs resestev = R

$=\frac{\text{V}_0\text{R}}{\sqrt{\text{R}^2+\omega^2\text{L}^2}}$

Voltage drop accurs, reduse,

$=1\text{X}_\text{L}$

$=\frac{(\omega\text{L)}}{\sqrt{\text{X}^2+\omega^2\text{L}^2}}$

2. Phase Difference,

$\tan\theta=\frac{\text{X}_\text{L}}{\text{R}}$

And resistance lead to the Inductance.

1. Peak Voltage Across.

1. Resistance R is $\text{V}_\text{R}=\text{I}_0\text{R}=\frac{\text{V}_0\text{R}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}$

2. Capacitor C is $\text{V}_\text{C}=\text{I}_0\text{X}_\text{C}=\frac{\text{I}_0\text{X}_\text{C}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}$

$\phi=\tan^{-1}\Big(\frac{\text{V}_\text{C}}{\text{V}_\text{R}}\Big)=\tan^{-1}\Big(\frac{\text{X}_\text{C}}{\text{R}}\Big)=$ Phase difference between V & I is ahead of V.