Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Alternating Current
Q1: A series AC circuit containing an inductor ( $20 \mathrm{mH}$ ), a capacitor ( $120 \mu \mathrm{F}$ ) and a resistor ( $60 \Omega$ ) is driven by an AC source of $24 \mathrm{~V} / 50 \mathrm{~Hz}$. The energy dissipated in the circuit in $60 \mathrm{~s}$ is
Options: (a) $3.39 \times 10^{3} \mathrm{~J}$
(b) $5.65 \times 10^{2} \mathrm{~J}$
(c) $2.26 \times 10^{3} \mathrm{~J}$
(d) $5.17 \times 10^{2} \mathrm{~J}$
Answer: (d) $5.17 \times 10^{2} \mathrm{~J}$
Solution: Impedance, $\mathrm{Z}=\left(\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}\right)^{1 / 2}$
$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=(2 \pi \mathrm{vL})$
$\mathrm{X}_{\mathrm{L}}=6.28 \times 50 \times 20 \times 10^{-3}=6.28 \Omega$
$\mathrm{X}_{\mathrm{C}}=(1 / \omega \mathrm{C})=1 /(2 \pi \mathrm{vC})=1 /\left(6.28 \times 120 \times 10^{-6} \times 50\right)=26.54 \Omega$
$\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}=26.54-6.28=20.26$
$\mathrm{Z}=\left((60)^{2}+(20.26)^{2}\right)^{1 / 2}$
$\mathrm{Z}^{2}=4010 \Omega^{2}$
Average power dissipated, $\operatorname{Pav}=\varepsilon_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \Phi$
$\mathrm{P}_{\mathrm{av}}=\varepsilon_{\text {rms }} \times\left(\varepsilon_{\text {rms }} / \mathrm{Z}\right) \times(\mathrm{R} / \mathrm{Z})$
$\mathrm{P}_{\mathrm{av}}=\left(\varepsilon_{\text {rms }}{ }^{2} / \mathrm{Z}^{2}\right) \times \mathrm{R}=\left[(24)^{2} / 4010\right] \times 60 \mathrm{~W}=8.62 \mathrm{~W}$
Energy dissipated in $60 \mathrm{~S}=8.62 \times 60=5.17 \times 10^{2} \mathrm{~J}$
Topic: Alternating Current
Q2: In an AC generator, a coil with \(\mathbf{N}\) turns, all of the same area \(A\) and total resistance \(R\), rotates with frequency \(\omega\) in a magnetic field \(B\). The maximum value of emf generated in the coil is
Options: (a) NAB
(b) NABR
(c) NAB \omega
(d) NABRc
Answer: (c) In an a.c. generator, maximum emf $=\mathbf{N A B} \omega$
Topic: Alternating Current
Q3: The phase difference between the alternating current and emf is \(\pi / 2\). Which of the following cannot be the constituent of the circuit?
Options: (a) LC
(b) L alone
(c) C alone
(d) R, L
Answer: (d) R, L
Solution: $\mathrm{R}$ and $\mathrm{L}$ cause phase difference to lie between 0 and $\pi / 2$ but never 0 and $\pi / 2$ at extremities
Topic: Alternating Current
Q4: Alternating current cannot be measured by D.C ammeter because
Options: (a) A.C cannot pass through D.C ammeter
(b) A.C changes direction
(c) The average value of current for the complete cycle is zero
(d) D.C. ammeter will get damaged
