Physics Board Question Paper Solution: February 2020
Physical Constants
- h = \(6.63 \times 10^{-34}\) Js
- c = \(3 \times 10^8\) m/s
- \(\pi = 3.142\)
- g = \(9.8\) m/s²
- \(\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2\)
- \(\mu_0 = 4\pi \times 10^{-7}\) Wb/A·m
Section - A
Q.1. Select and write correct answers of the following questions: [10 Marks]
i. If the length of a potentiometer wire is increased by keeping constant potential difference across the wire, then _______.
Explanation: Potential gradient \(K = V/L\). If L increases, K decreases. Since balancing length \(l = E/K\), if K decreases, \(l\) increases.
ii. If a coil of metal wire is kept stationary in a uniform magnetic field, then_______.
Explanation: For induction, there must be a change in magnetic flux. Since both the coil and field are stationary/uniform, flux is constant.
iii. In rotational motion of a rigid body, all particles move with _______.
Explanation: Angular velocity \(\omega\) is constant for the whole body, but linear velocity \(v = r\omega\) changes with distance \(r\) from the axis.
iv. A standing wave is produced on a string fixed at one end and free at other. The length of string must be an _______.
Explanation: \(L = (2n-1)\frac{\lambda}{4}\).
v. Herapathite (iodo sulphate of quinine) is used in the production of _______.
vi. A diffraction pattern is obtained by making blue light incident on a narrow slit. If blue light is replaced by red light, then the diffraction bands _______.
Explanation: Fringe width \(\beta \propto \lambda\). Since \(\lambda_{red} > \lambda_{blue}\), the bands become broader.
vii. Two droplets coalesce in a single drop. In this process _______.
Explanation: Total surface area decreases, so surface energy decreases. The difference is released as heat.
viii. In hydrogen atom, electron jumps from the 3rd orbit to the 1st orbit. The change in angular momentum is _______.
Explanation: \(\Delta L = \frac{h}{2\pi}(n_2 - n_1) = \frac{6.63\times 10^{-34}}{2(3.142)}(3 - 1) \approx 2.11 \times 10^{-34}\) Js.
ix. A fixed volume of iron is drawn into a wire of length ‘L’. The extension ‘x’ produced in the wire by a constant force ‘F’ is proportional to _______.
Explanation: \(Y = \frac{FL}{Ax}\). Since Volume \(V = AL\), \(A = V/L\). Thus \(Y = \frac{FL^2}{Vx} \Rightarrow x \propto L^2\).
x. Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between two successive maximum intensities will be _______.
Explanation: Beat frequency \(N = |n_1 - n_2| = 4\) Hz. Period of beats \(T = 1/N = 1/4\) s.
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Q.2. Answer the following questions : [8 Marks]
i. What is attenuation in communication system?
ii. What is isothermal process?
iii. What does the negative sign indicate in Lenz’s law?
iv. Define uniform circular motion.
v. At which position, the total energy of a particle executing linear S.H.M. is purely potential?
vi. State the name of the visible series in hydrogen spectrum.
vii. At what height the acceleration due to gravity is 25% of that at the surface of the Earth, in terms of radius of the Earth.
\(\frac{1}{4} = \left(\frac{R}{R+h}\right)^2 \Rightarrow \frac{1}{2} = \frac{R}{R+h} \Rightarrow R+h = 2R \Rightarrow h = R\).
**Answer: At height h = R (Radius of Earth).**
viii. Capacity of a parallel capacitor with dielectric constant 5 is 40 \(\mu\)F. Calculate the capacity of the same capacitor when dielectric material is removed.
\(C_0 = \frac{40}{5} = 8 \mu\text{F}\).
**Answer: 8 \(\mu\)F.**
Section - B
Attempt any EIGHT questions [16 Marks]
Q.3. Define: (a) Threshold frequency (b) Photoelectric work function
(a) Threshold frequency: It is the minimum frequency of incident radiation required to start the emission of photoelectrons from a photosensitive material.
(b) Photoelectric work function: It is the minimum amount of energy required by an electron to escape from the metal surface at zero velocity.
