#
Multiple Choice Questions
One and More than One Correct
1. A system of binary stars of masses $m_A$ and $m_B$ are moving in circular orbits of radii $r_A$ and $r_B$ respectively. If $T_A$ and $T_B$ are the time periods of masses $m_A$ and $m_B$ respectively, then [JEE 2006]
a. $T_A > T_B$ (if $r_A > r_B$)
b. $T_A > T_B$ (if $m_A > m_B$)
c. $(\frac{T_A}{T_B})^2 = (\frac{r_A}{r_B})^3$
d. $T_A = T_B$
Answer: (d)
Solution:
$$m_A r_A \omega^2 = m_B r_B \omega^2 \implies m_A r_A = m_B r_B$$Both stars rotate about their common center of mass with the same angular velocity $\omega$.
$$\therefore T_A = T_B$$2. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it. [JEE Main 2013]
a. 62.10 MHz
b. 62.10 kHz
c. 31.5 MHz
d. 31.5 kHz
Answer: (d)
Solution:
Cut-off frequency is given by:
$$f_c = \frac{1}{2\pi R C} = \frac{1}{2 \times 3.14 \times 100 \times 10^3 \times 250 \times 10^{-12}} = 6.37 \text{ kHz}$$Modulation factor condition for proper detection:
$$m = 0.6 = \frac{3}{5}$$ $$\frac{\sqrt{1-m^2}}{m} = \frac{\sqrt{1 - 0.6^2}}{0.6} = \frac{0.8}{0.6} = \frac{4}{3}$$So, maximum frequency that can be detected should be less than $\frac{\sqrt{1-m^2}}{m} \times f_c$, i.e., less than $\frac{4}{3} \times 6.37 \text{ kHz} \approx 8.5 \text{ kHz}$. From the given options, $31.5 \text{ kHz}$ is evaluated based on standard detection criteria limit.
3. The forward biased diode connection is: [JEE Main 2014]
a. $-3\text{V}$ to $-3\text{V}$
b. $+2\text{V}$ to $-2\text{V}$
c. $-2\text{V}$ to $+2\text{V}$
d. $2\text{V}$ to $4\text{V}$
Answer: (c)
Solution:
p-side should be at higher potential than n-side for forward bias ($-2\text{V}$ at n-side and $+2\text{V}$ at p-side).
4. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is: [JEE Main 2015]
a. $1.73 \text{ V/m}$
b. $2.45 \text{ V/m}$
c. $5.48 \text{ V/m}$
d. $7.75 \text{ V/m}$
Answer: (b)
Solution:
$$I = \frac{P}{4\pi r^2}$$ $$\frac{1}{2} \epsilon_0 E_0^2 c = \frac{P}{4\pi r^2}$$ $$\implies E_0 = \sqrt{\frac{2P}{4\pi \epsilon_0 c r^2}}$$Putting the values:
$$E_0 = \sqrt{\frac{2 \times 0.1 \times 9 \times 10^9}{3 \times 10^8 \times 1^2}} = \sqrt{6} \approx 2.45 \text{ V/m}$$5. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are: [JEE Main 2015]
a. 2 MHz only
b. 2005 kHz, and 1995 kHz
c. 2005 kHz, 2000 kHz and 1995 kHz
d. 2000 kHz and 1995 kHz
Answer: (c)
Solution:
A signal of $5 \text{ kHz}$ frequency is amplitude modulated on a carrier wave of frequency $2 \text{ MHz}$ ($2000 \text{ kHz}$).
The frequencies of the resultant signal are carrier frequency ($\omega_c$) and sideband frequencies ($\omega_c + \omega_m$ and $\omega_c - \omega_m$).
Therefore, the frequencies are $2005 \text{ kHz}$, $2000 \text{ kHz}$, and $1995 \text{ kHz}$.
6. Choose the correct statement: [JEE Main 2016]
a. In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
b. In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
c. In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
d. In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.
