Christopher Cabellon

 NUMERICAL PROBLEMS F.Sc. Physics, Chapter # 12: ELECTROSTATICS 12.1 Compare magnitudes of electrical and gravitational forces exerted on an object (mass = 10.0 g, charge = 20.0 µC) by an identical object that is placed 10.0 cm from the first. (𝐺 = 6. 67 × 10−11 𝑁𝑚2𝑘𝑔−2) Given Data: Masses 𝑚1 = 𝑚2 = 𝑚 = 10 𝑔 = 0.01 𝑘𝑔, Charges 𝑞1 = 𝑞2 = 𝑞 = 20 µ𝐶 = 20 × 10−6𝐶, Distance 𝑟 = 10 𝑐𝑚 = 0.1 𝑚 To Determine: 𝐹𝑒 =? 𝐹𝑔 g1g22 Calculations: 𝐹𝑒 = (𝑘 2 ) = 𝑘𝑞 𝑞= 𝑘𝑞2 = 9×109×(20×10−6) = 5.4 × 10 r 1 214 𝐹𝑔(𝐺𝑚1𝑚2)𝐺𝑚1𝑚2𝐺𝑚26.67×10−11×(0.01)2 r2 12.2 Calculate vectorially the net electrostatic force on q as shown in the figure. Given Data: Charges 𝑞1 = 1 µ𝐶 = 1 × 10−6𝐶, 𝑞2 = −1 µ𝐶 = −1 × 10−6𝐶, 𝑞 = 4 µ𝐶 = 4 × 10−6𝐶 To Determine: Total Force on 𝑞 F =? Calculations: From Fig. tan 𝜃 = 0.8 ⟹ 𝜃 = tan−1 0.8 = 53° () 0.60.6 −6−6 Force Exerted by Charge 𝑞 on 𝑞: 𝐹 = 𝑘 𝑞𝑞1 = 9 × 109 × 4×10 ×1×10= 36 × 10−3 𝑁 11𝑟2(1)2 −6−6 Force Exerted by Charge 𝑞 on 𝑞: 𝐹 = 𝑘 𝑞𝑞2 = 9 × 109 × 4×10 ×1×10= 36 × 10−3 𝑁 22𝑟2(1)2 Now 𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 = 𝐹1 cos 𝜃 + 𝐹2 cos 𝜃 = 36 × 10−3 × cos 53° + 36 × 10−3 × cos(2𝜋 − 53°) = 0.043 𝑁 And 𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 = 𝐹1 sin 𝜃 + 𝐹2 sin 𝜃 = 36 × 10−3 × sin 53° + 36 × 10−3 × sin(2𝜋 − 53°) = 0 𝑁 Magnitude of Resultant Force 𝐹 = √𝐹2 + 𝐹2 = √(0.043)2 + (0)2 = 0.043 𝑁 𝑥𝑦 0 Direction of Resultant Force tan 𝜃 = 𝐹𝑦 ⟹ 𝜃 = tan−1 () = 0° (Resultant is along x-axis) 𝐹𝑥0.043 Resultant Force ⃗F→ = 0.043 ı̂ 12.3 A point charge q = − 8. 0 × 10−8𝐶 is placed at the origin. Calculate electric field at a point 2.0 m from the origin on the z-axis. Given Data: Charge 𝑞 = − 8.0 × 10−8𝐶, Distance 𝑟 = 2 𝑚, Direction: z-axis 𝑟̂ = 𝑘̂ To Determine: Electric Field 𝐸 =? −8 Calculations: ⃗E→ = 𝑘 𝑞 𝑘̂ = 9 × 109 × (− 8.0 × 10 ) 𝑘̂ = (−1.8 × 102 𝑘̂)𝑁𝐶−1 𝑟2(2)2 F.Sc. Physics, (1st Year), Complete Physics Notes CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-complete-physics- notes.html 12.4 Determine the electric field at the position r→ = (4 ı̂ + 3ĵ )𝑚 caused by a point charge q = 5. 0 × 10−6 𝐶 placed at origin. Given Data: Position Vector r→ = (4 ı̂ + 3 ̂ )𝑚, Charge 𝑞 = 5.0 × 10−6 𝐶 To Determine: Electric Field ⃗E→ =? Calculations: ⃗E→ = 𝑘 𝑞 𝑟̂ − − − − (1) 𝑟2 r→4 ı̂ + 3 ̂ 𝑁𝑜𝑤 𝑟 = |𝑟| = √42 + 32 = 5, 𝑎𝑛𝑑 𝑟̂ == |𝑟|5 Equation (1) becomes: 𝑞5.0 × 10−64 ı̂ + 3 ̂ ⃗E→ = 𝑘𝑟̂ = 9 × 109 ××= 360 × (4 ı̂ + 3 )̂ = 1440 ı̂ + 1080 ̂ 𝑟2(5)25 12.5 Two point charges, q1 = −1. 0 × 10−6𝐶 and q2 = 4. 0 × 10−6𝐶, are separated by a distance of 3.0 m. Find and justify the zero-field location. Given Data: Charges 𝑞1 = −1.0 × 10−6𝐶, 𝑞2 = 4.0 × 10−6𝐶, Let Distance between Charges 𝑟 = 3 𝑚 To Determine: Zero Field Location Calculations: Let P is zero field location, then at point P (distant x from 𝑞1): Electric Field due to 𝑞1 = Electric Field due to 𝑞2 ⟹ 𝐸1 = 𝐸2 Let Distance of P from 𝑞1 = |AP| = 𝑟1 = 𝑥 and Distance of P from 𝑞2 = |BP| = 𝑟2 = 𝑥 + 3 −6−6 Now consider 𝐸 = 𝐸 ⟹ 𝑘 𝑞1 = 𝑘 𝑞2 ⟹ 𝑞1 = 𝑞2 ⟹ 1.0×10= 4.0×10⟹ 1 =4 12𝑟2𝑟2𝑟2𝑟2𝑥2(𝑥+3)2𝑥2(𝑥+3)2 1212 ⟹ (𝑥 + 3)2 = 4𝑥2 ⟹ 𝑥2 + 6𝑥 + 9 = 4𝑥2 ⟹ 3𝑥2 − 6𝑥 − 9 = 0 ⟹ 𝑥2 − 2𝑥 − 3 = 0 ⟹ 𝑥2 − 3𝑥 + 𝑥 − 3 = 0 ⟹ 𝑥(𝑥 − 3) + 1(𝑥 − 3) = 0 ⟹ (𝑥 − 3)(𝑥 + 1) = 0 ⟹ 𝑥 − 3 = 0 𝑜𝑟 𝑥 + 1 = 0 ⟹ 𝑥 = 3 𝑜𝑟 𝑥 = −1 (𝑁𝑜𝑡 𝑃𝑜𝑠𝑠i𝑏𝑙𝑒) So the correct answer is 𝑥 = 3 12.6 Find the electric field strength required to hold suspended a particle of mass 1. 0 × 10−6 𝑘𝑔 and charge 1. 0 µ𝐶 between two plates 10.0 cm apart. Given Data: Mass 𝑚 = 1.0 × 10−6 𝑘𝑔, Charge 𝑞 = 1.0 µ𝐶 = 1.0 × 10−6 𝐶 , Distance between Plates 𝑑 = 10 𝑐𝑚 = 0.1 𝑚 To Determine: Electric Field Strength 𝐸 =? Calculations: For present case: Electrical Force = Gravitational Force ⟹ 𝑞𝐸 = 𝑚𝑔 𝑚𝑔1.0 × 10−6 × 9.8 ⟹ 𝐸 === 9.8 𝑁𝐶−1 𝑞1.0 × 10−6 12.7 A particle having a charge of 20 electrons on it falls through a potential difference of 100 volts. Calculate the energy acquired by it in electron volts (eV). Given Data: Charge 𝑞 = 20𝑒, Potential Difference ∆𝑉 = 100 𝑉 To Determine: Energy Acquired =? Calculations: Energy Acquired = 𝑞∆𝑉 = (20𝑒)(100 𝑉) = 2000 𝑒𝑉 = 2 × 103𝑒𝑉 12.8 In Millikan’s experiment, oil droplets are introduced into the space between two flat horizontal plates, 5.00 mm apart. The plate voltage is adjusted to exactly 780 V so that the droplet is held stationary. The plate voltage is switched off and the selected droplet is observed to fall a measured distance of 1.50 mm in 11.2 s. Given that the density of the oil used is 900 kg m-3, and the viscosity of air at laboratory temperature is 1. 