Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Communication System
Q9: A signal of frequency 20 kHz and peak voltage of 5 volts is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 volts. Choose the correct statement.
Options: (a) Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz
(b) Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz
(c) Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz
(d) Modulation index = 5, side frequency bands are at 21.2 kHz and 18.8 kHz

Answer: Answer: (b) Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz
Solution: Modulation index, m = (V_m / V_c) = (5 / 25) = 0.2
Frequency of carrier wave, f_c = 1.2 × 10^3 kHz = 1200 kHz
Frequency of modulate wave = 20 kHz
f1 = f_c - f_m = 1200 - 20 = 1180 kHz
f2 = f_c + f_m = 1200 + 20 = 1220 kHz
Topic: Communication System
Q10: An audio signal consists of two distinct sounds: one a human speech signal in the frequency band of 200 Hz to 2700 Hz, while the other is a high-frequency music signal in the frequency band of 10200 Hz to 15200 Hz. The ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech is
Options: (a) 2
(b) 5
(c) 6
(d) 3

Answer: Answer: (c) 6
Solution: Band width for both signals = 15200 Hz - 200 Hz = 15000 Hz
Band width for human speed 2700 Hz - 200 Hz = 2500 Hz
The ratio = 15000 / 2500 = 6
Topic: Communication System
Q11: In an amplitude modulator circuit, the carrier wave is given by, C(t)=4 sin(20000πt), while modulating signal is given by, m(t)=2 sin(2000πt). The values of modulation index and lower sideband frequency are
Options: (a) 0.4 and 10 kHz
(b) 0.5 and 9 kHz
(c) 0.3 and 9 kHz
(d) 0.5 and 10 kHz

Answer: Answer: (b) 0.5 and 9 kHz
Solution: Given C(t) = 4 sin(20000πt) ⇒ A_c = 4
m(t) = 2
Topic: Elasticity
Q1: A compressive force, $F$ is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by $\Delta T$. The net change in its length is zero. Let $L$ be the length of the rod, $A$ is its area of cross-section. $Y$ is Young's modulus, and $lpha$ is its coefficient of linear expansion. Then, $F$ is equal to
Options: a. $\mathrm{L}^{2} \mathrm{Y} lpha \Delta \mathrm{T}$
b. $\mathrm{AY} / lpha \Delta \mathrm{T}$
c. AY $lpha \Delta \mathrm{T}$
d. LAY $lpha \Delta \mathrm{T}$
Answer: c
Solution: Thermal expansion, $\Delta \mathrm{L}=\mathrm{L} lpha \Delta \mathrm{T}$\ Let $\Delta \mathrm{L}'$ be the compression produced by applied force\ $\mathrm{Y}=\mathrm{FL} / \mathrm{A} \Delta \mathrm{L}' \Rightarrow \mathrm{F}=\mathrm{YA} \Delta \mathrm{L}' / \mathrm{L}$\ Net change in length $=0 \Rightarrow \Delta \mathrm{L}'=$\ From (1),(2) and (3)\ $\mathrm{F}=\mathrm{YAx}(\mathrm{L} lpha \Delta \mathrm{T}) / \mathrm{L}=\mathrm{YA} lpha \Delta \mathrm{T}$
Topic: Elasticity
Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 \mathrm{~N}$ to the lower end. The weight stretches the wire by $1 \mathrm{~mm}$. Then the elastic energy stored in the wire is
Options: a. $0.2 \mathrm{~J}$
b. $10 \mathrm{~J}$
c. $20 \mathrm{~J}$
d. $0.1 \mathrm{~J}$
Answer: d
Solution: Elastic energy per unit volume $=1 / 2 \mathrm{x}$ stress $\mathrm{x}$ strain\ Elastic Energy $=1 / 2 \mathrm{x}$ stress $\mathrm{x}$ strain $\mathrm{x}$ volume\ $=1 / 2 	imes F / A 	imes(\Delta L / L) 	imes(A L)$\ $=1 / 2 	imes$ F $\Delta \mathrm{L}$\ $=1 / 2 	imes 200 	imes 10^{-3}$\ Elastic Energy $=0.1 \mathrm{~J}$
Topic: Elasticity
Q3: A rod of length $L$ at room temperature and uniform area of cross-section $A$, Is made of a metal having a coefficient of linear expansion $lpha$. It is observed that an external compressive force $F$ is applied to each of its ends, prevents any change in the length of the rod when its temperature rises by $\Delta T \mathrm{~K}$. Young's modulus, $Y$ for this metal is
Options: a.F/A $lpha \Delta \mathrm{T}$
b.F/A $lpha(\Delta \mathrm{T}-273)$
c. F/2A $lpha \Delta$
d. $2 \mathrm{~F} / \mathrm{A} lpha \Delta \mathrm{T}$
Answer: a
Solution: Young's Modulus $\mathrm{Y}=$ stress/strain $=(\mathrm{F} / \mathrm{A}) /(\Delta l / l)$\ Substituting the coefficient of linear expansion\ $lpha=\Delta l /(l \Delta \mathrm{T})$\ $\Delta l / l=lpha \Delta \mathrm{T}$\ $\mathrm{Y}=(\mathrm{F} / \mathrm{A} lpha \Delta \mathrm{T})$
Topic: Elasticity
Q4: Young's moduli of two wires A and B are in the ratio 7:4. Wire A is $2 \mathrm{~m}$ long and has radius $R$. Wire $B$ is $1.5 \mathrm{~m}$ long and has a radius of $2 \mathrm{~mm}$. If the two wires stretch by the same length for a given load, then the value of $R$ is close $t$
Options: a. $1.5 \mathrm{~mm}$
b. $1.9 \mathrm{~mm}$
c. $1.7 \mathrm{~mm}$
d. $1.3 \mathrm{~mm}$
Answer: c
Solution: $\Delta_{1}=\Delta_{2}$\ $\left(\mathrm{Fl}{1} / \pi \mathrm{r}{1}^{2} \mathrm{y}{1}ight)=\left(\mathrm{Fl}{2} / \pi \mathrm{r}{2}^{2} \mathrm{y}{2}ight)$\ $2 /(\mathrm{R}^{2} 	imes 7)=1.5 /(2^{2} 	imes 4)$\ $\mathrm{R}=1.75 \mathrm{~mm}$
Topic: Elasticity
Q5: The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a $400 \mathrm{~N}$ load without exceeding its elastic limit?
Options: a. $1 \mathrm{~mm}$
b. $1.15 \mathrm{~mm}$
c. $0.90 \mathrm{~mm}$
d. $1.36 \mathrm{~mm}$
Answer: b
Solution: Stress $=\mathrm{F} / \mathrm{A}$\ Stress $=400 	imes 4 / \pi \mathrm{d}^{2}$\ $=379 	imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$\ $\mathrm{d}^{2}=(400 	imes 4) /(379 	imes 10^{6} \pi)$\ $\mathrm{d}=1.15 \mathrm{~mm}$
Topic: Elasticity
Q6: A uniform cylindrical rod of length $L$ and radius $r$, is made from a material whose Young's modulus of Elasticity equals $Y$. When this rod is heated by temperature $T$ and simultaneously subjected to a net longitudinal compressional force $F$, its length remains unchanged. The coefficient of volume expansion, of the material of the rod is nearly equal to