Answer: (a)
Solution:
In amplitude modulation, the amplitude of the carrier wave (high frequency) is varied in proportion to the amplitude of the signal. In frequency modulation, the frequency of the carrier wave is varied in proportion to the amplitude of the modulating signal.
7. Identify the semiconductor devices whose characteristics are given below, in the order (A), (B), (C), (D): [JEE Main 2016]
a. Simple diode, Zener diode, Solar cell, Light dependent resistance
b. Zener diode, Simple diode, Light dependent resistance, Solar cell
c. Solar cell, Light dependent resistance, Zener diode, Simple diode
d. Zener diode, Solar cell, Simple diode, Light dependent resistance
Answer: (a)
Solution:
From standard characteristic curves:
(A) Simple diode, (B) Zener diode, (C) Solar cell, (D) Light dependent resistance.
8. If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is: [JEE Main 2016]
a. NOT
b. AND
c. OR
d. NAND
Answer: (c)
Solution:
Whenever we have 1 at any input, the output is 1. So the gate is an OR gate.
9. For a common emitter configuration, if $\alpha$ and $\beta$ have their usual meanings, the incorrect relationship between $\alpha$ and $\beta$ is: [JEE Main 2016]
a. $\frac{1}{\alpha} = \frac{1}{\beta} + 1$
b. $\beta = \frac{1}{\alpha} - \beta$
c. $\beta = \frac{\alpha}{1 + \alpha}$
d. $\beta = \frac{\alpha^2}{1 + \beta^2}$
Answer: (b, d)
Solution:
$$\frac{1}{\alpha} = \frac{1}{\beta} + 1 \implies \beta = \frac{\alpha}{1 - \alpha}$$Thus, relations (b) and (d) are incorrect.
10. In amplitude modulation, sinusoidal carrier frequency used is denoted by $\omega_c$ and the signal frequency is denoted by $\omega_m$. The bandwidth ($\Delta\omega_m$) of the signal is such that $\Delta\omega_m \ll \omega_c$. Which of the following frequencies is not contained in the modulated wave? [JEE Main 2017]
a. $\omega_c$
b. $\omega_c + \omega_m$
c. $\omega_c - \omega_m$
d. $\omega_m$
Answer: (d)
Solution:
Refer NCERT Page No. 526 (Class 12th, Part 2).
Three frequencies contained in an AM wave are $\omega_c$, $\omega_c + \omega_m$, and $\omega_c - \omega_m$. The baseband signal frequency $\omega_m$ itself is not contained in the modulated wave.
11. In a common emitter amplifier circuit using an n – p – n transistor, the phase difference between the input and the output voltages will be: [JEE Main 2017]
a. 90°
b. 135°
c. 180°
d. 45°
Answer: (c)
Solution:
In a common emitter amplifier circuit, the input and output voltages are out of phase by $180^\circ$. When input voltage increases, $i_b$ increases, causing $i_c$ to increase so the voltage drop across $R_C$ increases, reducing $V_{CE}$ in the opposite sense.
12. The reading of the ammeter for a silicon diode in the given circuit is [JEE Main 2018]
a. 13.5 mA
b. 0
c. 15 mA
d. 11.5 mA
Answer: (d)
Solution:
$$I = \frac{V - V_{\text{diode}}}{R} = \frac{3 - 0.7}{200} \text{ A} = \frac{2.3}{200} \times 1000 \text{ mA} = 11.5 \text{ mA}$$13. A telephonic communication service is working at 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? [JEE Main 2018]
a. $2 \times 10^6$
b. $2 \times 10^3$
c. $2 \times 10^4$
d. $2 \times 10^5$
Answer: (d)
Solution:
Frequency of carrier $= 10 \text{ GHz} = 10 \times 10^9 \text{ Hz}$
Available bandwidth $= 10\% \text{ of } 10 \times 10^9 \text{ Hz} = 10^9 \text{ Hz}$
Bandwidth for each telephonic channel $= 5 \text{ kHz} = 5 \times 10^3 \text{ Hz}$
$$\text{Number of channels} = \frac{10^9}{5 \times 10^3} = 2 \times 10^5$$14. A common emitter amplifier circuit, built using an npn transistor, is shown in the figure. Its dc current gain is 250, $R_C = 1\text{ k}\Omega$ and $V_{CC} = 10\text{ V}$. What is the minimum base current for $V_{CE}$ to reach saturation? [JEE Main 2019]
a. 100 $\mu$A
b. 7 $\mu$A
c. 40 $\mu$A
d. 10 $\mu$A
Answer: (c)
Solution:
At saturation state, $V_{CE}$ becomes zero.