0 × 10−6 𝑁𝑚−2𝑠, calculate: a) The mass, andb) The charge on the droplet (Assuming g=9.8ms-2) Given Data: Distance between Plates 𝑑 = 5 𝑚𝑚 = 5 × 10−3𝑚, Potential Difference 𝑉 = 780 𝑉 Distance Covered ∆𝑠 = 1.50 𝑚𝑚 = 1.50 × 10−3𝑚, Time 𝑡 = 11.2 𝑠, Density 𝜌 = 900 kg m−3, Viscosity 5 = 1.0 × 10−6 𝑁𝑚−2𝑠 To Determine: Mass of Droplet 𝑚 =?, Charge on Droplet 𝑞 =? Calculations: (a) Mass of Droplet 𝑚 = 4 𝜋𝑟3𝜌 − − − − (1) 3 −3 Terminal Velocity 𝑣 = ∆𝑠 = 1.50×10= 0.1339 × 10−3 𝑚𝑠−1 𝑡𝑡11.2 For a body moving with terminal velocity: 𝑚𝑔 = 6𝜋5𝑟𝑣 ⟹ 4 𝜋𝑟3𝜌𝑔 = 6𝜋5𝑟𝑣∴ 𝐵𝑦 (1) 3𝑡 295𝑣𝑡95𝑣𝑡9 × 1.0 × 10−6 × 0.1339 × 10−3−6 ⟹ 𝑟 =⟹ 𝑟 = √= √= 1.1 × 10 𝑚 2𝑔𝜌2𝑔𝜌2 × 9.8 × 900 Equation (1) becomes: 𝑚 = 4 𝜋𝑟3𝜌 = 4 × 3.14 × (1.1 × 10−6)3 × 900 = 5.018 × 10−15𝑘𝑔 33 −15−3 (b) Charge 𝑞 = 𝑚g𝑑 = 5.018×10×9.8×5×10= 3.15 × 10−19𝐶 𝑉780 12.9 A proton placed in a uniform electric field of 5000 NC-1 directed to right is allowed to go a distance of 10.0 cm from A to B. Calculate (a) Potential difference between the two points (b) Work done by the field (c) The change in P.E. of proton (d) The change in K.E. of the proton (e) Its velocity (mass of proton is 1. 67 × 10−27 𝑘𝑔) Given Data: Electric Field 𝐸 = 5000 NC−1, Distance covered 𝑑 = 10 𝑐𝑚 = 0.1 𝑚, Charge on proton 𝑞 = 1.6 × 10−19 𝐶 = 1𝑒, Mass of proton 𝑚 = 1.67 × 10−27𝑘𝑔 To Determine: (a) Potential Difference 𝑉 =?, (b) Work Done W =?, (c) Change in P.E. ∆𝑈 =? (d) Change in K.E. ∆𝐾 =?, (e) Velocity 𝑣 =? Calculations: (a) As 𝐸 = − ∆𝑉 ⟹ 𝑉 = −𝐸𝑑 = −5000 × 0.1 = −500 𝑉 𝑑 (b) As ∆𝑉 = 𝖶 ⟹ W = 𝑞∆𝑉 = 1𝑒 × 500 𝑉 = −500 𝑒𝑉 𝑞 (c) ∆𝑈 = −W = −500 𝑒𝑉∴ −ve sign indicate decrease of P. E. (d) By work-energy principle: ∆𝐾 = W = 500 𝑒𝑉 (e) As ∆𝐾 = 1 𝑚𝑣2 ⟹ 𝑣2 = 2×∆𝐾 ⟹ 𝑣 = √2×∆K = √2×500 eV=2×500× 1.6 × 10−19𝐽 2𝑚m1.67 × 10−27kg√1.67 × 10−27𝑘g ⟹ 𝑣 = 3.097 × 105 𝑚𝑠−1 12.10 Using zero reference point at infinity, determine the amount by which a point charge of 4. 0 × 10−8 𝐶 alters the electric potential at a point 1.2 m away, when (a) Charge is positive (b) Charge is negative Given Data: Charge 𝑞 = 4.0 × 10−8 𝐶, Distance 𝑟 = 1.2 𝑚 To Determine: (a) Electric Potential when charge is positive 𝑉+ =?, (b) Electric Potential when charge is negative 𝑉− =?, −8 Calculations: (a) 𝑉 = 1 𝑞 = 9 × 109 × 4.0×10= 300 𝑉 +4𝜋𝗌0 𝑟1.2 −8 (b) 𝑉 = 1 −𝑞 = 9 × 109 × (−4.0×10 ) = −300 𝑉 −4𝜋𝗌0 𝑟1.2 12.11 In Bohr's atomic model of hydrogen atom, the electron is in an orbit around the nuclear proton at a distance of 5. 29 × 10−11 𝑚 with a speed of 2. 18 × 106 𝑚𝑠−1. (𝑒 = 1. 60 × 10−19𝐶, mass of electron = 9. 10 × 10−31𝑘𝑔). Find (a) The electric potential that a proton exerts at this distance (b) Total energy of the atom in eV (c) The ionization energy for the atom in eV Given Data: Distance 𝑟 = 5.29 × 10−11 𝑚 , Speed 𝑣 = 2.18 × 106 𝑚𝑠−1, Charge of Electron 𝑒 = 1.60 × 10−19𝐶, Mass of Electron = 9.10 × 10−31𝑘𝑔 To Determine: (a) Electric Potential due to proton 𝑉 =?, (b) Total Energy of atom 𝐸1 =? (c) Ionization Energy of the atom =? −19 Calculations: (a) 𝑉 = 1 𝑞 = 9 × 109 × 1.60×10= 27.22 𝑉 4𝜋𝗌0 𝑟5.29 ×10−11 2 (b) From theory of atomic spectra, the energy of electron in nth orbit: 𝐸 = − 𝑘𝑒− − − − (1) 𝑛2 𝑟𝑛 2 For present case 𝑛 = 1, so equation (1) takes the form: 𝐸 = − 𝑘𝑒2 = − 9×109×(1.60×10−19) 12 𝑟12×5.29 ×10−11 2.18 × 10−18 = −2.18 × 10−18𝐽 = −= −13.6 𝑒𝑉 1.60 × 10−19 (c) As electron possess 13.6 eV energy in the ground state of a H-atom. So, if we want to ionize such H- atom, we must supply 13.6 eV. Hence, the ionization energy of H-atom in ground state is 13.6 eV 12.12 The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of a 750 µF capacitor is 330 V. Determine the energy that is used to produce the flash. Given Data: Potential Difference 𝑉 = 330 𝑉, Capacitance 𝐶 = 750 𝑝𝐹 = 750 × 10−6𝐹 To Determine: Energy 𝑈 =? Calculations: As 𝑈 = 1 𝐶𝑉2 = 0.5 × 750 × 10−6 × (330)2 = 40.83 𝐽 2 12.13 A capacitor has a capacitance of 2. 5 × 10−8 𝐹. In the charging process, electrons are removed from one plate and placed on the other one. When the potential difference between the plates is 450 V, how many electrons have been transferred?(𝑒 = 1. 