Q.4. Distinguish between step-up and step-down transformer.
| Step-up Transformer | Step-down Transformer |
|---|---|
| Increases the AC voltage ($V_s > V_p$). | Decreases the AC voltage ($V_s < V_p$). |
| Decreases the current strength ($I_s < I_p$). | Increases the current strength ($I_s > I_p$). |
| Number of turns in secondary coil is greater than primary ($N_s > N_p$). | Number of turns in primary coil is greater than secondary ($N_p > N_s$). |
Q.5. Write a short note on sky wave propagation.
- Sky wave propagation (or ionospheric propagation) is a mode of radio wave propagation used for long-distance communication in the frequency range of 3 MHz to 30 MHz.
- In this mode, radio waves transmitted from the earth are reflected back to the earth by the ionosphere (layers of charged particles in the atmosphere).
- The critical frequency is the highest frequency that can be reflected back by the ionosphere for vertical incidence.
- It allows communication over distances larger than the line-of-sight.
Q.6. Obtain the relation between the magnitude of linear acceleration and angular acceleration in circular motion.
For a particle moving in a circular path of radius \(r\), linear velocity is \(v = r\omega\).
Differentiating with respect to time:
\(\frac{dv}{dt} = \frac{d}{dt}(r\omega) = r\frac{d\omega}{dt}\) (since \(r\) is constant).
Here, \(\frac{dv}{dt}\) is the tangential acceleration \(a_t\) and \(\frac{d\omega}{dt}\) is the angular acceleration \(\alpha\).
Therefore, \(a_t = r\alpha\).
In vector form: \(\vec{a} = \vec{\alpha} \times \vec{r}\).
Q.7. Deduce Boyle’s law using the expression for pressure exerted by the gas.
The pressure exerted by an ideal gas is given by \(P = \frac{1}{3} \frac{Nm}{V} v_{rms}^2\), where N is number of molecules, m is mass of one molecule, V is volume.
\(PV = \frac{1}{3} N m v_{rms}^2\)
Multiply and divide by 2:
\(PV = \frac{2}{3} N \left(\frac{1}{2} m v_{rms}^2\right)\)
Since Kinetic Energy \(\text{K.E.} = \frac{1}{2} m v_{rms}^2\) is directly proportional to absolute temperature \(T\), if temperature is constant, K.E. is constant.
Therefore, for a fixed mass of gas at constant temperature, \(PV = \text{constant}\).
\(P \propto \frac{1}{V}\). This is Boyle's Law.
Q.8. State any two postulates of Bohr’s theory of hydrogen atom.
- The electron revolves around the nucleus in circular orbits. The necessary centripetal force is provided by the electrostatic force of attraction between the nucleus and the electron.
- The electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of \(h/2\pi\). i.e., \(L = mvr = \frac{nh}{2\pi}\).
- (Extra) Energy is radiated only when an electron jumps from a higher energy orbit to a lower energy orbit.
Q.9. State the factors on which resolving power of microscope depends. How can it be increased?
Resolving Power (R.P.) formula: \(R.P. = \frac{2\mu \sin\theta}{\lambda}\) or \(\frac{2 \text{N.A.}}{\lambda}\).
Factors:
- Wavelength of light used (\(\lambda\)).
- Refractive index of the medium between the objective and the object (\(\mu\)).
- Half-angle of the cone of light entering the objective (\(\theta\)).
To Increase R.P.:
- Decrease the wavelength (\(\lambda\)) (e.g., use blue light or UV).
- Increase the refractive index (\(\mu\)) (e.g., use oil immersion objective).
Q.10. The photoelectric work function for a metal surface is \(3.84 \times 10^{-19}\) J. If the light of wavelength 5000 Å is incident on the surface of the metal, will there be photoelectric emission?
Given:
Work Function \(\phi_0 = 3.84 \times 10^{-19}\) J
Wavelength \(\lambda = 5000 \, \text{\AA} = 5 \times 10^{-7}\) m
Calculation:
Energy of incident photon \(E = \frac{hc}{\lambda}\)
\(E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-7}}\)
\(E = \frac{19.89 \times 10^{-26}}{5 \times 10^{-7}} = 3.978 \times 10^{-19}\) J
Comparison:
Since Incident Energy (\(3.978 \times 10^{-19}\) J) > Work Function (\(3.84 \times 10^{-19}\) J).
Answer: Yes, there will be photoelectric emission.