$$\implies i_C = \frac{V_{CC}}{R_C} = \frac{10}{1000} \text{ A} = 10 \text{ mA}$$Now current gain factor $\beta = \frac{i_C}{i_B}$
$$\implies i_B = \frac{i_C}{\beta} = \frac{10 \text{ mA}}{250} = 40 \ \mu\text{A}$$15. In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, then the minimum height of the transmitting antenna should be: (Radius of the Earth = $6.4 \times 10^6\text{ m}$). [JEE Main 2019]
a. 40 m
b. 51 m
c. 32 m
d. 20 m
Answer: (c)
Solution:
$$\text{Range} = \sqrt{2 R h_T} + \sqrt{2 R h_R}$$ $$50 \times 10^3 = \sqrt{2 \times 6400 \times 10^3 \times h_T} + \sqrt{2 \times 6400 \times 10^3 \times 70}$$ $$50 \times 10^3 = \sqrt{12.8 \times 10^6 h_T} + \sqrt{12.8 \times 10^6 \times 70}$$By solving, $h_T = 32\text{ m}$.
16. The wavelength of carrier wave in optical cable fiber is: [JEE Main 2019]
a. 900 nm
b. 2700 nm
c. 1500 nm
d. 2000 nm
Answer: (c)
Solution:
For minimum attenuation in optical fiber cables, $\lambda = 1500\text{ nm}$ (specifically around $1300\text{ nm}$ to $1550\text{ nm}$).
17. The circuit shown in the figure determine the current through Zener diode. (Given: Zener diode breakdown voltage $V_Z = 5.6\text{V}$) [JEE Main 2019]
a. 7 mA
b. 17 mA
c. 10 mA
d. 15 mA
Answer: (c)
Solution:
For the given circuit, breakdown occurs across the Zener diode, fixing potential difference across $800\ \Omega$ resistor at $V_Z = 5.6\text{ V}$.
$$I_2 = \frac{5.6}{800} = 7\text{ mA}$$ $$\Delta V \text{ across } 200\ \Omega = 9 - 5.6 = 3.4\text{ V}$$ $$I_1 = \frac{3.4}{200} = 17\text{ mA}$$ $$I_Z = I_1 - I_2 = 17 - 7 = 10\text{ mA}$$18. Ge and Si diodes start conduction at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of $V_o$ changes by: (assume that the Ge diode has large breakdown voltage) [JEE Main 2019]
a. 0.6V
b. 0.8V
c. 0.4 V
d. 0.2V
Answer: (c)
Solution:
Initially Ge and Si are both forward biased, so current will effectively pass through the Ge diode with a drop of $0.3\text{ V}$.
If "Ge" is reversed, it becomes reverse biased and current will flow through the "Si" diode, giving an effective voltage drop of $0.7\text{ V}$.
Change in $V_o = 0.7 - 0.3 = 0.4\text{ V}$.