Answer: (c) The average value of current for the complete cycle is zero
Solution: The average value of A.C for the complete cycle is zero. Hence A.C cannot be measured by D.C ammeter
Topic: Alternating Current
Q5: A power transmission line feeds input power at $2300 \mathrm{~V}$ to a step-down transformer with its primary windings having 4000 turns. The output power is delivered at $230 \mathrm{~V}$ by the transformer. If the current in the primary of the transformer is $5 \mathrm{~A}$ and its efficiency is $90 \%$, the output current would be
Options: (a) $25 \mathrm{~A}$
(b) $50 \mathrm{~A}$
(c) $45 \mathrm{~A}$
(d) $35 \mathrm{~A}$
Answer: (c) 45 A
Solution: Given $\epsilon_{\mathrm{p}}=2300 \mathrm{~V}, \mathrm{~N}_{\mathrm{p}}=4000$
$\epsilon_{\mathrm{s}}=230 \mathrm{~V}$,
$\mathrm{I}_{\mathrm{p}}=5 \mathrm{~A}$,
$\eta=90 \%=0.9$
$\eta=\mathrm{P}_{\mathrm{o}} / \mathrm{P}_{\mathrm{i}}=\left(\epsilon_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}\right) /\left(\epsilon_{\mathrm{p}} \mathrm{I}_{\mathrm{p}}\right)$
$\mathrm{I}_{\mathrm{s}}=\eta \epsilon_{\mathrm{p}} \mathrm{I}_{\mathrm{p}} / \epsilon_{\mathrm{s}}=(0.9 \times 2300 \times 5) / 230=45 \mathrm{~A}$
Topic: Alternating Current
Q6: A circuit connected to an ac source of emf $e=e_{0} \sin (100 t)$ with $t$ in seconds, gives a phase difference of $\pi / 4$ between the emf $\mathrm{e}$ and current $\mathrm{I}$. Which of the following circuits will exhibit this?
Options: (a) $\mathrm{RC}$ circuit with $\mathrm{R}=1 \mathrm{k} \Omega$ and $\mathrm{C}=10 \mu \mathrm{F}$
(b) $\mathrm{RL}$ circuit with $\mathrm{R}=1 \mathrm{k} \Omega$ and $\mathrm{L}=10 \mathrm{mH}$
(c) $\mathrm{RC}$ circuit with $\mathrm{R}=1 \mathrm{k} \Omega$ and $\mathrm{C}=1 \mu \mathrm{F}$
(d) $\mathrm{RL}$ circuit with $\mathrm{R}=1 \mathrm{k} \Omega$ and $\mathrm{L}=1 \mathrm{mH}$
Answer: (a) $RC$ circuit with $R=1 \mathrm{k} \Omega$ and $C=10 \mu \mathrm{F}$
Solution: $\mathrm{X}_{\mathrm{c}}=\mathrm{R}$
$1 / \omega \mathrm{C}=\mathrm{R}$
$\omega=100 \mathrm{rad} / \mathrm{s}$
Therefore, $1 / 100=\mathrm{RC}$
Substituting values from option (a)
$\mathrm{R}=1 \mathrm{k} \Omega=10^{3} \Omega$
$\mathrm{C}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}$
$\mathrm{RC}=10^{3} \times 10 \times 10^{-6}=1 / 100$
Therefore, option (a) is correct
Topic: Alternating Current
Q7: In an a.c. circuit, the instantaneous e.m.f. and current is given by $e=100 \sin 30 t, i=20 \sin (30 t-\pi / 4)$. In one cycle of A.C, the average power consumed by the circuit and the wattless current are, respectively
Options: (a) 50,10
(b) $1000 / \sqrt{2}, 10$