60 × 10−19𝐶) Given Data: Capacitance 𝐶 = 2.5 × 10−8 𝐹, Potential Difference 𝑉 = 450 𝑉, Charge 𝑞 = 𝑒 = 1.60 × 10−19𝐶 To Determine: Total Number of Electrons Transferred 𝑛 =? Calculations: For a capacitor 𝑞 = 𝐶𝑉 − − − (1), From Quantization of Charges 𝑞 = 𝑛𝑒 − − − (2) −8 Comparing (1) and (2): 𝑛𝑒 = 𝐶𝑉 ⟹ 𝑛 = 𝐶𝑉 = 2.5 ×10 ×450 = 7.03 × 1013 𝑒1.60×10−19 F.Sc. Physics, (1st Year), Multiple Choice Questions (MCQs) CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-1st-year-multiple- choice.html NUMERICAL PROBLEMS F.Sc. Physics, Chapter # 13: CURRENT ELECTRICY 13.1 How many electrons pass through an electric bulb in one minute if the 300 mA current is passing through it? Given Data: Current 𝐼 = 300 𝑚𝐴 = 300 × 10−3𝐴, Time 𝑡 = 1 𝑚i𝑛 = 60 𝑠 To Determine: No. of Electrons 𝑛 =? Calculations: As Current 𝐼 = 𝑞 ⟹ 𝐼 = 𝑛𝑒∴ 𝑞 = 𝑛𝑒(𝑄𝑢𝑎𝑛𝑡i𝑧𝑎𝑡i𝑜𝑛 𝑜ƒ 𝐶ℎ𝑎𝑟𝑔𝑒) 𝑡𝑡 𝐼 × 𝑡300 × 10−3 × 60 ⟹ 𝑛 === 1.125 × 1020 𝑒1.6 × 10−19 13.2 A charge of 90 C passes through a wire in 1 hour and 15 minutes. What is the current in the wire? Given Data: Charge 𝑞 = 90 𝐶, Time 𝑡 = 1 ℎ𝑜𝑢𝑟 𝑎𝑛𝑑 15 𝑚i𝑛𝑢𝑡𝑒𝑠 = 75 𝑚i𝑛𝑢𝑡𝑒𝑠 = 75 × 60𝑠 To Determine: Current 𝐼 =? Calculations: Current 𝐼 = 𝑞 =90= 0.02 𝐴 = 20 𝑚𝐴 𝑡75×60 13.3 Find the equivalent resistance of the circuit (Fig.), total current drawn from the source and the current through each resistor. Given Data: emf 𝑉 = 6 𝑉, Resistances 𝑅1 = 6 Ω, 𝑅2 = 6 Ω, 𝑅3 = 3 Ω, To Determine: (a) Equivalent Resistance 𝑅𝑒𝑞 =?, (b) Total Current Drawn from Source 𝐼 =?, (c) Current Through 𝑅1 = 𝐼1 =?, (d) Current Through 𝑅2 = 𝐼2 =?, (e) Current Through 𝑅3 = 𝐼3 =?, Calculations: (a) (a) Let 𝑅′ is Equivalent resistance of 𝑅 and 𝑅 , then 1 = 1 + 1 = 1 + 1 = 1 ⟹ 𝑅′ = 3 Ω 12𝑅𝘍𝑅1𝑅2663 As 𝑅′ and 𝑅3 are connected in series, therefore: 𝑅𝑒𝑞 = 𝑅′ + 𝑅3 = 3 + 3 = 6 Ω (b) Total Current Drawn from Source 𝐼 = 𝑉 = 6 = 1 𝐴 𝑅𝑒g6 (c) Potential Difference across 𝑅′: 𝑉′ = 𝐼𝑅′ = 1 × 3 = 3 𝑉 𝘍 Current Through 𝑅 = 𝐼 = 𝑉 = 3 = 0.5 𝐴 11𝑅16 𝘍 (d) Current Through 𝑅 = 𝐼 = 𝑉 = 3 = 0.5 𝐴 22𝑅26 (e) Current Through 𝑅3 = 𝐼3 = 𝐼1 + 𝐼2 = 0.5 + 0.5 = 1 𝐴 13.4 A rectangular bar of iron is 2.0 cm by 2.0 cm in cross section and 40 cm long. Calculate its resistance if the resistivity of iron is 11 x 10-8 Ωm. Given Data: Cross-Sectional Area 𝐴 = 2𝑐𝑚 × 2𝑐𝑚 = 4 𝑐𝑚2 = 4 × 10−4 𝑚2, Length 𝑙 = 40 𝑐𝑚 = 40 × 10−2𝑚, Resistivity 𝜌 = 11 × 10−8 Ωm To Determine: Resistance 𝑅 =? −8−2 Calculations: As 𝑅 = 𝜌𝐿 = 11×10 ×40×10= 1.1 × 10−4Ω 𝐴4×10−4 13.5 The resistance of an iron wire at 0 oC is 1x104 Ω. What is the resistance at 500 oC if the temperature coefficient of resistance of iron is 5.2 x 10-3 K-1? Given Data: Resistance at 0 𝑅𝑜 = 1 × 104 Ω, Temperature 𝑇 = 500 Temperature Coefficient of Resistance 𝛼̅ = 5.2 × 10−3 K−1 To Determine: Resistance of Iron at 500 oC 𝑅𝑇 =? Calculations: From definition of Temperature Coefficient of Resistance 𝛼̅ = 𝑅𝑇−𝑅𝑜 ⟹ 𝑅 − 𝑅 = 𝛼̅𝑅 𝑇 𝑅𝑜𝑇𝑇𝑜𝑜 ⟹ 𝑅𝑇 = 𝑅𝑜 + 𝛼̅𝑅𝑜𝑇 = 1 × 104 + 5.2 × 10−3 × 1 × 104 × 500 = 3.6 × 104Ω 13.6 Calculate terminal potential difference of each of cells in circuit of Fig. Given Data: emf of 1st battery s1 = 24 𝑉, Internal Resistance of 1st Battery 𝑟1 = 0.10 Ω, emf of 2nd battery s2 = 6 𝑉, Internal Resistance of 2nd Battery 𝑟2 = 0.90 Ω, External Resistance 𝑅 = 8 Ω To Determine: (a) Terminal Potential Difference of 1st Battery 𝑉𝑡,1 =?, (b) Terminal Potential Difference of 2nd Battery 𝑉𝑡,2 =? Calculations: As 𝑟1, 𝑅 and 𝑟2 are in series: Equivalent Resistance 𝑅𝑒𝑞 = 𝑟1 + 𝑅 + 𝑟2 = 0.1 + 8 + 0.9 = 9 Ω Effective emf s = s1 − s2 = 24 − 6 = 18 𝑉∴ 𝐵𝑜𝑡ℎ 𝑏𝑎𝑡𝑡𝑒𝑟i𝑒𝑠 𝑜𝑝𝑝𝑜𝑠i𝑛𝑔 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟 Current in Circuit 𝐼 = 𝗌 = 18 = 2 𝐴 𝑅𝑒g9 (a) 𝑉𝑡,1 = s1 − 𝐼𝑟1 = 24 − 2 × 0.1 = 23.8 𝑉 (b) 𝑉𝑡,2 = s2 − 𝐼𝑟2 = 6 + 2 × 0.9 = 7.8 𝑉 13.7 Find the current which flows in all the resistances of the circuit of Fig. Given Data: emf of 1st battery s1 = 9 𝑉, emf of 2nd battery s2 = 6 𝑉 Resistances 𝑅1 = 18 Ω, 𝑅2 = 12 Ω To Determine: (a) Current Through 𝑅1, (b) Current Through 𝑅2 Calculations: Let 𝐼1 and 𝐼2 are the current through Loop-1 and Loop-2 respectively Applying KVR for Loop-1: s1 − (𝐼1 − 𝐼2)𝑅1 = 0 ⟹ 9 − 18(𝐼1 − 𝐼2) = 0 ⟹ 18(𝐼1 − 𝐼2) = 9 ⟹ 𝐼1 − 𝐼2 = 0.5 ⟹ 𝐼1 = 0.5 + 𝐼2 − − − − (1) Applying KVR for Loop-2: s2 − (𝐼2 − 𝐼1)𝑅1 − 𝐼2𝑅2 = 0 ⟹ 6 − 18(𝐼2 − 𝐼1) − 12𝐼2 = 0 − − − − (2) Putting value of 𝐼1 from equation (1) in equation (2), we get: 6 − 18(𝐼2 − 0.5 − 𝐼2) − 12𝐼2 = 0 15 ⟹ 6 + 9 − 12𝐼2 = 0 ⟹ 12𝐼2 = 15 ⟹ 𝐼2 = 12 = 1.25 𝐴 Putting value of 𝐼2 in equation (1): 𝐼1 = 0.5 + 1.25 = 1.75 𝐴 (a) Current Through 𝑅1 = 𝐼1 − 𝐼2 = 1.75 − 1.25 = 0.5 𝐴 (b) Current Through 𝑅2 = 𝐼2 = 1.25 𝐴 13.8 Find the current and power dissipated in each resistance of the circuit shown in the Fig. Given Data: emf of 1st battery s1 = 6 𝑉, emf of 2nd battery s2 = 10 𝑉 Resistances 𝑅1 = 𝑅2 = 1 Ω, 𝑅3 = 2 Ω, 𝑅4 = 𝑅5 = 1 Ω, 𝑅6 = 2 Ω To Determine: (a) Current Through Resistances, (b) Power Dissipation in Resistances Calculations: Let 𝐼1 and 𝐼2 are the current through Loop-1 and Loop-2 respectively Applying KVR for Loop-1: s1 − 𝐼1𝑅1 − (𝐼1 − 𝐼2)𝑅3 − 𝐼1𝑅2 = 0 ⟹ 6 − 𝐼1 − 2(𝐼1 − 𝐼2) − 𝐼1 = 0 ⟹ 6 − 𝐼1 − 2𝐼1 + 2𝐼2 − 𝐼1 = 0 ⟹ 6 = 4𝐼1 − 