Options: a. $9 \mathrm{~F} /(\pi \mathrm{r}^{2} \mathrm{YT})$
b. $6 \mathrm{~F} /(\pi \mathrm{r}^{2} \mathrm{YT})$
c. $3 \mathrm{~F} /(\pi \mathrm{r}^{2} \mathrm{YT})$
d. $\mathrm{F} /(3 \pi \mathrm{r}^{2} \mathrm{YT})$
Answer: c
Solution: $\mathrm{Y}=\left(\mathrm{F} / \pi \mathrm{r}^{2}ight) 	imes \mathrm{L} / \Delta \mathrm{L}$\ $\Delta \mathrm{L}=\mathrm{F} / / \pi \mathrm{r}^{2} \mathrm{Y}-------(1)$\ Change in length due to temperature change\ $\Delta \mathrm{L}=\mathrm{L} lpha \Delta \mathrm{T}$\ From equa (1) and (2)\ $\mathrm{L} lpha \Delta \mathrm{T}=\mathrm{FL} / \mathrm{AY}$\ $lpha=\mathrm{F} / \mathrm{AY} \Delta \mathrm{T}$\ $lpha=\mathrm{F} / \pi \mathrm{r}^{2} \mathrm{YT}$\ Coefficient of volume expansion\ $3 \propto=3 \mathrm{~F} / \pi \mathrm{r}^{2} \mathrm{YT}$
Topic: Elasticity
Q7: The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
Options: (a) length $=200 \mathrm{~cm}$, diameter $=2 \mathrm{~mm}$
(b) length $=300 \mathrm{~cm}$, diameter $=3 \mathrm{~mm}$
(c) length $=50 \mathrm{~cm}$, diameter $=0.5 \mathrm{~mm}$
(d) length $=100 \mathrm{~cm}$, diameter $=1 \mathrm{~mm}$
Answer: c
Solution: Since all four wires are made from the same material Young's modulus will be the same.\ $\Delta \mathrm{L} \propto \mathrm{L} / \mathrm{D}^{2}$\ In (a) $\mathrm{L} / \mathrm{D}^{2}=200 /(0.2)^{2}=5 	imes 10^{3} \mathrm{~cm}^{-1}$\ In (b) $\mathrm{L} / \mathrm{D}^{2}=300 /(0.3)^{2}=3.3 	imes 10^{3} \mathrm{~cm}^{-1}$\ In (c) $\mathrm{L} / \mathrm{D}^{2}=50 /(0.5)^{2}=20 	imes 10^{3} \mathrm{~cm}^{-1}$\ In (d) $\mathrm{L} / \mathrm{D}^{2}=100 /(0.1)^{2}=10 	imes 10^{3} \mathrm{~cm}^{-1}$
Topic: Elasticity
Q8: A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains the same, the stress in the leg will change by a factor of
Options: (a) $1 / 9$
(b) 81
(c) $1 / 81$
(d) 9
Answer: d
Solution: Stress $=$ Force $/$ Area\ Stress $=$ Force $/ L^{2}$\ Now, dimensions increases by a factor of 9\ Now, $\mathrm{S}=$ (volume $\mathrm{x}$ density) $\mathrm{xg} / \mathrm{L}^{2}$\ $\mathrm{S}=\mathrm{L}^{3} 	imes ho \mathrm{g} / \mathrm{L}^{2}=\mathrm{L} ho \mathrm{g}$\ Stress S $\propto \mathrm{L}$\ $\mathrm{S}{2} / \mathrm{S}{1}=\mathrm{L}{2} \mathrm{~L}{1}=9 \mathrm{~L}{1} / \mathrm{L}{1}=9$
Topic: Elasticity
Q9. A solid sphere of radius $r$ made of a soft material of bulk modulus $K$ is surrounded by a liquid in a cylindrical container. A massless piston of the area a floats on the surface of the liquid, covering an entire cross-section of the cylindrical container. When a mass $m$ is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere is $\mathrm{Mg} / lpha \mathrm{AB}$. Find the value of $lpha$.