(c) $50 / \sqrt{2},0$
(d) 50,0
Answer: (b) 1000/ $\sqrt{2}, 10$
Solution: $\mathrm{P}_{\mathrm{avg}}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \theta$
$\mathrm{P}_{\text {avg }}=\left(\mathrm{V}_{0} / \sqrt{2}\right)\left(\mathrm{I}_{0} / \sqrt{2}\right) \cos \theta$
$=(100 / \sqrt{2})(20 / \sqrt{2}) \cos 45^{\circ}$
$P_{\text {avg }}=1000 / \sqrt{2}$ watt
Wattless current $=\mathrm{I}_{\mathrm{rms}} \sin \theta$
Wattless current $=\left(\mathrm{I}_{0} / \sqrt{2}\right) \sin \theta$
$=(20 / \sqrt{2}) \sin 45^{0}$
$=10 \mathrm{amp}$
Topic: Alternating Current
Q8: A coil having $\mathbf{n}$ turns and resistance $R \Omega$ is connected with a galvanometer of resistance 4R $\Omega$. This combination is moved in time $t$ seconds from a magnetic field $W_{1}$ weber to $W_{2}$ weber. The induced current in the circuit is
Options: (a) $-\left(\mathrm{W}_{2}-\mathrm{W}_{1}\right) / 5 \mathrm{Rnt}$
(b) $-\mathrm{n}\left(\mathrm{W}_{2}-\mathrm{W}_{1}\right) / 5 \mathrm{Rt}$
(c) $-\left(\mathrm{W}_{2}-\mathrm{W}_{1}\right) / \mathrm{Rnt}$
(d) $-\left(\mathrm{W}_{2}-\mathrm{W}_{1}\right) / 5 \mathrm{Rnt}$
Answer: (b) $-\mathbf{n}\left(\mathbf{W}_{2}-\mathbf{W}_{1}\right) / 5 R t$
Solution: The emf induced in the coil is $\mathrm{e}=-\mathrm{n}(\mathrm{d} \Phi / \mathrm{dt})$
Induced current, $\mathrm{I}=\mathrm{e} / \mathrm{R}^{\prime}=-\left(\mathrm{n} / \mathrm{R}^{\prime}\right)(\mathrm{d} \Phi / \mathrm{dt})$
Given, $\mathrm{R}^{\prime}=\mathrm{R}+4 \mathrm{R}=5 \mathrm{R}$
$\mathrm{d} \Phi=\mathrm{W}_{2}-\mathrm{W}_{1}$
$\mathrm{dt}=\mathrm{t}$
(here, $\mathrm{W}_{1}$ and $\mathrm{W}_{2}$ are flux associated with one turn)
Substituting the given values in equa(1) we get
$\mathrm{I}=(-\mathrm{n} / 5 \mathrm{R})(\mathrm{W}_{2}-\mathrm{W}_{1} / \mathrm{t})$
Topic: Alternating Current
Q9: An alternating voltage $V(t)=220 \sin 100 \pi t$ volts is applied to a purely resistive load of $50 \Omega$. The time taken for the current to rise from half of the peak value is
Options: (a) $5 \mathrm{~ms}$
(b) $2.2 \mathrm{~ms}$
(c) $7.2 \mathrm{~ms}$
(d) $3.3 \mathrm{~ms}$
Answer: (d) $3.3 \mathrm{~ms}$
Solution: As $v(t)=220 \sin 100 \pi t$
So $\mathrm{I}(\mathrm{t})=(220 / 50) \sin 100 \pi \mathrm{t}$
I.e., $\mathrm{I}=\mathrm{I}_{\mathrm{m}}=\sin 100 \pi \mathrm{t}$
For $\mathrm{I}=\mathrm{I}_{\mathrm{m}}$
$\mathrm{t}_{1}=(\pi / 2) \times(1 / 100 \pi)=(1 / 200) \mathrm{sec}$,
And for $\mathrm{I}=\mathrm{I}_{\mathrm{m}} / 2$
$\Rightarrow(\mathrm{I}_{\mathrm{m}} / 2)=\mathrm{I}_{\mathrm{m}} \sin \left(100 \pi \mathrm{t}_{2}\right)$
$\Rightarrow(\pi / 6)=100 \pi \mathrm{t}_{2}$