2𝐼2 ⟹ 3 = 2𝐼1 − 𝐼2 ⟹ 𝐼2 = 2𝐼1 − 3− − − − (1) Applying KVR for Loop-2: −s2 − 𝐼2𝑅6 − 𝐼2𝑅4 − (𝐼2 − 𝐼1)𝑅3 − 𝐼2𝑅5 = 0 ⟹ −10 − 2𝐼2 − 𝐼2 − 2(𝐼2 − 𝐼1) − 𝐼2 = 0 ⟹ −10 − 2𝐼2 − 𝐼2 − 2𝐼2 + 2𝐼1 − 𝐼2 = 0 ⟹ −10 − 6𝐼2 + 2𝐼1 = 0 ⟹ 2𝐼1 − 6𝐼2 = 10 ⟹ 𝐼1 − 3𝐼2 = 5 − − − − (2) Putting value of 𝐼2 from equation (1) in equation (2), we get: 𝐼1 − 3(2𝐼1 − 3) = 5 ⟹ 𝐼1 − 6𝐼1 + 9 = 5 ⟹ −5𝐼1 = −4 ⟹ 𝐼1 = 0.8 Putting this value in equation (1), we get: 𝐼2 = 2 × 0.8 − 3 = −1.4 𝐴 (a) Current Through 𝑅1: 𝐼1 = 0.8 𝐴 Current Through 𝑅2: 𝐼1 − 𝐼2 = 0.8 − (−1.4) = 2.2 𝐴 Current Through 𝑅3: 𝐼1 = 0.8 𝐴 Current Through 𝑅4: 𝐼2 = 1.4 𝐴 Current Through 𝑅5: 𝐼2 = 1.4 𝐴 Current Through 𝑅6: 𝐼2 = 1.4 𝐴 (b) Power Dissipation in 𝑅1: 𝑃1 = 𝐼2𝑅1 = (0.8)2 × 1 = 0.64 W 1 Power Dissipation in 𝑅2: 𝑃2 = (𝐼1 − 𝐼2)2𝑅2 = (2.2)2 × 2 = 9.68 W Power Dissipation in 𝑅3: 𝑃3 = 𝐼2𝑅3 = (0.8)2 × 1 = 0.64 W 1 Power Dissipation in 𝑅4: 𝑃4 = 𝐼2𝑅4 = (1.4)2 × 1 = 1.96 W 2 Power Dissipation in 𝑅5: 𝑃5 = 𝐼2𝑅5 = (1.4)2 × 1 = 1.96 W 2 Power Dissipation in 𝑅6: 𝑃6 = 𝐼2𝑅6 = (1.4)2 × 2 = 3.92 W 2 F.Sc. Physics, (1st Year), Exercise Short Questions CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/exercise-short-questions-fsc- physics.html F.Sc. Physics, (2nd Year), Exercise Short Questions CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/exercise-short-questions-fsc- physics_10.html NUMERICAL PROBLEMS F.Sc. Physics, Chapter # 14: ELECTROMAGNETISM 14.1 Find the value of the magnetic field that will cause a maximum force of 7. 0 × 10−3 N on a 20.0 cm straight wire carrying a current of 10.0 A. Given Data: Maximum Magnetic Force 𝐹𝑚 = 7.0 × 10−3 N, Length 𝐿 = 20.0 cm = 0.2 m, Current 𝐼 = 10.0 A To Determine: Magnetic Field Strength 𝐵 =? −3 Calculations: For maximum magnetic force: 𝐹 = 𝐼𝐿𝐵 ⟹ 𝐵 = 𝐹𝑚 = 7.0×10= 3.5 × 10−3𝑇 𝑚𝐼𝐿10.0×0.2 14.2 How fast must a proton move in a magnetic field of 2. 5 × 10−3 T such that the magnetic force is equal to its weight? Given Data: Magnetic Field 𝐵 = 2.5 × 10−3 T, Charge of Proton 𝑞 = 1.6 × 10−19 𝐶, Mass of Proton 𝑚 = 1.67 × 10−27 𝑘𝑔 To Determine: Speed of Proton 𝑣 =? Calculations: For present case: Magnetic Force = Weight ⟹ 𝑒𝑣𝐵 = 𝑚𝑔 ⟹ 𝑣 = 𝑚g 𝑒𝐵 −27 ⟹ 𝑣 =1.67×10×9.8= 4.5 × 10−5 𝑚𝑠−1 1.6×10−19×2.5×10−3 14.3 A velocity selector has a magnetic field of 0.30 T. If a perpendicular electric field of 10,000 Vm-1 is applied, what will be the speed of the particle that will pass through the selector? Given Data: Magnetic field 𝐵 = 0.30 𝑇, Electric Field 𝐸 = 10,000 Vm−1 To Determine: Velocity 𝑣 =? Calculations: 𝑣 = 𝐸 = 10,000 = 3.333 × 104 𝑚𝑠−1 𝐵0.30 14.4 A coil of 0.1 m x 0.1 m and of 200 turns carrying a current of 1.0 mA is placed in a uniform magnetic field of 0.1 T. Calculate the maximum torque that acts on the coil. Given Data: Area of Coil 𝐴 = 0.1 m × 0.1 m = 0.01 m2, Number of Turns 𝑁 = 200 Current 𝐼 = 1 𝑚𝐴 = 10−3𝐴, Magnetic Field 𝐵 = 0.1 𝑇 To Determine: Maximum Torque 𝑟𝑚 =? Calculations: 𝑟𝑚 = 𝑁𝐼𝐴𝐵 = 200 × 10−3 × 0.01 × 0.1 = 2 × 10−4𝐴 14.5 A power line 10.0 m high carries a current 200 A. Find the magnetic field of the wire at the ground. Given Data: Distance from Ground 𝑟 = 10.0 m , Current 𝐼 = 200 A To Determine: Magnetic Field at Ground 𝐵 =? −7 Calculations: By Ampere’s Law: 𝐵(2𝜋𝑟) = 𝜇 𝐼 ⟹ 𝐵 = 𝜇𝑜𝐼 = 4𝜋×10 ×200 = 4 × 10−6 𝑇 𝑜2𝜋𝑟2𝜋×10 F.Sc. Physics, (2nd Year), Complete Physics Notes CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-2nd-year-complete- physics.html 14.6 You are asked to design a solenoid that will give a magnetic field of 0.10 T, yet the current must not exceed 10.0 A. Find the number of turns per unit length that the solenoid should have. Given Data: Magnetic Field 𝐵 = 0.1 𝑇, Current 𝐼 = 10 𝐴, To Determine: Number of Turns per unit length 𝑛 =? Calculations: For Solenoid 𝐵 = 𝜇 𝑛𝐼 ⟹ 𝑛 = 𝐵 =0.1= 7.96 × 103 𝑜𝜇𝑜𝐼4𝜋×10−7×10 14.7 What current should pass through a solenoid that is 0.5 m long with 10000 turns of copper wire so that it will have a magnetic field of 0.4 T? Given Data: Length 𝐿 = 0.5 𝑇, Number of Turns 𝑁 = 10000, Magnetic Field 𝐵 = 0.4 𝑇 To Determine: Current 𝐼 =? Calculations: For Solenoid 𝐵 = 𝜇 𝑛𝐼 ⟹ 𝐵 = 𝜇𝑜𝑁𝐼 ⟹ 𝐼 = 𝐵𝐿 =0.4 ×0.5= 15.92𝐴 ≈ 16 𝐴 𝑜𝐿𝜇𝑜𝑁4𝜋×10−7×- A galvanometer having an internal resistance Rg = 15.0 Ω gives full scale deflection with current lg = 20.0 mA. It is to be converted into an ammeter of range 10.0 A. Find the value of shunt resistance Rs. Given Data: Internal Resistance of Galvanometer 𝑅g = 15.0 Ω, Current Range 𝐼 = 10 𝐴 Current for Full Scale Deflection 𝐼g = 20.0 mA = 20 × 10−3𝐴 To Determine: Shunt Resistance 𝑅𝑆 =? Calculations: For Ammeter 𝑅 = 