Options: a. 4
b. 5
c. 3
d. 2
Answer: c
Solution: Increase in pressure is $\Delta \mathrm{p}=\mathrm{Mg} / \mathrm{A}$\ Bulk modulus is $\mathrm{B}=\Delta \mathrm{p} /(\Delta \mathrm{V} / \mathrm{V})$\ $\Delta \mathrm{V} / \mathrm{V}=\Delta \mathrm{p} / \mathrm{B}=\mathrm{Mg} / \mathrm{AB}-----(1)$\ The volume of the sphere is $\mathrm{V}=(4 / 3) \pi \mathrm{R}^{3}$\ $\Delta \mathrm{V} / \mathrm{V}=3(\Delta \mathrm{R} / \mathrm{R})$\ From equation (1) we get\ $\mathrm{Mg} / \mathrm{AB}=3(\Delta \mathrm{R} / \mathrm{R})$\ Therefore $lpha=3$
Topic: Elasticity
Q10: A steel wire having a radius of $2.0 \mathrm{~mm}$, carrying a load of $4 \mathrm{~kg}$, is hanging from a ceiling. Given that $\mathrm{g}$ $=3.1 \pi \mathrm{m} / \mathrm{s}^{2}$, what will be the tensile stress that would be developed in the wire?
Options: a. $4.8 	imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
b. $3.1 	imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
c. $6.2 	imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
d. $5.2 	imes 10^{6} \mathrm{~N} / \mathrm{m}^{2}$
Answer: b
Solution: Tensile stress $=$ Force/Area\ Tensile stress $=(4)(3.1 \pi) / \pi\left(2 	imes 10^{-3}ight)^{2}$\ Tensile stress $=3.1 	imes 10^{6} \mathrm{Nm}^{-2}$
Topic: Elasticity
Q11: A steel rail of length $5 \mathrm{~m}$ and area of cross-section $40 \mathrm{~cm}^{2}$ is prevented from expanding along its length while the temperature rises by $10^{\circ} \mathrm{C}$. If the coefficient of linear expansion and Young's modulus of steel is $1.2 	imes 10^{-5} \mathrm{~K}^{-1}$ and $2 	imes 10^{11} \mathrm{Nm}^{-2}$ respectively, the force developed in the rail is approximately
Options: a. $2 	imes 10^{9} \mathrm{~N}$
b. $3 	imes 10^{-5} \mathrm{~N}$
c. $2 	imes 10^{7} \mathrm{~N}$
d. $1 	imes 10^{5} \mathrm{~N}$
Answer: d
Solution: $\mathrm{A}=40 \mathrm{~cm}^{2}=4 	imes 10^{-3} \mathrm{~m}^{2}$\ $\Delta \mathrm{T}=10^{\circ} \mathrm{C}$\ $\mathrm{Y}=2 	imes 10^{11} \mathrm{Nm}^{-2}$\ $lpha=1.2 	imes 10^{-5} \mathrm{~K}^{-1}$\ Force $=\mathrm{YA} lpha \Delta \mathrm{T}$\ Force $=\left(2 	imes 10^{11}ight)\left(4 	imes 10^{-3}ight)\left(1.2 	imes 10^{-5}ight)(10)=9.6 	imes 10^{4} \mathrm{~N}$\ Force $pprox 1 	imes 10^{5} \mathrm{~N}$
Topic: Elasticity
Q12: If $\mathrm{S}$ is the stress and $\mathrm{Y}$ is Young's Modulus of the material of the wire, the energy stored in the wire per unit volume is
Options: a. $2 \mathrm{Y} / \mathrm{S}$
b. $\mathrm{S} / 2 \mathrm{Y}$
c. $2 S^{2} Y$
d. $\mathrm{S}^{2} / 2 \mathrm{Y}$
Answer: d
Solution: Young's modulus, $\mathrm{Y}=$ Stress $/$ Strain\ $\Rightarrow$ Strain $=$ Stress $/ \mathrm{Y}=\mathrm{S} / \mathrm{Y}$\ Energy stored per unit volume $=1 / 2 \mathrm{x}$ stress $\mathrm{x}$ strain\ $=$ Stress $\mathrm{x}$ Stress $/ 2 \mathrm{Y}=\mathrm{S}^{2} / 2 \mathrm{Y}$ (since Young's modulus, $\mathrm{Y}=$ Stress $/$ Strain)
Topic: Elasticity
Q13Two wires are made of the same material and have the same volume. However, wire 1 has a cross-sectional area $A$ and wire 2 has cross-sectional area $3 A$. If the length of the wire 1 increases by $\Delta x$ on applying force $F$, how much force is needed to stretch wire 2 by the same amount?