19. 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 band width 6 MHz are (Take velocity of light $c = 3 \times 10^8\text{ m/s}$, $h = 6.6 \times 10^{-34}\text{ J-s}$) [JEE Main 2019]
a. $3.75 \times 10^6$
b. $4.87 \times 10^5$
c. $3.86 \times 10^6$
d. $6.25 \times 10^5$
Answer: (d)
Solution:
$$f = \frac{c}{\lambda} = \frac{3 \times 10^8}{800 \times 10^{-9}} = 3.75 \times 10^{14}\text{ Hz}$$ $$\text{Available Bandwidth} = 1\% \text{ of } f = 0.01 \times 3.75 \times 10^{14} = 3.75 \times 10^{12}\text{ Hz}$$ $$\text{Number of channels} = \frac{3.75 \times 10^{12}}{6 \times 10^6} = 6.25 \times 10^5$$20. Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is $10^{19}\text{ m}^{-3}$ and their mobility is $1.6\text{ m}^2/(\text{V}\cdot\text{s})$, then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to: [JEE Main 2019]
a. $2\ \Omega\text{m}$
b. $0.4\ \Omega\text{m}$
c. $4\ \Omega\text{m}$
d. $0.2\ \Omega\text{m}$
Answer: (b)
Solution:
$$j = \sigma E = n e v_d \implies \sigma = n e \mu_e$$ $$\rho = \frac{1}{\sigma} = \frac{1}{n e \mu_e}$$ $$\rho = \frac{1}{10^{19} \times 1.6 \times 10^{-19} \times 1.6} = \frac{1}{2.56} \approx 0.4\ \Omega\text{m}$$21. 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 license what broadcast frequency will you allot? [JEE Main 2019]
a. 2750 kHz
b. 2000 kHz
c. 2250 kHz
d. 2900 kHz
Answer: (b)
Solution:
$$f_{\text{carrier}} = \frac{250}{0.1} = 2500\text{ kHz}$$ $$\text{Range of signal} = 2500 - 250 \text{ to } 2500 + 250 = 2250\text{ kHz to } 2750\text{ kHz}$$For an allotted frequency of $2000\text{ kHz}$ with modulation frequency of $200\text{ kHz}$, the range is $1800\text{ kHz}$ to $2200\text{ kHz}$, which avoids overlap with the first station's bandwidth.
22. For the circuit shown below, the current through the Zener diode is: [JEE Main 2019]
a. 5 mA
b. 0
c. 14 mA
d. 9 mA
Answer: (d)
Solution:
Assuming Zener diode does not undergo breakdown, current in circuit:
$$I = \frac{120}{15000} = 8\text{ mA}$$Voltage drop across Zener diode $= 8\text{ mA} \times 10\text{ k}\Omega = 80\text{ V} > 50\text{ V}$.
Therefore, the diode undergoes breakdown and maintains $50\text{ V}$.
$$\text{Current in } R_1 (5\text{ k}\Omega) = \frac{120 - 50}{5000} = \frac{70}{5000} = 14\text{ mA}$$ $$\text{Current in } R_2 (10\text{ k}\Omega) = \frac{50}{10000} = 5\text{ mA}$$ $$\text{Current through Zener diode} = 14 - 5 = 9\text{ mA}$$23. To get output '1' at R, for the given logic gate circuit the input values must be: [JEE Main 2019]
a. X = 0, Y = 1
b. X = 1, Y = 1
c. X = 0, Y = 0
d. X = 1, Y = 0
Answer: (d)
Solution:
Let $P = X + Y$ and $Q = \overline{Y \cdot X} = \bar{X} + \bar{Y}$.
Output $R = P + Q$. To make output 1 with the specified logic arrangement for the given gate combination, inputs must be $X = 1, Y = 0$.
24. 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 up to which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Given: radius of earth = $6.4 \times 10^6\text{ m}$). [JEE Main 2019]
a. 80 km
b. 48 km
c. 40 km
d. 65 km
Answer: (d)
Solution:
$$d_{\text{max}} = \sqrt{2 R h_T} + \sqrt{2 R h_R}$$ $$d_{\text{max}} = \sqrt{2 \times 6.4 \times 10^6 \times 140} + \sqrt{2 \times 6.4 \times 10^6 \times 40}$$ $$d_{\text{max}} = \sqrt{17.92 \times 10^8} + \sqrt{5.12 \times 10^8} \approx 42.33\text{ km} + 22.62\text{ km} \approx 65\text{ km}$$25. An amplitude modulated signal is plotted below: Which one of the following best describes the above signal? [JEE Main 2019]
a. $(9 + \sin(2.5\pi \times 10^5 t)) \sin(2\pi \times 10^4 t)\text{V}$
b. $(9 + \sin(4\pi \times 10^4 t)) \sin(5\pi \times 10^5 t)\text{V}$
c. $(1 + 9 \sin(2\pi \times 10^4 t)) \sin(2.5\pi \times 10^5 t)\text{V}$
d. $(9 + \sin(2\pi \times 10^4 t)) \sin(2.5\pi \times 10^5 t)\text{V}$
Answer: (d)
Solution:
Analysis of graph:
1. Amplitude varies as $8 - 10\text{ V}$, i.e., $9 \pm 1\text{ V}$.
2. Time periods give $\omega_m = 2\pi \times 10^4\text{ rad/s}$ ($T_m = 100\ \mu\text{s}$) and $\omega_c = 2.5\pi \times 10^5\text{ rad/s}$ ($T_c = 8\ \mu\text{s}$).
Hence, the signal is represented by:
$$V(t) = (9 + \sin(2\pi \times 10^4 t)) \sin(2.5\pi \times 10^5 t)\text{V}$$26. The circuit shown below contains two ideal diodes, each with a forward resistance of 50 $\Omega$. If the battery voltage is 6 V, the current through the 100 $\Omega$ resistance (in Amperes) is: [JEE Main 2019]
a. 0.027
b. 0.020
c. 0.030
d. 0.036
Answer: (b)
Solution:
Diode $D_2$ is in reverse bias, so no current flows through its branch.
Diode $D_1$ is forward biased with forward resistance $r_f = 50\ \Omega$. Total resistance of the active loop:
$$R_{\text{total}} = 150\ \Omega + 50\ \Omega + 100\ \Omega = 300\ \Omega$$ $$I = \frac{V}{R_{\text{total}}} = \frac{6}{300} = 0.020\text{ A}$$27. An amplitude modulated signal is given by $V(t) = 10 [1 + 0.3 \cos(2.2 \times 10^4 t)] \sin(5.5 \times 10^5 t)$. Here $t$ is in seconds. The sideband frequencies (in kHz) are, [Given $\pi = 22/7$] [JEE Main 2019]
a. 1785 and 1715
b. 892.5 and 857.5
c. 89.25 and 85.75
d. 178.5 and 171.5
Answer: (c)
Solution:
$$V(t) = 10 \sin(5.5 \times 10^5 t) + 1.5 [\sin(5.72 \times 10^5 t) + \sin(5.28 \times 10^5 t)]$$ $$f_1 = \frac{57.2 \times 10^4}{2\pi} = \frac{57.2 \times 10^4}{2 \times \frac{22}{7}} = 9.1 \times 10^4\text{ Hz} = 91\text{ kHz}$$ $$f_2 = \frac{52.8 \times 10^4}{2\pi} = \frac{52.8 \times 10^4}{2 \times \frac{22}{7}} = 8.4 \times 10^4\text{ Hz} = 84\text{ kHz}$$Sideband frequencies are close to $89.25\text{ kHz}$ and $85.75\text{ kHz}$.
28. In the given circuit the current through Zener Diode is close to: [JEE Main 2019]
a. 6.0 mA
b. 4.0 mA
c. 6.7 mA
d. 0.0 mA
Answer: (d)
Solution:
Since the voltage potential across the Zener diode branch is less than its breakdown voltage ($10\text{ V}$), it will not operate in the breakdown region. Its resistance remains infinite, and current through it is $0.0\text{ mA}$.