$\Rightarrow \mathrm{t}_{2}=(1 / 600) \mathrm{s}$
$\therefore \mathrm{t}_{\text {req }}=(1 / 200)-(1 / 600)=(2 / 600)=(1 / 300) \mathrm{s}=3.3 \mathrm{~ms}$
Topic: Alternating Current
Q10: An arc lamp requires a direct current of $10 \mathrm{~A}$ at $80 \mathrm{~V}$ to function. If it is connected to a $220 \mathrm{~V}$ (RMS), 50 $\mathrm{Hz}$ AC supply, the series inductor needed for it to work is close to
Options: (a) $0.065 \mathrm{H}$
(b) $80 \mathrm{H}$
(c) $0.08 \mathrm{H}$
(d) $0.044 \mathrm{H}$
Answer: (a) $0.065 \mathrm{H}$
Solution: $\mathrm{I}=10 \mathrm{~A}$
$\mathrm{V}=80 \mathrm{~V}$
$\mathrm{R}=8 \Omega$
$10=220 /\left(8^{2}+\mathrm{X}_{\mathrm{L}}{ }^{2}\right)^{1 / 2}$
$64+\mathrm{X}_{\mathrm{L}}{ }^{2}=484$
$X_{L}=\sqrt{ } 420$
$2 \pi \times 50 \mathrm{~L}=\sqrt{ } 420$
$\mathrm{L}=\sqrt{ } 420 / 100 \pi$
$\mathrm{L}=0.065 \mathrm{H}$
Topic: Alternating Current
Q11: In a series, LCR circuit $R=200 \Omega$ and the voltage and the frequency of the main supply is $220 \mathrm{~V}$ and $50 \mathrm{~Hz}$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $30^{\circ}$. The power dissipated in the LCR circuit is
Options: (a) $242 \mathrm{~W}$
(b) $305 \mathrm{~W}$
(c) $210 \mathrm{~W}$
(d) Zero
Answer: (a) $242 \mathrm{~W}$
Solution: Tan $\Phi=\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{c}}\right) / \mathrm{R}$
$\operatorname{Tan} 30^{\circ}=\mathrm{X}_{\mathrm{c}} / \mathrm{R}=\mathrm{X}_{\mathrm{c}}=\mathrm{R} / \sqrt{3}$
$\operatorname{Tan} 30^{\circ}=\mathrm{X}_{\mathrm{L}} / \mathrm{R}=\mathrm{X}_{\mathrm{L}}=\mathrm{R} / \sqrt{3}$
$\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{c}} \Rightarrow$ Condition for resonance
So $\Phi=0^{0}$\n$\mathrm{P}=\mathrm{VI} \cos 0^{0}$
$\mathrm{P}=\mathrm{V}^{2} / \mathrm{R}=(220)^{2} / 200=242 \mathrm{~W}$
Topic: Capacitor
Q1: A parallel plate capacitor with plates of area $1 \mathrm{~m}^{2}$ each are at a separation of $0.1 \mathrm{~m}$. If the electric field between the plates is $100 \mathrm{~N} \mathrm{C}^{-1}$, the magnitude of charge on each plate is
Options: (a) $8.85 \times 10^{-10} \mathrm{C}$
(b) $7.85 \times 10^{-10} \mathrm{C}$
(c) $9.85 \times 10^{-10} \mathrm{C}$
(d) $6.85 \times 10^{-10} \mathrm{C}$
Answer: (a) $8.85 \times 10^{-10} \mathrm{C}$
Solution: The electric field between two plates is
$\mathrm{E}=\sigma / \varepsilon_{0}=\mathrm{q} / \mathrm{A} \varepsilon_{0} \Rightarrow \mathrm{q}=\mathrm{EA} \varepsilon_{0}$
$\mathrm{q}=(100)(1)\left(8.85 \times 10^{-12}\right)=8.85 \times 10^{-10} \mathrm{C}$
Topic: Capacitor