Options: a. F
b. $4 F$
c. $6 F$
d. $9 F$
Answer: d
Solution: For the same material, Young's modulus is the same and it is given that the volume is the same and the area of the cross-section for the wire $\mathrm{L}{1}$ is and that of $\mathrm{L}{2}$ is $3 A$\ $V=V_{1}=V_{2}$\ $V=A 	imes L_{1}=3 A 	imes L_{2} \Rightarrow L_{2}=L_{1} / 3$\ $Y=(F / A) /(\Delta L / L)$\ $F_{1}=YA(\Delta L_{1} / L_{1})$\ $F_{2}=Y 3 A(\Delta L_{2}, L_{2})$\ Given $\Delta L_{1}=\Delta L_{2}=\Delta x$ (for the same extension)\ $F_{2}=Y 3 A(\Delta x /(L_{1} / 3))=9 (YA \Delta x / L_{1})=9 F_{1}=9 F$
Topic: Elasticity
Q14 A wire elongates by $l \mathrm{~mm}$ when a load $W$ is hanged from it. If the wire goes over a pulley and two weighs $W$ each is hung at the two ends, the elongation of the wire will be (in $\mathrm{mm}$ )
Options: a. $l / 2$
b. l
c. 2
d. Zero
Answer: b
Solution: $Y=(	ext{Force} 	imes L) /(A 	imes l)=WL / Al$\ $l=WL / AY$\ Due to the arrangement of the pulley, the length of wire is $L / 2$ on each side and so the elongation will be $l / 2$. For both sides, elongation $=l$\ Answer: (b) $l$
Topic: Electromagnetic Waves
Q1: The mean intensity of radiation on the surface of the Sun is about $10^{8} \mathrm{~W} / \mathrm{m}^{2}$. The RMS value of the corresponding magnetic field is closest to
Options: a. $10^{-2} \mathrm{~T}$
b. $1 \mathrm{~T}$
c. $10^{-4} \mathrm{~T}$
d. $10^{2} \mathrm{~T}$
Answer: c
Solution: Mean intensity $\mathrm{I}=(\mathrm{B}_{\text {rms }}^{2} / \mu_{0}) \mathrm{c}$\ $\mathrm{B}_{\text {rms }}^{2}=(10^{8} \times 4 \pi \times 10^{-7}) /(3 \times 10^{8})$\ $\mathrm{B}_{\mathrm{rms}} \approx 10^{-4} \mathrm{~T}$\ Answer: (c) $10^{-4} \mathrm{~T}$
Topic: Electromagnetic Waves
Q2: A plane electromagnetic wave travels in free space along the $x$-direction. The electric field component of the wave at a particular point of space and time is $E=6 \mathrm{~V} \mathrm{~m}^{-1}$ along the $y$-direction. Its corresponding magnetic field component, $B$ would be
Options: a. $6 \times 10^{-8} \mathrm{~T}$ along the $\mathrm{x}$-direction
b. $2 \times 10^{-8} \mathrm{~T}$ along the $\mathrm{y}$-direction
c. $2 \times 10^{-8} \mathrm{~T}$ along the $\mathrm{z}$-direction
d. $6 \times 10^{-8} \mathrm{~T}$ along the $\mathrm{z}$-direction
Answer: c
Solution: The direction of electromagnetic wave travelling is given by\ $$\bar{k}=\bar{E} \times \bar{B}$$\ As the wave is travelling along $\mathrm{x}$-direction and $\mathrm{E}$ is along the $\mathrm{y}$-direction.\ So B must point towards the z-direction.\ The magnetic field, $\mathrm{B}=\mathrm{E} / \mathrm{c}=6 /(3 \times 10^{8})=2 \times 10^{-8} \mathrm{~T}$\ Answer: (c) $2 \times 10^{-8} \mathrm{~T}$ along the z-direction
Topic: Electromagnetic Waves
Q3: $50 \mathrm{~W} / \mathrm{m}^{2}$ energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy ( $\mathbf{2 5 \%}$ ) is reflected from the surface and the rest is absorbed. The force exerted on $1 \mathrm{~m}^{2}$ surface area will be close to $\left(\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right.$ )