𝐼𝑔𝑅𝑔 = 20×10−3×15 = 0.03 Ω 𝑆𝐼−𝐼𝑔10−20×10−3 14.9 The resistance of a galvanometer is 50.0 Ω and reads full scale deflection with a current of 2.0 mA. Show by a diagram how to convert this galvanometer into voltmeter reading 200 V full scale. Given Data: Internal Resistance of Galvanometer 𝑅g = 50.0 Ω, Voltage Range 𝑉 = 200 𝑉 Current for Full Scale Deflection 𝐼g = 2.0 mA = 2 × 10−3𝐴 To Determine: Series High Resistance 𝑅ℎ =? Calculations: For voltmeter 𝑅 = 𝑉 − 𝑅 =200 − 50 = 99950 Ω ℎ𝐼𝑔g2×10−3 F.Sc. Physics, (2nd Year), Multiple Choice Questions (MCQs) CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-2nd-year-multiple- choice.html 14.10 The resistance of a galvanometer coil is 10.0 Ω and reads full scale with a current of 1.0 mA. What should be the values of resistances R1, R2 and R3 to convert this galvanometer into a multi-range ammeter of 100, 10.0 and 1.0 A. Given Data: Internal Resistance of Galvanometer 𝑅g = 10.0 Ω, Current for Full Scale Deflection 𝐼g = 1.0 mA = 1 × 10−3𝐴, (a) Current Range 𝐼1 = 100 𝐴, (b) Current Range 𝐼2 = 10 𝐴, (c) Current Range 𝐼3 = 1 𝐴 To Determine: (a) Shunt Resistance for 100 A Current Range 𝑅1 =? (b) Shunt Resistance for 10 A Current Range 𝑅2 =?, (c) Shunt Resistance for 1A Current Range 𝑅3 =? Calculations: (a) 𝑅 = 𝐼𝑔𝑅𝑔 = 1×10−3×10 = 0.0001 Ω 1𝐼1−𝐼𝑔100−1×10−3 (b) 𝑅 = 𝐼𝑔𝑅𝑔 = 1×10−3×10 = 0.001 Ω 2𝐼2−𝐼𝑔10−1×10−3 (c) 𝑅 = 𝐼𝑔𝑅𝑔 = 1×10−3×10 = 0.01 Ω 3𝐼3−𝐼𝑔1−1×10−3 NUMERICAL PROBLEMS F.Sc. Physics, Chapter # 15: ELECTROMAGNETIC INDUCTION 15.1 An emf of 0.45 V is induced between the ends of a metal bar moving through a magnetic field of 0.22 T. What field strength would be needed to produce an emf of 1.5 V between the ends of the bar, assuming that all other factors remain the same? Given Data: Case 1: Induced emf s1 = 0.45 V , Magnetic field strength 𝐵1 = 0.22 T Case 2: Induced emf s2 = 1.5 V , To Determine: Magnetic field strength 𝐵2 =? Calculations: For case 1: s1 = 𝑣𝐵1𝐿 sin 𝜃 − − − − (1), For case 2: s2 = 𝑣𝐵2𝐿 sin 𝜃 − − − − (2) Dividing (1) and (2): 𝗌2 = 𝑣𝐵2𝐿 sin 𝜃 ⟹ 𝗌2 = 𝐵2 ⟹ 𝐵 = 𝗌2 × 𝐵 = 1.5 × 0.22 = 0.73 𝑇 𝗌1𝑣𝐵1𝐿 sin 𝜃𝗌1𝐵12𝗌110.45 15.2 The flux density B in a region between the pole faces of a horse-shoe magnet is 0.5 Wbm-2 directed vertically downward. Find the emf induced in a straight wire 5.0 cm long, perpendicular to B when it is moved in a direction at an angle of 60° with the horizontal with a speed of 100 cms-1. Given Data: Flux Density 𝐵 = 0.5 Wbm−2 , Length 𝐿 = 5.0 cm = 5 × 10−2𝑚, Speed 𝑣 = 100 cms−1 = 1 ms−1, Angle with Horizontal 𝜃′ = 60° To Determine: Induced emf s =? Calculations: Angle with B: 𝜃 = 90° − 60° = 30° By expression of Motional emf s = 𝑣𝐵𝐿 sin 𝜃 = 1 × 0.5 × 5 × 10−2 × sin 30° = 1.25 × 10−2𝑉 15.3 A coil of wire has 10 loops. Each loop has an area of 1.5 x 10-3 m2. A magnetic field is perpendicular to the surface of each loop at all times. If the magnetic field is changed from 0.05T to 0.06 T in 0.1 s, find the average emf induced in the coil during this time. Given Data: Number of Loops 𝑁 = 10, Area of Loop 𝐴 = 1.5 × 10−3 m2, Time ∆𝑡 = 0.1 𝑠 Initial Magnetic Field 𝐵i = 0.05T , Final Magnetic Field 𝐵ƒ = 0.06T , Angle with normal 𝜃 = 0° To Determine: Induced emf s =? Calculations: Change in Magnetic Field ∆𝐵 = 𝐵ƒ − 𝐵i = 0.06 = 0.05 = 0.01 𝑇 −3 By Faraday’s Law s = 𝑁 ∆Ø = 𝑁 (∆𝐵)𝐴 cos 𝜃 = 10 × (0.01 )×1.5 ×10 ×cos 0 = 1.5 × 10−3 V ∆𝑡∆𝑡0.1 15.4 A circular coil has 15 turns of radius 2 cm each. The plane of the coil lies at 40° to a uniform magnetic field of 0.2 T. If the field is increased by 0.5 T in 0.2 s, find the magnitude of the induced emf. Given Data: Number of Loops 𝑁 = 15, Radius of Circular Loop 𝑟 = 2 cm = 0.02, Time ∆𝑡 = 0.2 𝑠 Initial Magnetic Field 𝐵i = 0.2 T , Final Magnetic Field 𝐵ƒ = 0.5 T, Angle of plane with B: 𝜃′ = 40° To Determine: Induced emf s =? Calculations: Change in Magnetic Field ∆𝐵 = 𝐵ƒ − 𝐵i = 0.5 = 0.2 = 0.3 𝑇, Angle of normal to plane with B 𝜃 = 90° − 40° = 50°, Area of Loop 𝐴 = 𝜋𝑟2 = 3.14 × (0.02)2 =- 𝑚2 By Faraday’s Law s = 𝑁 ∆Ø = 𝑁 (∆𝐵)𝐴 cos 𝜃 = 15 × (0.3 )×-×cos 50° = 1.8 × 10−2 V ∆𝑡∆𝑡0.2 15.5 Two coils are placed side by side. An emf of 0.8 V is observed in one coil when the current is changing at the rate of 200 As-1 in the other coil. What is the mutual inductance of the coils? Given Data: Induced emf s = 0.8 V, Rate of Change of current ∆𝐼𝑝 = 200 As−1 𝑠∆𝑡 To Determine: Mutual Inductance 𝑀 =? Calculations: From definition of Mutual Inductance 𝑀 = 𝗌𝑠 = 0.8 = 4 × 10−3𝐻 = 4 𝑚𝐻 ∆𝐼𝑝200 ( ∆𝑡 ) 15.6 A pair of adjacent coils has a mutual inductance of 0.75 H. If the