Q.11. A parallel L-C circuit comprises of a 5H inductor and 5\(\mu\)F capacitor. Calculate the resonant frequency of the circuit.
Given: \(L = 5\) H, \(C = 5 \mu\text{F} = 5 \times 10^{-6}\) F.
Formula: Resonant frequency \(f_r = \frac{1}{2\pi\sqrt{LC}}\)
Calculation:
\(\sqrt{LC} = \sqrt{5 \times 5 \times 10^{-6}} = \sqrt{25 \times 10^{-6}} = 5 \times 10^{-3}\)
\(f_r = \frac{1}{2 \times 3.142 \times 5 \times 10^{-3}} = \frac{1}{0.03142} = \frac{1000}{31.42}\)
Answer: \(f_r \approx 31.83\) Hz
Q.12. A wire length 1 m and mass 2 g is in unison with a tuning fork of frequency 300 Hz. Calculate the tension produced in the wire.
Given: \(L = 1\) m, Mass \(M = 2\) g \(= 0.002\) kg, \(n = 300\) Hz.
Linear density \(m = \frac{M}{L} = \frac{0.002}{1} = 0.002\) kg/m.
Formula: \(n = \frac{1}{2L}\sqrt{\frac{T}{m}}\)
Calculation:
\(300 = \frac{1}{2(1)}\sqrt{\frac{T}{0.002}}\)
\(600 = \sqrt{\frac{T}{0.002}}\)
Squaring both sides:
\(360000 = \frac{T}{0.002}\)
\(T = 360000 \times 0.002 = 720\) N
Answer: Tension \(T = 720\) N.
Q.13. Energy of 1000 J is spent to increase the angular speed of a wheel from 20 rad/s to 30 rad/s. Calculate the moment of inertia of the wheel.
Given: Work (Change in KE) \(W = 1000\) J, \(\omega_1 = 20\) rad/s, \(\omega_2 = 30\) rad/s.
Formula: \(W = \frac{1}{2}I(\omega_2^2 - \omega_1^2)\)
Calculation:
\(1000 = \frac{1}{2}I(30^2 - 20^2)\)
\(2000 = I(900 - 400)\)
\(2000 = I(500)\)
\(I = \frac{2000}{500} = 4\) kg·m²
Answer: Moment of Inertia \(I = 4\) kg·m².
Q.14. Energy of an electron in the second Bohr orbit is –3.4 eV. Calculate the energy of an electron in the third Bohr orbit.
Given: \(E_2 = -3.4\) eV.
Formula: \(E_n \propto \frac{1}{n^2}\). Also \(E_n = \frac{E_1}{n^2}\).
Using \(E_2\): \(-3.4 = \frac{E_1}{2^2} \Rightarrow E_1 = -3.4 \times 4 = -13.6\) eV.
Now for 3rd orbit (\(n=3\)):
\(E_3 = \frac{E_1}{3^2} = \frac{-13.6}{9}\)
Answer: \(E_3 = -1.51\) eV.
Section - C
Attempt any EIGHT questions [24 Marks]
Q.15. State Newton’s law of gravitation. Obtain the relation between universal gravitational constant and gravitational acceleration on the surface of the earth.
Newton's Law of Gravitation: Every particle of matter attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Relation derivation:
Consider Earth as a sphere of mass \(M\) and radius \(R\). An object of mass \(m\) is on the surface.
According to Newton's law: \(F = \frac{GMm}{R^2}\)
According to Newton's 2nd law, the force due to gravity is weight: \(F = mg\)
Equating both: \(mg = \frac{GMm}{R^2}\)
\(g = \frac{GM}{R^2}\)
Q.16. Define linear S.H.M. Obtain differential equation of linear S.H.M.
Definition: Linear Simple Harmonic Motion is defined as the linear periodic motion of a body, in which the restoring force (or acceleration) is always directed towards the mean position and its magnitude is directly proportional to the displacement from the mean position.
Differential Equation:
Force \(f \propto -x \Rightarrow f = -kx\)
By Newton's 2nd law, \(f = ma = m \frac{d^2x}{dt^2}\)
\(\therefore m \frac{d^2x}{dt^2} = -kx\)
\(\frac{d^2x}{dt^2} + \frac{k}{m}x = 0\)
Put \(\frac{k}{m} = \omega^2\)
\(\frac{d^2x}{dt^2} + \omega^2 x = 0\)
Q.17. Explain Doppler effect in sound. State any two applications of Doppler effect.