Q2: The parallel combination of two air filled parallel plate capacitors of capacitance $\mathbf{C}$ and $\mathbf{n C}$ is connected to a battery of voltage, $\mathrm{V}$. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is
Options: (a) $\mathrm{V}$
(b) $(\mathrm{n}+1) \mathrm{V} /(\mathrm{K}+\mathrm{n})$
(c) $(\mathrm{nV}) /(\mathrm{K}+\mathrm{n})$
(d) $\mathrm{V} /(\mathrm{K}+\mathrm{n})$
Answer: (b) $(\mathrm{n}+1) \mathrm{V} /(\mathrm{K}+\mathrm{n})$
Solution: For parallel combination, $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}+\mathrm{nC}=\mathrm{C}(\mathrm{n}+1)$
Charge on capacitor, $\mathrm{q}=\mathrm{C}_{\mathrm{eq}} \mathrm{V}=\mathrm{CV}(\mathrm{n}+1)$
Now, after removing the battery, dielectric material is placed.
Then, $\mathrm{C}_{\mathrm{eq}}=\mathrm{KC}+\mathrm{nC}=\mathrm{C}(\mathrm{K}+\mathrm{n})$
New potential difference, $\mathrm{V}^{\prime}=\mathrm{q} / \mathrm{C}_{\mathrm{eq}}=\mathrm{CV}(\mathrm{n}+1) / \mathrm{C}(\mathrm{K}+\mathrm{n})=(\mathrm{n}+1) \mathrm{V} /(\mathrm{K}+\mathrm{n})$
Topic: Capacitor
Q3: A parallel plate capacitor is made by stacking $\mathbf{n}$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is $\mathrm{C}$ then the resultant capacitance is
Options: (a) $\mathrm{C}$
(b) $\mathrm{nC}$
(c) $(n-1) C$
(d) $(\mathrm{n}+1) \mathrm{C}$
Answer: (c) $(n-1) C$
Solution: $\mathrm{n}$ plates connected alternately give rise to $(\mathrm{n}-1$ ) capacitors connected in parallel
Resultant capacitance $=(n-1) C$.
Topic: Capacitor
Q4: Capacitance (in $F$ ) of a spherical conductor with radius $1 \mathrm{~m}$ is
Options: (a) $1.1 \times 10^{-10}$
(b) $10^{-6}$
(c) $9 \times 10^{-9}$
(d) $10^{-3}$
Answer: (a) $1.1 \times 10^{-10}$
Solution: $\mathrm{C}=4 \pi \varepsilon_{0} \mathrm{R}=1 / 9 \times 10^{9}=1.1 \times 10^{-10} \mathrm{~F}$
Topic: Capacitor
Q5: If there are $\mathbf{n}$ capacitors in parallel connected to $\mathbf{V}$ volt source, then the energy stored is equal to
Options: (a) $\mathrm{CV}$
(b) $(1 / 2) \mathrm{nCV}^{2}$
(c) $\mathrm{CV}^{2}$
(d) $(1 / 2 \mathrm{n}) \mathrm{CV}^{2}$
Answer: (b) $(1 / 2) n C V^{2}$
Solution: Total capacity $=\mathrm{nC}$
$\therefore$ Energy $=1 / 2 \mathrm{nCV}{ }^{2}$
Topic: Capacitor
Q6: A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}$. The separation between its plates is $\mathrm{d}$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_{1}=3$ and thickness $d / 3$ while the other one has dielectric constant $k_{2}=6$ and thickness $2 \mathrm{~d} / 3$. The capacitance of the capacitor is now
Options: (a) $20.25 \mathrm{pF}$
(b) $1.8 \mathrm{pF}$
(c) $45 \mathrm{pF}$
(d) $40.5 \mathrm{pF}$
Answer: (d) $40.5 \mathrm{pF}$
Solution: $$\mathrm{K}_{1}=3 \quad \mathrm{~K}_{2}=3$$