current in the primary changes from 0 to 10 A in 0.025 s, what is the average induced emf in the secondary? What is the change in flux in it if the secondary has 500 turns? Given Data: Mutual Inductance 𝑀 = 0.75 H, Change of current in Primary Coil ∆𝐼𝑝 = 10 − 0 = 10 𝐴 Time ∆𝑡 = 0.025 s, Number of Turns of Secondary Coil 𝑁𝑠 = 500 To Determine: (a) Average Induced emf s𝑠 =?, (b) Change in Flux in Secondary Coil ∆Φ =? Calculations:(a) From definition of Mutual Inductance 𝑀 = 𝗌𝑠 ⟹ s = 𝑀 ∆𝐼𝑝) = 0.75 × 10 = 7.5 V ∆𝐼𝑝𝑠( ∆𝑡 ( ∆𝑡 ) (b) By Faraday’s Law: s = 𝑁 ∆Ø ⟹ ∆Φ = 𝗌𝑠×∆𝑡 = 7.5×0.025 = 1.5 × 10−2 W𝑏 𝑠𝑠 ∆𝑡𝑁𝑠500 15.7 A solenoid has 250 turns and its self inductance is 2.4 mH. What is the flux through each turn when the current is 2 A? What is the induced emf when the current changes at 20 As-1? Given Data: Number of Turns 𝑁 = 250, Self Inductance 𝐿 = 2.4 mH = 2.4 × 10−3 H, Current 𝐼 = 2 𝐴, Rate of Chang of current ∆𝐼 = 20 As−1 ∆𝑡 To Determine: (a) Flux Φ =? (b) Induced emf s =? −3 Calculations: (a) For the case of a coil: 𝑁Ф = 𝐿𝐼 ⟹ Ф = 𝐿𝐼 = 2.4×10 ×2 = 1.92 × 10−5W𝑏 𝑁250 (b) As Self Inductance 𝐿 = 𝗌 ⟹ s = 𝐿 (∆𝐼) = 2.4 × 10−3 × 20 = 48 × 10−3 = 48 𝑚𝐻 (∆𝐼)∆𝑡 ∆𝑡 15.8 A solenoid of length 8.0 cm and cross sectional area 0.5 cm2 has 520 turns. Find the self-inductance of the solenoid when the core is air. If the current in the solenoid increases through1.5A in 0.2 s, find the magnitude of induced emf in it. (µo= 4π x 10-7 WbA-1m-1) Given Data: Length of Solenoid 𝑙 = 8.0 𝑐𝑚 = 8 × 10−2 𝑚, Number of Turns 𝑁 = 520 Cross-Sectional Area 𝐴 = 0.5 cm2 = 0.5 × 10−4 m2, Change in Current ∆𝐼 = 1.5 𝐴, Time ∆𝑡 = 0.2 𝑠 To Determine: (a) Self Inductance 𝐿 =?, (b) Induced emf s =? 2 Calculations: (a) For the case of a coil: 𝑁Ф = 𝐿𝐼 ⟹ 𝐿 = 𝑁Φ = 𝑁𝐵𝐴 = 𝑁(𝜇𝑜𝑛𝐼)𝐴 = 𝑁 (𝜇 𝑁) 𝐴 = 𝜇𝑜𝑁 𝐴 𝐼𝐼𝐼𝑜 𝑙𝑙 𝜇𝑜𝑁2𝐴4𝜋 × 10−7 × (520)2 × 0.5 × 10−4 ⟹ 𝐿 === 2.12 × 10−4 𝐻 𝑙8 × 10−2 (b) As Self Inductance 𝐿 = 𝗌 ⟹ s = 𝐿 (∆𝐼) = 2.12 × 10−4 × 1.5) = 1.6 × 10−3 𝑉 ∆𝐼( ( )∆𝑡0.2 ∆𝑡 15.9 When current through a coil changes from 100 mA to 200 mA in 0.005 s, an induced emf of 40 mV is produced in the coil. (a) What is the self-inductance of the coil? (b) Find the increase in the energy stored in the coil. Given Data: Initial Current 𝐼i = 100 𝑚𝐴 = 0.1 𝐴, Final Current 𝐼ƒ = 200 𝑚𝐴 = 0.2 𝐴, Time ∆𝑡 = 0.005 s, Induced emf s = 40 𝑚𝑉 = 40 × 10−3 𝑉, To Determine: (a) Self Inductance 𝐿 =?, (b) Increase in Energy Stored ∆𝑈 =? Calculations: Change in Current ∆𝐼 = 0.2 − 0.1 = 0.1 𝐴 −3 (a) As Self Inductance 𝐿 = 𝗌 = 40×10= 2 × 10−3 𝐻 (∆𝐼)( 0.1 ) ∆𝑡0.005 −3 (b) ∆𝑈 = 𝑈 − 𝑈 = 1 𝐿𝐼2 − 1 𝐿𝐼2 = 𝐿 (𝐼2 − 𝐼2) = 2×10[(0.2)2 − (0.1)2] = 0.03 × 10−3 𝐽 ƒi2ƒ2i2 ƒi2 15.10 Like any field, the earth's magnetic field stores energy. Find the magnetic energy stored in a space where strength of earth's field is 7 x 10-5 T, if the space occupies an area of 10x108 m2 and has a height of 750 m. Given Data: Magnetic Field Strength 𝐵 = 7 × 10−5 T, Area 𝐴 = 10 × 108 m2, Height 𝑙 = 750 m To Determine: Magnetic Energy 𝑈𝑚 =? 2 Calculations: 𝑈= 𝐵2 (𝐴𝑙) = (7×10−5) (10 × 108 × 750) = 1.46 × 109𝐽 𝑚2𝜇𝑜2×4𝜋×10−7 15.11 A square coil of side 16 cm has 200 tums and rotates in a uniform magnetic field of magnitude 0.05 T. If the peak emf is 12 V, what is the angular velocity of the coil? Given Data: Length of side of square loop 𝑙 = 16 cm = 0.16 𝑚, Number of turns 𝑁 = 200, Magnetic field strength 𝐵 = 0.05 T, Peak emf s𝑜 = 12 𝑉 To Determine: Angular Velocity 𝜔 =? Calculations: For square 𝐴 = 𝑙2 = (0.16 )2 = 0.0256 𝑚2 As s = 𝑁𝜔𝐴𝐵 ⟹ 𝜔 = 𝗌𝑜 =12= 46.87 𝑟𝑎𝑑 𝑜𝑁𝐴𝐵200×0.0256×0.05𝑠 15.12 A generator has a rectangular coil consisting of 360 turns. The coil rotates at 420 rev per min in 0.14 T magnetic field. The peak value of emf produced by the generator is 50 V. If the coil is 5.0 cm wide, find the length of the side of the coil. Given Data: Number of turns 𝑁 = 360, Angular Velocity 𝜔 = 420 rev = 420 2 rad = 14𝜋 rad, min60 ss Magnetic field strength 𝐵 = 0.14 T, Peak emf s𝑜 = 50 𝑉, Breadth of Coil 𝑏 = 5.0 cm = 0.05 m To Determine: Length of Coil 𝑙 =? Calculations: As s𝑜 = 𝑁𝜔𝐴𝐵 = 𝑁𝜔(𝑙 × 𝑏)𝐵∴ 𝐴 = 𝑙 × 𝑏 (𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑐𝑜i𝑙) s𝑜 50 ⟹ 𝑙 === 0.45 𝑚 = 45 𝑐𝑚 𝑁𝜔𝑏𝐵360 × 14𝜋 × 0.05 × 0.14 15.13 It is desired to make an a.c. generator that can produce an emf of maximum value 5 kV with 50 Hz frequency. A coil of area 1 m2 having 200 turns is used as armature. What should be the magnitude of the magnetic field in which the coil rotates? Given Data: Peak emf s𝑜 = 5 𝑘𝑉 = 5000 𝑉, Frequency ƒ = 50 𝐻𝑧, Area 𝐴 = 1 m2, Turn 𝑁 = 200 To Determine: Magnetic Field