Doppler Effect: The apparent change in the frequency of sound heard by an observer, due to relative motion between the source of sound and the observer, is called the Doppler effect.
Applications:
- Speed Gun: Used by traffic police to measure the speed of speeding vehicles.
- RADAR: Used to detect the position and speed of aircrafts.
- Astrophysics: Used to determine the speed of stars/galaxies (Red shift/Blue shift).
Q.18. State and prove the principle of parallel axes in rotational motion.
Statement: The moment of inertia of a body about any axis (\(I_O\)) is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass (\(I_C\)) and the product of its mass (\(M\)) and the square of the perpendicular distance (\(h\)) between the two parallel axes.
Formula: \(I_O = I_C + Mh^2\)
Proof Outline: Consider a particle of mass \(dm\) at point P. Distance from axis through C is \(CP=r\), from axis through O is \(OP\). Let distance between axes be \(h\). Using geometry and integration \(\int OP^2 dm = \int CP^2 dm + h^2 \int dm + \text{cross term (which becomes 0)}\), we arrive at the result.
Q.19. Explain the concept of a parallel plate capacitor. State its any ‘two’ applications.
Concept: A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance. The space between the plates is filled with a dielectric medium. When one plate is charged (+Q) and the other is grounded, a potential difference is created, allowing it to store charge.
Applications:
- In radio circuits for tuning (LC circuits).
- In power supplies for smoothing (filtering) rectified current.
- In camera flash units to store energy.
Q.20. Define: (a) Young’s modulus (b) Bulk modulus (c) Poisson’s ratio
(a) Young's modulus (Y): The ratio of longitudinal stress to longitudinal strain within the elastic limit.
(b) Bulk modulus (K): The ratio of volume stress (or hydraulic stress) to volume strain within the elastic limit.
(c) Poisson's ratio (\(\sigma\)): The ratio of lateral strain to longitudinal strain within the elastic limit.
Q.21. Explain energy distribution spectrum of a black body radiation in terms of wavelength.
Lummer and Pringsheim studied energy distribution of black body radiation:
- At a given temperature, energy is not uniformly distributed among all wavelengths.
- Radiant power increases with wavelength, reaches a maximum for a particular wavelength (\(\lambda_{max}\)), and then decreases.
- As temperature increases, the peak of the curve shifts towards shorter wavelengths (Wien's Displacement Law: \(\lambda_{max} T = \text{constant}\)).
- The area under the curve represents the total energy emitted per unit area per unit time, which increases with \(T^4\) (Stefan's Law).
Q.22. In a biprism experiment, light of wavelength 5200 Å is used to obtain an interference pattern on the screen. The fringewidth changes by 1.3 mm when the screen is moved towards the biprism by 50 cm. Calculate the distance between the two virtual images of the slit.
Given: \(\lambda = 5200 \, \text{\AA} = 5.2 \times 10^{-7}\) m.
Change in \(D\) (\(\Delta D\)) = 50 cm = 0.5 m.
Change in \(\beta\) (\(\Delta \beta\)) = 1.3 mm = \(1.3 \times 10^{-3}\) m.
Formula: \(\beta = \frac{\lambda D}{d} \Rightarrow \Delta \beta = \frac{\lambda \Delta D}{d}\)
Calculation:
\(d = \frac{\lambda \Delta D}{\Delta \beta}\)
\(d = \frac{5.2 \times 10^{-7} \times 0.5}{1.3 \times 10^{-3}}\)
\(d = \frac{2.6 \times 10^{-7}}{1.3 \times 10^{-3}} = 2 \times 10^{-4}\) m
Answer: Distance between virtual images \(d = 0.2\) mm.
Q.23. A simple pendulum of length 1 m has mass 10 g and oscillates freely with amplitude of 5 cm. Calculate its potential energy at extreme position.
Given: \(L = 1\) m, \(m = 10\) g = 0.01 kg, \(A = 5\) cm = 0.05 m.
At extreme position, Total Energy = Potential Energy.
Formula: \(P.E. = \frac{1}{2} m \omega^2 A^2\). Also \(\omega^2 = g/L\).