$$\mathrm{C}=\varepsilon_{0} \mathrm{~A} / \mathrm{d}=9 \times 10^{-12} \mathrm{~F}$$

With dielectric, $$\mathrm{C}=\varepsilon_{0} \mathrm{Ak} / \mathrm{d}$$

$$\mathrm{C}_{1}=\varepsilon_{0} \mathrm{~A} 3 /(\mathrm{d} / 3)=9 \mathrm{C}$$

$$\mathrm{C}_{2}=\varepsilon_{0} \mathrm{~A} 6 /(2 \mathrm{~d} / 3)=9 \mathrm{C}$$

$$\mathrm{C}_{\text {total }}=\mathrm{C}_{1} \mathrm{C}_{2} /(\mathrm{C}_{1}+\mathrm{C}_{2})$$ as they are in series

$$\mathrm{C}_{\text {total }}=(9 \mathrm{C} \times 9 \mathrm{C}) / 18 \mathrm{C}=(9 / 2) \mathrm{C}=(9 / 2)(9 \times 10^{-12} \mathrm{~F})$$

$$\mathrm{C}_{\text {total }}=40.5 \mathrm{pF}$$
Topic: Capacitor
Q7: A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is
Options: (a) zero
(b) $1 / 2(\mathrm{~K}-1) \mathrm{CV}^{2}$
(c) $\mathrm{CV}^{2}(\mathrm{~K}-1) / \mathrm{K}$
(d) $(\mathrm{K}-1) \mathrm{CV}^{2}$
Answer: (a) zero
Solution: The potential energy of a charged capacitor

$$\mathrm{U}_{\mathrm{i}}=\mathrm{q}^{2} / 2 \mathrm{C}$$\where $$\mathrm{U}_{\mathrm{i}}$$ is the initial potential energy.

If a dielectric slab is slowly introduced, the energy $$=\mathrm{q}^{2} / 2 \mathrm{KC}$$

Once it is taken out, again the energy increases to the old value. Therefore after it is taken out, the potential energy comes back to the old value.

Total work done $$=$$ zero

Answer: (a) zero
Topic: Capacitor
Q8: A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be
Options: (a) $1 / 2$
(b) 1
(c) 2
(d) $1 / 4$
Answer: (a) $1 / 2$
Solution: Let $$\mathrm{E}$$ be emf of the battery

Work done by the battery $$\mathrm{W}=\mathrm{CE}^{2}$$

Energy stored in the capacitor

$$\mathrm{U}=1 / 2 \mathrm{CE}^{2}$$

$$\mathrm{U} / \mathrm{W}=1 / 2 \mathrm{CE}^{2} / \mathrm{CE}^{2}=1 / 2$$

Answer: (a) $1 / 2$
Topic: Capacitor
Q9: A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor
Options: (a) decreases
(b) remains unchanged
(c) becomes infinite
(d) increases
Answer: (b) remains unchanged
Solution: With air as dielectric, $$\mathrm{C}=\varepsilon_{0} \mathrm{~A} / \mathrm{d}$$

With space partially filled, $$\mathrm{C}^{\prime}=\varepsilon_{0} \mathrm{~A} /(\mathrm{d}-\mathrm{t})=\varepsilon_{0} \mathrm{~A} / \mathrm{d}=\mathrm{C}$$

Aluminium is a good conductor. Its sheet introduced between the plates of a capacitor is of negligible thickness. Therefore, capacity remains unchanged.

Answer: (b) remains unchanged
Topic: Capacitor
Q10: If the electric flux entering and leaving an enclosed surface respectively is $\Phi_{1}$ and $\Phi_{2}$, the electric charge inside the surface will be
Options: (a) $(\Phi_{2}-\Phi_{1}) \varepsilon_{0}$
(b) $(\Phi_{1}+\Phi_{2}) / \varepsilon_{0}$
(c) $(\Phi_{2}-\Phi_{1}) / \varepsilon_{0}$
(d) $(\Phi_{1}+\Phi_{2}) \varepsilon_{0}$
Answer: (a) $(\Phi_{2}-\Phi_{1}) \varepsilon_{0}$
Solution: According to Gauss theorem,

$(\Phi_{2}-\Phi_{1})=\mathrm{Q} / \varepsilon_{0}$

$\Rightarrow \mathrm{Q}=(\Phi_{2}-\Phi_{1}) \varepsilon_{0}$

The flux enters the enclosure if one has a negative charge $(-\mathrm{q}_{2})$ and flux goes out if one has a + ve charge $(+\mathrm{q}_{1})$. As one does not know whether $\Phi_{1}>\Phi_{2}, \Phi_{2}>\Phi_{1}, \mathrm{Q}=\mathrm{q} 1 \sim \mathrm{q} 2$