Strength 𝐵 =? Calculations: As s = 𝑁𝜔𝐴𝐵 ⟹ 𝐵 = 𝗌𝑜 =𝗌𝑜=5000= 0.08 𝑇 𝑜𝑁𝜔𝐴𝑁(2𝜋ƒ)𝐴200(2𝜋×50)×1 15.14 The back emf in a motor is 120 V when the motor is turning at 1680 rev per min. What is the back emf when the motor turns 3360 rev per min? Given Data: Case 1: Back emf s = 120 𝑉, Angular Velocity 𝜔 = 1680 rev 1min Case 2: Angular Velocity 𝜔 = 3360 rev min To Determine: Back emf for Case-2: s2 =? Calculations: For Case-1: s1 = 𝑁𝜔1𝐴𝐵 − − − − (1), For Case-2: s2 = 𝑁𝜔2𝐴𝐵 − − − − (2) Dividing eq (1) and (2): 𝗌2 = 𝑁𝜔2𝐴𝐵 ⟹ 𝗌2 = 𝜔2 ⟹ s = 𝜔2 × s = 3360 × 120 = 240 𝑉 𝗌1𝑁𝜔1𝐴𝐵𝗌1𝜔12𝜔111680 15.15 A D.C motor operates at 240 V and has a resistance of 0.5 Ω. When the motor is running at normal speed, the armature current is 15 A. Find the back emf in the armature. Given Data: Operating Voltage 𝑉 = 240 V, Resistance 𝑟 = 0.5 Ω, Current 𝐼 = 15 𝐴 To Determine: Back emf s =? Calculations: As s = 𝑉 − 𝐼𝑟 = 240 − 15 × 0.5 = 232.5 𝑉 15.16 A copper ring has a radius of 4.0 cm and resistance of 1.0 mΩ. A magnetic field is applied over the ring, perpendicular to its plane. If the magnetic field increases from 0.2 T to 0.4 T in a time interval of 5 x 10-3 s, what is the current in the ring during this interval? Given Data: Radius of Ring 𝑟 = 4 𝑐𝑚 = 0.04 𝑚, Resistance 𝑟 = 1.0 mΩ = 10−3Ω, Loops 𝑁 = 1 Initial Magnetic Field 𝐵i = 0.2 𝑇, Final Magnetic Field 𝐵ƒ = 0.4 𝑇, Time ∆𝑡 = 5 × 10−3 s To Determine: Current in the Loop 𝐼 =? Calculations: Change in Magnetic Field ∆𝐵 = 𝐵ƒ − 𝐵i = 0.4 − 0.2 = 0.2 𝑇, Area of Loop 𝐴 = 𝜋𝑟2 22 By Faraday’s Law, Induced emf s = 𝑁 ∆Ø = 𝑁 (∆B)𝐴 = 𝑁 (∆B)𝜋𝑟 = 1 × (0.2)3.14×(0.04) = 0.20096 𝑉 ∆𝑡∆𝑡∆𝑡5×10−3 Induced Current 𝐼 = 𝗌 = 0.20096 = 200.96 ≈ 201 𝐴 𝑅10−3 15.17 A coil of 10 turns and 35 cm2 area is in a perpendicular magnetic field of 0.5 T. The coil is pulled out of the field in 1.0 s. Find the induced emf in the coil as it is pulled out of the field. Given Data: Number of Turns 𝑁 = 10, Area of Coil 𝐴 = 35 cm2 = 35 × 10−4 m2 Initial Magnetic Field 𝐵i = 0.5 T, Final Magnetic Field 𝐵ƒ = 0 𝑇, Time ∆𝑡 = 1 𝑠 To Determine: Induced emf s =? Calculations: Change in Magnetic Field ∆𝐵 = 𝐵ƒ − 𝐵i = 0 − 0.5 = −0.5 𝑇 −4 By Faraday’s Law, s = 𝑁 ∆Ø = 𝑁 (∆B)𝐴 = 10 × (0.5)×35×10= 1.75 × 10−2 𝑉 ∆𝑡∆𝑡1 15.18 An ideal step down transformer is connected to main supply of 240 V. It is desired to operate a 12 V, 30 W lamp. Find the current in the primary and the transformation ratio? Given Data: Primary Voltage 𝑉𝑃 = 240 V, Secondary Voltage 𝑉𝑠 = 12 𝑉, Output Power 𝑃𝑠 = 30 W To Determine: (a) Current in Primary Coil 𝐼 =?, (b) Transformation Ratio 𝑁𝑠 =? 𝑃𝑁𝑃 Calculations: (a) As for ideal transformer, Power Input 𝑃𝑃 = Power Output 𝑃𝑠 ⟹ 𝑉𝑃𝐼𝑃 = 𝑃𝑠 𝑃𝑠30 ⟹ 𝐼𝑃 = 𝑉 = 240 = 0.125 𝑉 𝑃 (b) As Transformation Ratio 𝑁𝑠 = 𝑉𝑆 = 12 = 1 𝑁𝑃𝑉𝑃24020 F.Sc. Physics, (1st Year), Complete Physics Notes CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-complete-physics- notes.html F.Sc. Physics, (2nd Year), Complete Physics Notes CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-2nd-year-complete- physics.html NUMERICAL PROBLEMS F.Sc. Physics, Chapter # 16: ALTERNATING CURRENT Q # 1. An alternating current is represented by the equation 𝐼 = 20 sin(100𝜋𝑡). Compute its frequency and the maximum and the rms value of current. Given Data: Equation of current 𝐼 = 20 sin(100𝜋𝑡) − − − (1) To Determine: Frequency ƒ =?, Maximum Current 𝐼𝑜 =?, RMS Current 𝐼𝑅𝑀𝑆 =? Calculations: General Equation of A.C. 𝐼 = 𝐼𝑜 sin(2𝜋ƒ𝑡) − − − (2) Comparing eq (1) and (2), we have: i.2𝜋ƒ𝑡 = 100𝜋𝑡 ⟹ ƒ = 50 𝐻𝑧 ii.𝐼𝑜 = 20 𝐴 iii.𝐼= 𝐼𝑜 = 20 = 14 𝐴 𝑅𝑀𝑆√2√2 Q # 2. A solenoidal A.C. has a maximum value of 15 A. What is its rms values? If the time is recorded from the instant the current is zero and is becoming positive, what is the instantaneous value of current after 1/300 s, given the frequency is 50 Hz. Given Data: Maximum Current 𝐼 = 15 𝐴, Time 𝑡 = 1 𝑠, Frequency ƒ = 50 𝐻𝑧 𝑜300 To Determine: RMS Current 𝐼𝑅𝑀𝑆 =?, Instantaneous Alternating Current 𝐼 =? Calculations: i.𝐼= 𝐼𝑜 = 15 = 10.6 𝐴 𝑅𝑀𝑆√2√2 ii.Equation of Instantaneous Alternating Current 𝐼 = 𝐼 sin(2𝜋ƒ𝑡) = 15 sin (2𝜋 × 50 × 1 ) = 12.96 𝐴 𝑜300 Q # 3. Find the values of the current and inductive reactance when A.C. voltage of 220 V at 50 Hz is passed through an inductor of 10 H. Given Data: Voltage 𝑉 = 220 𝑉, Frequency ƒ = 50 𝐻𝑧, Inductance 𝐿 = 10 𝐻 To Determine: Inductive Reactance X𝐿 =?, Current 𝐼 =? Calculations: i.X𝐿 = 𝜔𝐿 = 2𝜋ƒ𝐿 = 2 × 3.14 × 50 × 10 = 3140 Ω ii.