Options: a. $20 \times 10^{-8} \mathrm{~N}$
b. $10 \times 10^{-8} \mathrm{~N}$
c. $35 \times 10^{-8} \mathrm{~N}$
d. $15 \times 10^{-8} \mathrm{~N}$
Answer: a
Solution: Given energy density $=50 \mathrm{~W} / \mathrm{m}^{2}$\ Here, change in momentum $\Delta p=p_{f}-p_{i}$\ $\Delta \mathrm{p}=\left(-\mathrm{p}_{\mathrm{i}} / 4\right)-\mathrm{p}_{\mathrm{i}}$\ $\Delta \mathrm{p}=-5 \mathrm{p}_{\mathrm{i}} / 4 \because \mathrm{p}_{\mathrm{i}}=\mathrm{E} / \mathrm{c}=50 \mathrm{~W} / \mathrm{s} /(3 \times 10^{8})$\ $\Delta \mathrm{p} / \Delta \mathrm{t}=\mathrm{F}=\left|-5 \mathrm{p}_{\mathrm{i}} / 4\right|=(5 / 4) \times\left(50 /(3 \times 10^{8}=20.8 \times 10^{-8} \mathrm{~N} \approx 20 \times 10^{-8} \mathrm{~N}\right.$\ Answer: (a) $20 \times 10^{-8} \mathrm{~N}$
Topic: Electromagnetic Waves
Q4: A red LED emits light at 0.1 watts uniformly around it. The amplitude of the electric field of the light at a distance of $1 \mathrm{~m}$ from the diode is
Options: a. $5.48 \mathrm{~V} / \mathrm{m}$
b. $7.75 \mathrm{~V} / \mathrm{m}$
c. $1.73 \mathrm{~V} / \mathrm{m}$
d. $2.45 \mathrm{~V} / \mathrm{m}$
Answer: d
Solution: The intensity of light, $\mathrm{I}=\mathrm{u}_{\mathrm{av}} \mathrm{c}$\ Also, $\mathrm{I}=\mathrm{P} / 4 \pi \mathrm{r}^{2}$ and $\mathrm{u}_{\mathrm{av}}=1 / 2 \varepsilon_{0} \mathrm{E}_{0}{ }^{2}$\ $\therefore \mathrm{P} / 4 \pi \mathrm{r}^{2}=(1 / 2) \varepsilon_{0} \mathrm{E}_{0}{ }^{2} \mathrm{c}$\ Or $\quad E_{0}=\sqrt{\frac{2 P}{4 \pi \epsilon_{0} r^{2} c}}$\ Here, $\mathrm{P}=0.1 \mathrm{~W}, \mathrm{r}=1 \mathrm{~m}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$\ $1 / 4 \pi \varepsilon_{0}=9 \times 10^{9} \mathrm{~N} \mathrm{C}^{-2} \mathrm{~m}^{2}$\ $$ E_{0}=\sqrt{\frac{2 \times 0.1 \times 9 \times 10^{9}}{1^{2} \times 3 \times 10^{8}}} =\sqrt{6}$$\ $=2.45 \mathrm{~V} \mathrm{~m}^{-1}$\ Answer: (d) $2.45 \mathrm{~V} / \mathrm{m}$
Topic: Electromagnetic Waves
Q5: Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists
Options: a. (i) (ii) (iii) (iv)
b. (iv) (iii) (ii) (i)
c. (i) (ii) (iv) (iii)
d. (iii) (ii) (i) (iv)
Answer: a
Solution: Infrared waves are used to treat muscular strain. Radio waves are used for broadcasting. X-rays are used to detect the fracture of bones. Ultraviolet rays are absorbed by the ozone layer of the atmosphere.\ Answer: (a)
Topic: Electromagnetic Waves
Q6: During the propagation of electromagnetic waves in a medium
Options: a. both electric and magnetic energy densities are zero
b. electric energy density is double of the magnetic energy density
c. electric energy density is half of the magnetic energy density
d. electric energy density is equal to the magnetic energy density
Answer: d