Answer: (a) $(\Phi_{2}-\Phi_{1}) \varepsilon_{0}$
Topic: Capacitor
Q14: Voltage rating of a parallel plate capacitor is $500 \mathrm{~V}$. Its dielectric can withstand a maximum electric field of $10^{6} \mathrm{~V} \mathrm{~m}^{-1}$. The plate area is $10^{-4} \mathrm{~m}^{2}$. What is the dielectric constant if the capacitance is $15 \mathrm{pF}$ ? (given $\varepsilon_{0}=8.86 \times 10^{-12} C^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )
Options: (a) 3.8
(b) 8.5
(c) 6.2
(d) 4.5
Answer: (b) 8.5
Solution: $\mathrm{C}=\mathrm{K} \varepsilon_{0} \mathrm{~A} / \mathrm{d}$ or $\mathrm{K}=\mathrm{CV} / \varepsilon_{0} \mathrm{AE}_{\max }$

$\mathrm{K}=(15 \times 10^{-12} \times 500) /(8.86 \times 10^{-12} \times 10^{-4} \times 10^{6})=8.5$

Answer: (b) 8.5
Topic: Capacitor
Q15:In free space, a particle $A$ of charge $1 \mu \mathrm{C}$ is held fixed at a point $\mathrm{P}$. Another particle $B$ of the same charge and mass $4 \mathrm{~g}$ is kept at a distance of $1 \mathrm{~mm}$ from $P$. If $B$ is released, then its velocity at a distance of 9 $\mathrm{mm}$ from $P$ is

(take $1 / 4 \pi \varepsilon_{0}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$ )

(a) $3.0 \times 10^{4} \mathrm{~m} / \mathrm{s}$

(b) $1.0 \mathrm{~m} / \mathrm{s}$

(c) $1.5 \times 10^{2} \mathrm{~m} / \mathrm{s}$

(d) $2.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$

\section*{Solution}
Using work energy theorem

$\mathrm{W}_{\mathrm{E}}=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}=(1 / 2) \mathrm{mv}^{2}$ or $\mathrm{kq}_{1} \mathrm{q}_{2}[\left(1 / \mathrm{r}_{1}\right)-\left(1 / \mathrm{r}_{2}\right)]=(1 / 2) \mathrm{mv}^{2}$

$$
\begin{aligned}
& \mathrm{W}_{\mathrm{E}}=\left(9 \times 10^{9}\right) \times\left(1 \times 10^{-6}\right)^{2}\left\{\left(1 / 10^{-3}\right)-\left(1 /\left(9 \times 10^{-3}\right)\right\}=(1 / 2) \mathrm{mv}^{2}\right. \\
& \Rightarrow\left(9 \times 10^{9} \times 2 \times 10^{-12}\right) /\left(4 \times 10^{-6}\right)=\mathrm{v}^{2} \Rightarrow \mathrm{v}=2.0 \times 10^{3} \mathrm{~ms}^{-1}
\end{aligned}
$$

Answer: (d) $2.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$
Answer: (d) $2.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$
Solution: Using work energy theorem

$\mathrm{W}_{\mathrm{E}}=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}}=(1 / 2) \mathrm{mv}^{2}$ or $\mathrm{kq}_{1} \mathrm{q}_{2}[\left(1 / \mathrm{r}_{1}\right)-\left(1 / \mathrm{r}_{2}\right)]=(1 / 2) \mathrm{mv}^{2}$

$$
\begin{aligned}
& \mathrm{W}_{\mathrm{E}}=\left(9 \times 10^{9}\right) \times\left(1 \times 10^{-6}\right)^{2}\left\{\left(1 / 10^{-3}\right)-\left(1 /\left(9 \times 10^{-3}\right)\right\}=(1 / 2) \mathrm{mv}^{2}\right. \\
& \Rightarrow\left(9 \times 10^{9} \times 2 \times 10^{-12}\right) /\left(4 \times 10^{-6}\right)=\mathrm{v}^{2} \Rightarrow \mathrm{v}=2.0 \times 10^{3} \mathrm{~ms}^{-1}
\end{aligned}
$$

Answer: (d) $2.0 \times 10^{3} \mathrm{~m} / \mathrm{s}$
Topic: Communication System
Q1: The wavelength of the carrier waves in a modern optical fibre communication network is close to
Options: (a) 600 nm
(b) 2400 nm
(c) 1500 nm
(d) 900 nm