So, \(P.E. = \frac{1}{2} m \left(\frac{g}{L}\right) A^2\)
Calculation:
\(P.E. = \frac{1}{2} \times 0.01 \times \left(\frac{9.8}{1}\right) \times (0.05)^2\)
\(P.E. = 0.5 \times 0.01 \times 9.8 \times 0.0025\)
\(P.E. = 0.049 \times 0.0025 = 1.225 \times 10^{-4}\) J
Answer: Potential Energy = \(1.225 \times 10^{-4}\) J.
Q.24. If the difference in the velocities of light in glass and water is \(0.25 \times 10^8\) m/s, calculate the velocity of light in air. Given that refractive index of glass and water with respect to air are 1.5 and 4/3 respectively.
Given: \(\mu_g = 1.5 = 3/2\), \(\mu_w = 4/3\).
\(v_w - v_g = 0.25 \times 10^8\).
Formula: \(v = c/\mu\).
Calculation:
\(\frac{c}{\mu_w} - \frac{c}{\mu_g} = 0.25 \times 10^8\)
\(c \left( \frac{1}{4/3} - \frac{1}{3/2} \right) = 0.25 \times 10^8\)
\(c \left( \frac{3}{4} - \frac{2}{3} \right) = 0.25 \times 10^8\)
\(c \left( \frac{9-8}{12} \right) = 0.25 \times 10^8\)
\(c \left( \frac{1}{12} \right) = 0.25 \times 10^8\)
\(c = 12 \times 0.25 \times 10^8 = 3 \times 10^8\) m/s
Answer: Velocity of light in air = \(3 \times 10^8\) m/s.
Q.25. In a Circus, a motor-cyclist having mass of 50 kg moves in a spherical cage of radius 3 m. Calculate the least velocity with which he must pass the highest point without losing contact. Also calculate his angular speed at the highest point.
Given: \(m = 50\) kg (not needed for v), \(r = 3\) m, \(g = 9.8\) m/s².
(1) Least velocity at highest point:
\(v_{min} = \sqrt{rg}\)
\(v_{min} = \sqrt{3 \times 9.8} = \sqrt{29.4}\)
\(v_{min} \approx 5.42\) m/s.
(2) Angular speed:
\(\omega = \frac{v}{r} = \frac{5.42}{3}\)
\(\omega \approx 1.81\) rad/s.
Q.26. Two resistances X and Y in the two gaps of a meter-bridge gives a null point dividing the wire in the ratio 2:3. If each resistance is increased by 30 \(\Omega\), the null point divides the wire in the ratio 5:6, calculate the value of X and Y.
Case 1:
\(\frac{X}{Y} = \frac{l_X}{l_Y} = \frac{2}{3} \Rightarrow X = \frac{2}{3}Y\) --- (1)
Case 2:
Resistances become \(X+30\) and \(Y+30\). Ratio is 5:6.
\(\frac{X+30}{Y+30} = \frac{5}{6}\)
\(6(X+30) = 5(Y+30)\)
\(6X + 180 = 5Y + 150\)
\(6X - 5Y = -30\) --- (2)
Substitute (1) in (2):
\(6(\frac{2}{3}Y) - 5Y = -30\)
\(4Y - 5Y = -30\)
\(-Y = -30 \Rightarrow Y = 30 \, \Omega\).
From (1):
\(X = \frac{2}{3}(30) = 20 \, \Omega\).
Answer: \(X = 20 \, \Omega, Y = 30 \, \Omega\).
Section - D
Attempt any THREE questions [12 Marks]
Q.27. Show that all harmonics are present in case of an air column vibrating in a pipe open at both ends.
Fundamental Mode (1st Harmonic):
\(L = \lambda_1 / 2 \Rightarrow \lambda_1 = 2L\).
Freq \(n_1 = v/\lambda_1 = v/2L\).
Second Mode (2nd Harmonic):
\(L = \lambda_2 \Rightarrow \lambda_2 = L\).
Freq \(n_2 = v/L = 2(v/2L) = 2n_1\).
Third Mode (3rd Harmonic):
\(L = 3\lambda_3 / 2 \Rightarrow \lambda_3 = 2L/3\).
Freq \(n_3 = v/(2L/3) = 3(v/2L) = 3n_1\).
Conclusion:
The frequencies are in the ratio \(n_1 : n_2 : n_3 = 1 : 2 : 3\).