Solution: Solution: In an em wave, energy is equally divided between the electric and the magnetic fields.\ Answer: (d) The electric energy density is equal to the magnetic energy density
Topic: Electromagnetic Waves
Q7: The magnetic field in a travelling electromagnetic wave has a peak value of $20 \mathrm{nT}$. The peak value of electric field strength is
Options: a. $12 \mathrm{~V} / \mathrm{m}$
b. $3 \mathrm{~V} / \mathrm{m}$
c. $6 \mathrm{~V} / \mathrm{m}$
d. $9 \mathrm{~V} / \mathrm{m}$
Answer: c
Solution: In an electromagnetic wave, the peak value of the electric field $\left(\mathrm{E}_{0}\right)$ and peak value of magnetic field $\left(\mathrm{B}_{0}\right)$ are related by\ $\mathrm{E}_{0}=\mathrm{B}_{0} \mathrm{C}$\ $\mathrm{E}_{0}=\left(20 \times 10^{-9} \mathrm{~T}\right)\left(3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\right)=6 \mathrm{~V} / \mathrm{m}$\ Answer: (c) 6 V/m
Topic: Electromagnetic Waves
Q8: An electromagnetic wave of frequency $=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with permittivity $=4.0$. Then
Options: a. wavelength is doubled and the frequency remains unchanged
b. wavelength is doubled and frequency becomes half
c. wavelength is halved and frequency remains unchanged
d. wavelength and frequency both remain unchanged
Answer: c
Solution: Solution: In an em wave, energy is equally divided between the electric and the magnetic fields.\ Answer: (d) The electric energy density is equal to the magnetic energy density
Topic: Electromagnetic Waves
Q9: Electromagnetic waves are transverse in nature is evident by
Options: a. polarization
b. interference
c. reflection
d. diffraction
Answer: a
Solution: Answer: (a) Polarization proves the transverse nature of electromagnetic waves.
Topic: Electromagnetic Waves
Q10: The energy associated with electric field is $\left(U_{E}ight)$ and with magnetic field is $\left(U_{B}ight)$ for an electromagnetic wave in free space. Then
Options: a. $U_{E}>U_{B}$
b. $U_{E}=U_{B} / 2$
c. $U_{E}=U_{B}$
d. $U_{E}<U_{B}$
Answer: c
Solution: Answer: (c) $\mathbf{U}_{\mathrm{E}}=\mathbf{U}_{\mathrm{B}}$
Topic: Electromagnetic Waves
Q11: A $27 \mathrm{~mW}$ laser beam has a cross-sectional area of $10 \mathrm{~mm}^{2}$. The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space $\varepsilon_{0}=9 \times 10^{-12}$ SI units, speed of light $\mathbf{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
Options: a. $2 \mathrm{kV} / \mathrm{m}$
b. $0.7 \mathrm{kV} / \mathrm{m}$
c. $1 \mathrm{kV} / \mathrm{m}$
d. $1.4 \mathrm{kV} / \mathrm{m}$
Answer: d
Solution: The intensity of the electromagnetic wave is given by\ $\mathrm{I}=\operatorname{Power}(\mathrm{P}) / \operatorname{Area}(\mathrm{A})=1 / 2 \varepsilon_{0} \mathrm{E}^{2} \mathrm{C}$\ $$ \begin{aligned} & E=\sqrt{\frac{2 P}{A E_{0} c}}=\sqrt{\frac{2 \times 27 \times 10^{-3}}{10 \times 10^{-6} \times 9 \times 10^{-12} \times 3 \times 10^{8}}} \ & E=\sqrt{0.2 \times 10^{7}}=\sqrt{2 \times 10^{6}} \end{aligned} $$\ $\mathrm{E}=1.4 \mathrm{kV} / \mathrm{m}$\ \section*{Answer: (d) $1.4 \mathrm{kV} / \mathrm{m}$}