Answer: Answer: (c) 1500 nm
Solution: Fibre optics communication is mainly conducted in a wavelength range from 1260 nm to 1625 nm
Topic: Communication System
Q2: The physical sizes of the transmitter and receiver antenna in a communication system are
Options: (a) inversely proportional to the modulation frequency
(b) proportional to the carrier frequency
(c) independent of both carrier and modulation frequency
(d) inversely proportional to the carrier frequency

Answer: Answer: (d) inversely proportional to the carrier frequency
Solution: The physical size of the transmitter and receiver antenna is inversely proportional to the carrier frequency.
Topic: Communication System
Q3: A telephonic communication service is working at a carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?
Options: (a) 2 × 10^3
(b) 2 × 10^4
(c) 2 × 10^5
(d) 2 × 10^6

Answer: Answer: (c) 2 × 10^5
Solution: Frequency of carrier wave = 10 × 10^9 Hz
Available bandwidth 10% of 10 × 10^9 Hz = 10^9 Hz
Bandwidth for each telephonic channel 5 kHz = 5 × 10^3 Hz
Number of channels = 10^9 / (5 × 10^3) = 2 × 10^5
Topic: Communication System
Q4: A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Given: radius of earth = 6.4 × 10^6 m )
Options: (a) 65 km
(b) 48 km
(c) 40 km
(d) 80 km

Answer: Answer (a) 65 km
Solution: Maximum distance upto which signal can be broadcasted is
d_max = √(2 R h_T) + √(2 R h_R)
where h_T and h_R are heights of transmitter tower and height of receiver respectively.
Putting all values, we get
d_max = √(2 × 6.4 × 10^6) [√140 + √40]
d_max = 65 km
Topic: Communication System
Q5: A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal are
Options: (a) 2005 kHz, 2000 kHz, and 1995 kHz
(b) 2000 kHz and 1995 kHz
(c) 2 MHz only
(d) 2005 kHz and 1995 kHz

Answer: Answer: (a) 2005 kHz, 2000 kHz, and 1995 kHz
Solution: Given, f_m = 5 kHz, f_c = 2 MHz = 2000 kHz
The frequencies of the resultant signal are f_c + f_m = (2000 + 5) kHz = 2005 kHz, f_c = 2000 kHz, and f_c - f_m = (2000 - 5) kHz = 1995 kHz
Topic: Communication System
Q6: Consider telecommunication through optical fibres. Which of the following statements is not true?
Options: (a) Optical fibres can be of graded refractive index
(b) Optical fibres are subject to electromagnetic interference from outside
(c) Optical fibres have extremely low transmission loss
(d) Optical fibres may have a homogeneous core with suitable cladding

Answer: Answer: (b) Optical fibres are not subject to electromagnetic interference from outside
Topic: Communication System
Q7: The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for a licence, what broadcast frequency will you allot?
Options: (a) 2750 kHz
(b) 2900 kHz
(c) 2250 kHz
(d) 2000 kHz

Answer: Answer: (d) 2000 kHz
Solution: 10% of f_c = 250 kHz
Hence, range of signal = (2500 ± 250 kHz) = 2250 kHz to 2750 kHz
10% of 2000 kHz = 200 kHz
Range is 1800 kHz to 2200 kHz
Hence, allocated broadcast frequency will be 2000 kHz
Topic: Communication System
Q8: In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of bandwidth 6 MHz is (Take velocity of light c=3×10^8 m/s, h=6.6×10^(-34) Js)
Options: (a) 3.75 × 10^6
(b) 4.87 × 10^5
(c) 6.25 × 10^5
(d) 3.86 × 10^6

Answer: Answer: (c) 6.25 × 10^5
Solution: f = C / λ = (3 × 10^8) / (8 × 10^-7) = 3.75 × 10^14 Hz
1% of f = 3.75 × 10^17 Hz = 3.75 × 10^6 MHz
Number of channels = (3.75 × 10^6) / 6
Number of channels = 6.25 × 10^5