This shows that both odd and even harmonics are present.
Q.28. What is a rectifier? With the help of a neat circuit diagram explain the working of a null wave rectifier.
Note: "Null wave" appears to be a typo in the question paper/OCR. The standard topics are Half Wave or Full Wave. Based on 4 marks, the solution for **Full Wave Rectifier** is provided below.
Rectifier: A device which converts alternating current (A.C.) into direct current (D.C.).
Full Wave Rectifier Working:
1. It uses two diodes ($D_1$ and $D_2$) and a center-tapped transformer.
2. During the positive half cycle of AC, $D_1$ is forward biased (conducts) and $D_2$ is reverse biased (does not conduct). Current flows through load $R_L$.
3. During the negative half cycle, $D_2$ becomes forward biased and $D_1$ is reverse biased. Current again flows through $R_L$ in the same direction.
4. Thus, output current flows during both half cycles, resulting in higher efficiency than half-wave.
Q.29. What is capillarity? Give any two applications of capillarity. Calculate the work done in blowing a soap bubble of radius 0.1 m. (Surface tension of soap solution = 30 dyne/cm)
Capillarity: The phenomenon of rise or fall of a liquid inside a capillary tube when it is dipped in the liquid is called capillarity.
Applications:
1. Oil rising up the wick of a lamp.
2. Sap rising from roots to the top of a tree.
Numerical:
Given: \(R = 0.1\) m = 10 cm. \(T = 30\) dyne/cm.
Soap bubble has 2 free surfaces.
Increase in Area \(\Delta A = 2 \times (4\pi R^2 - 0) = 8\pi R^2\).
Work Done \(W = T \times \Delta A = T \times 8\pi R^2\).
\(W = 30 \times 8 \times 3.142 \times (10)^2\)
\(W = 240 \times 3.142 \times 100\)
\(W = 24000 \times 3.142 \approx 75408\) ergs.
Or in Joules: \(7.54 \times 10^{-3}\) J.
Q.30. State the advantages and disadvantages of a moving coil galvanometer. [Numerical Part below]
Advantages:
1. It is very sensitive.
2. Not affected by stray magnetic fields (strong internal field).
Disadvantages:
1. Cannot measure AC directly.
2. Suspension fiber may break easily.
Numerical:
Given: \(N=10\), \(A = 12 \times 8\) cm² = \(96 \times 10^{-4}\) m², \(I = 125 \mu\text{A} = 125 \times 10^{-6}\) A, \(B = 10^{-2}\) T, \(C\) (Twist constant) \(= 12 \times 10^{-9}\) Nm/degree.
Formula: \(NIAB = C\theta \Rightarrow \theta = \frac{NIAB}{C}\)
Calculation:
\(\theta = \frac{10 \times (125 \times 10^{-6}) \times (96 \times 10^{-4}) \times 10^{-2}}{12 \times 10^{-9}}\)
\(\theta = \frac{125 \times 96 \times 10^{-11}}{12 \times 10^{-9}}\)
\(\theta = \frac{125 \times 8 \times 10^{-11}}{10^{-9}}\)
\(\theta = 1000 \times 10^{-2} = 10\) degrees.
Answer: Deflection \(\theta = 10^{\circ}\).
Q.31. Derive an expression for magnitude of magnetic dipole moment of a revolving electron. [Numerical Part below]
Derivation:
Current \(I = e/T\), where \(T = 2\pi r/v\). So \(I = \frac{ev}{2\pi r}\).
Orbital Magnetic Moment \(\mu = I \times A = I \times \pi r^2\).
\(\mu = \frac{ev}{2\pi r} \times \pi r^2 = \frac{evr}{2}\).
Numerical:
Circular coil: \(N=300\), \(d=14\) cm \(\Rightarrow r=7\) cm \(= 0.07\) m, \(I=15\) A.
Magnetic Dipole Moment \(M = NIA\).
Area \(A = \pi r^2 = \frac{22}{7} \times (0.07)^2 = \frac{22}{7} \times 0.0049 = 22 \times 0.0007 = 0.0154\) m².
\(M = 300 \times 15 \times 0.0154\)
\(M = 4500 \times 0.0154 = 69.3\) Am².
Answer: \(M = 69.3\) Am².
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