Topic: Electromagnetic Waves
Q12: The RMS value of the electric field of the light coming from the sun is $720 \mathrm{~N} / \mathrm{C}$. The average total energy density of the electromagnetic wave is
Options: a) $3.3 \times 10^{-3} \mathrm{~J} / \mathrm{m}^{3}$
b) $4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^{3}$
c) $6.37 \times 10^{-9} \mathrm{~J} / \mathrm{m}^{3}$
d) $81.35 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}$
Answer: b
Solution: $\mathrm{U}=\left(8.85 \times 10^{-12}\right) \times(720)^{2}=4.58 \times 10^{-6} \mathrm{Jm}^{-1}$\ \section*{Answer (b) $4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^{3}$}
Topic: Electromagnetic Waves
Q13: Which of the following are not electromagnetic waves?
Options: a) cosmic rays
b) gamma rays
c) $\beta$-rays
d) X-rays
Answer: c
Solution: Answer: (c) $\beta$-rays are not electromagnetic waves
Topic: Electromagnetic Waves
Q14: If a source of power $4 \mathrm{~kW}$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called
Options: a) microwaves
b) y-rays
c) X-rays
d) ultraviolet rays
Answer: c
Solution: Solution\ \section*{Answer: (c) X-rays}
Topic: Electromagnetic Waves
Q15: An EM wave from the air enters a medium. The electric fields are\ $$\begin{aligned} \vec{E}_{1} & =E_{01} \hat{x} \cos \left[2 \pi f\left(\frac{z}{c}-t\right)\right] \quad \text { in air and } \\ \vec{E}_{2} & =E_{02} \hat{x} \cos [k(2 z-c t)] \end{aligned}$$\ in medium, where the wavenumber $k$ and frequency $f$ refer to their values in air. The medium is non-magnetic. If $\varepsilon_{\mathrm{r} 1}$ and $\varepsilon_{\mathrm{r} 2}$ refer to relative permittivities of air and medium respectively, which of the following options is correct?
Options: a) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=4$
b) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=2$
c) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1 / 4$
d) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1 / 2$
Answer: c
Solution: During refraction, frequency remains unchanged, whereas the wavelength gets changed\ $\mathrm{k}^{\prime}=2 \mathrm{k}$ (From equations)\ $2 \pi / \lambda^{\prime}=2\left(2 \pi / \lambda_{0}\right)=\lambda^{\prime}=\lambda_{0} / 2$\ Since, $v=c / 2$\ $$ \frac{1}{\sqrt{\mu_{0} \epsilon_{r 2}}}=\frac{1}{2} \times \frac{1}{\sqrt{\mu_{0} \epsilon_{r 1}}} $$\ $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1 / 4$\ \section*{Answer: (c) $\varepsilon_{\mathrm{r} 1} / \varepsilon_{\mathrm{r} 2}=1 / 4$}
Topic: Kinematics 2D
Q1: A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction.
Options: due north
30° east of north
30° west of north
60° east of north
Answer: (1) due north
Solution: Let the time taken to cross the river be t
t=d/V_s cosθ
For the time t to be minimum, cosθ should be maximum
θ should be 0
Therefore swimmer should swim due north