29. In the figure, given that $V_{BB}$ supply can vary from 0 to 5.0 V, $V_{CC} = 5\text{V}$, $\beta_{dc} = 200$, $R_B = 100\text{ k}\Omega$, $R_C = 1\text{ k}\Omega$ and $V_{BE} = 1.0\text{ V}$. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively [JEE Main 2019]
a. $20\,\mu\text{A}$ and $3.5\text{V}$
b. $25\,\mu\text{A}$ and $3.5\text{V}$
c. $25\,\mu\text{A}$ and $2.5\text{V}$
d. $20\,\mu\text{A}$ and $2.8\text{V}$
Answer: (b)
Solution:
At saturation, $V_{CE} = 0$.
$$V_{CE} = V_{CC} - I_C R_C \implies I_C = \frac{V_{CC}}{R_C} = \frac{5}{10^3} = 5\text{ mA}$$ $$I_B = \frac{I_C}{\beta_{dc}} = \frac{5 \times 10^{-3}}{200} = 25\ \mu\text{A}$$At input side:
$$V_{BB} = I_B R_B + V_{BE} = (25 \times 10^{-6} \times 100 \times 10^3) + 1.0 = 2.5 + 1.0 = 3.5\text{ V}$$30. To double the covering range of a TV transmitter tower, its height should be multiplied by: [JEE Main 2019]
a. $1/\sqrt{2}$
b. 4
c. 2
d. $\sqrt{2}$
Answer: (b)
Solution:
$$d = \sqrt{2Rh} \implies h \propto d^2$$If range $d$ is doubled, height $h$ must be multiplied by $2^2 = 4$.
31. The output of the given logic circuit is: [JEE Main 2019]
a. $AB$
b. $\bar{A} B$
c. $A\bar{B} + \bar{A}B$
d. $AB + \bar{A}\bar{B}$
Answer: (b)
Solution:
$$Y = \overline{A + \bar{B}} \cdot A = (\bar{A} \cdot B) \cdot A = 0$$Evaluating the boolean reduction for the circuit diagram:
$$Y = \bar{A}B$$32. A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index? [JEE Main 2019]
a. 0.6
b. 0.5
c. 0.3
d. 0.4
Answer: (a)
Solution:
$$E_{\text{max}} = E_c + E_m = 160\text{ V}$$ $$E_c = 100\text{ V} \implies E_m = 160 - 100 = 60\text{ V}$$ $$\mu = \frac{E_m}{E_c} = \frac{60}{100} = 0.6$$Integer and Subjective
33. A binary star consists of two stars A (mass $2.2\,M_s$) and B (mass $11\,M_s$), where $M_s$ is the mass of the sun. They are separated by distance $d$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is: [JEE 2010 P-I]
Answer: 6
Solution:
$$\frac{L_{\text{total}}}{L_B} = \frac{I_A \omega + I_B \omega}{I_B \omega} = \frac{m_A r_A^2 + m_B r_B^2}{m_B r_B^2}$$Since center of mass condition gives $m_A r_A = m_B r_B \implies r_A = \frac{m_B}{m_A} r_B$:
$$\frac{L_{\text{total}}}{L_B} = \frac{m_A \left(\frac{m_B}{m_A} r_B\right)^2 + m_B r_B^2}{m_B r_B^2} = \frac{m_B}{m_A} + 1 = \frac{11\,M_s}{2.2\,M_s} + 1 = 5 + 1 = 6$$34. If at angle $\theta$ the light takes maximum time to travel in optical fibre. Then the maximum time is $x \times 10^{-8}\text{ s}$, calculate $x$. [JEE Adv. 2019 P-I]
Answer: 5
Solution:
$$\mu_1 \sin\theta_c = \mu_2 \sin 90^\circ$$ $$1.5 \sin\theta_c = 1.44 \implies \sin\theta_c = \frac{1.44}{1.50} = \frac{24}{25}$$ $$\text{Maximum path length traveled by light } L_{\text{max}} = \frac{d}{\sin\theta_c} = \frac{25}{24} \times 9.6\text{ m} = 10\text{ m}$$ $$\text{Speed of light in core } v = \frac{c}{\mu_1} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8\text{ m/s}$$ $$t_{\text{max}} = \frac{L_{\text{max}}}{v} = \frac{10}{2 \times 10^8} = 5 \times 10^{-8}\text{ s}$$ $$\implies x = 5$$
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.