𝐼 = 𝑉 = 220 = 0.07 𝐴 K𝐿3140 Q # 4. An inductor has an inductance of 1/𝜋 H and resistance of 2000 Ω. A 50 Hz A.C. is supplied to it. Calculate the reactance and impedance offered by the circuit. Given Data: Inductance 𝐿 = 1 𝐻, Resistance 𝑅 = 2000 Ω, Frequency ƒ = 50 𝐻𝑧 𝜋 To Determine: Inductive Reactance X𝐿 =?, Impedance 𝑍 =? Calculations: i.X = 𝜔𝐿 = 2𝜋ƒ𝐿 = 2𝜋 × 50 × 1 = 100 Ω 𝐿𝜋 ii.𝑍 = √𝑅2 + X2 = √(2000)2 + (100)2 = 2002.5 Ω 𝐿 Q # 5. An inductor of pure inductance 3/𝜋 H is connected in series with a resistance of 40 Ω. Find (i) the peak value of current (ii) the rms value, and (iii) the phase difference between the current and the applied voltage 𝑉 = 350 sin(100 𝜋𝑡). Given Data: Inductance 𝐿 = 3 𝐻, Resistance 𝑅 = 40 Ω, 𝜋 Equation of Instantaneous Voltage 𝑉 = 350 sin(100 𝜋𝑡) − − − (1) To Determine: Peak Value of Current 𝐼𝑜 =?, RMS Current 𝐼𝑅𝑀𝑆 =?, Phase Difference 𝜑 =? Calculations: General Equation of Instantaneous Voltage 𝑉 = 𝑉𝑜 sin(2𝜋ƒ𝑡) − − − (2) Comparing equations (1) and (2), we have: 𝑉𝑜 = 350 𝑉, 2𝜋ƒ𝑡 = 100 𝜋𝑡 ⟹ ƒ = 50 𝐻𝑧 Inductive Reactance X = 𝜔𝐿 = 2𝜋ƒ𝐿 = 2𝜋 × 50 × 3 = 300 Ω 𝐿𝜋 Impedance 𝑍 = √𝑅2 + X2 = √(40)2 + (300)2 = 302.6 Ω 𝐿 i.𝐼 = 𝑉𝑜 = 350 = 1.16 𝐴 𝑜Z302.6 ii.𝐼= 𝐼𝑜 = 1.16 = 0.81 𝐴 𝑅𝑀𝑆√2√2 iii.𝜑 = tan−1 K𝐿) = tan−1 300) = 82.4° (( 𝑅40 Q # 6. A 10 mH, 20 Ω coil is connected across 240 V and 180/𝜋 Hz source. How much power does it dissipate? Given Data: Inductance 𝐿 = 10 𝑚𝐻 = 0.01 𝐻, Resistance 𝑅 = 20 Ω, Peak Value of Voltage 𝑉𝑜 = 240 𝑉, Frequency ƒ = 180 𝐻𝑧 𝜋 To Determine: Power Dissipation 𝑃 =? Calculations: Inductive Reactance X = 𝜔𝐿 = 2𝜋ƒ𝐿 = 2𝜋 × 180 × 0.01 = 3.6 Ω 𝐿𝜋 Impedance 𝑍 = √𝑅2 + X2 = √(20)2 + (3.6)2 = 20.32 Ω 𝐿 Peak Value of Current 𝐼 = 𝑉𝑜 = 240 = 11.81 𝐴 𝑜Z20.32 Phase Difference 𝜑 = tan−1 (K𝐿) = tan−1 3.6) = 10.20° ( 𝑅20 Power Dissipation 𝑃 = 𝑉𝑜𝐼𝑜 cos 𝜑 = - ) cos(10.20°) = 2789.6 Ω Q # 7. Find the value of the current flowing through a capacitance 0. 5 𝜇𝐹 when connected to a source of 150 V at 50 Hz. Given Data: Capacitance 𝐶 = 0.5 𝜇𝐹 = 0.5 × 10−6 𝐹, Peak Value of Voltage 𝑉𝑜 = 150 𝑉, Frequency ƒ = 50 𝐻𝑧 To Determine: Current 𝐼 =? Calculations: Capacitive Reactance X = 1 =1=1= 6369.4 Ω 𝐶𝜔𝐶2𝜋ƒ𝐶2×3.14×50×0.5×10−6 Current 𝐼 = 𝑉𝑜 = 150 = 0.024 𝐴 K𝐶6369.4 Q # 8. An alternating source of emf 12 V and frequency 50 Hz is applied to a capacitor of capacitance 3 𝜇𝐹 in series with a resistor of resistance of 1 𝑘Ω. Calculate the phase angle. GivenData:𝑒𝑚ƒ 𝐸 = 12 𝑉,Frequencyƒ = 50 𝐻𝑧,Capacitance𝐶 = 0.3 𝜇𝐹 = 0.3 × 10−6 𝐹, Resistance 𝑅 = 1 𝑘Ω = 1000 Ω To Determine: Phase Angle 𝜑 =? Calculations: Capacitive Reactance X = 1 =1=1= 1062 Ω 𝐶𝜔𝐶2𝜋ƒ𝐶2×3.14×50×0.3×10−6 𝜑 = tan−1 X𝐶−1 1062 () = tan() = 46.7° 𝑅1000 Q # 9. What is the resonance frequency of a circuit which include a coil of inductance 2.5 H and a capacitance 40 𝜇𝐹? Given Data: Inductance 𝐿 = 2.5 𝐻, Capacitance 𝐶 = 40 𝜇𝐹 = 40 × 10−6 𝐹 To Determine: Resonant Frequency ƒ𝑅 =? Calculations: ƒ =1=1= 15.9 𝐻𝑧 𝑅2𝜋√𝐿𝐶2×3.14×√2.5×40×10−6 Q # 10. An inductor of inductance 150 𝜇𝐻 is connected in parallel with a variable capacitor whose capacitance can be changed from 500 𝑝𝐹 to 20 𝑝𝐹. Calculate the maximum frequency and minimum frequency for which the circuit can be tuned. Given Data: Inductance 𝐿 = 150 𝜇𝐻 = 150 × 10−6𝐻, Initial Capacitance 𝐶1 = 500 𝑝𝐹 = 500 × 10−12 𝐹, Final Capacitance 𝐶2 = 20 𝑝𝐹 = 20 × 10−12 𝐹 To Determine: Minimum Frequency ƒ1 =?, Maximum Frequency ƒ2 =?, Calculations: 11 Minimum Frequency ƒ1 === 0.58 × 106 𝐻𝑧 = 0.58 𝑀𝐻𝑧 2𝜋√𝐿𝐶12 × 3.14 × √150 × 10−6 × 500 × 10−12 11 Maximum Frequency ƒ2 === 2.91 × 106 𝐻𝑧 = 2.91 𝑀𝐻𝑧 2𝜋√𝐿𝐶22 × 3.14 × √150 × 10−6 × 20 × 10−12 NUMERICAL PROBLEMS F.Sc. Physics, Chapter # 17: PHYSICS OF SOLIDS 17.1 A 1.25 cm diameter cylinder is subjected to a load of 2500 kg. Calculate the stress on the bar in mega pascals. Given Data: Diameter of Cylinder 𝑑 = 1.25 cm , Load 𝑚 = 2500 kg To Determine: Stress 𝜎 =? Calculations: As 𝜎 = 𝐹 = 𝑚g = 4𝑚g = 4×2500×9.8 = 199.7 × 106 Pa = 199.7 MPa ≈ 200 MPa 𝐴(𝜋𝑑2)𝜋𝑑23.14×(1.25)2 4 17.2 A 1.0 m long copper wire is subjected to stretching force and its length increases by 20 cm. Calculate the tensile strain and the percent elongation which the wire undergoes. Given Data: Length 𝑙 = 1.0 m , Elongation ∆𝑙 = 20 cm = 0.2 m, To Determine: (a) Tensile Strain =?, (b) % Elongation =? Calculations: (a) Tensile Strain = ∆𝑙 = 0.2 = 0.2 𝑙1 (b) % Elongation = ∆𝑙 × 100 = 0.2 × 100 = 20 % 𝑙1 17.3 A wire 2.5 m long and cross-section area 10−5 𝑚2 is stretched 1.5 mm by a force of 100 N in the elastic region. Calculate (i) the strain (ii) Young's modulus (iii) the energy stored in the wire. Given Data: Length 𝑙 = 2.5 m, Cross-Section Area 𝐴 = 10−5 𝑚2, Force 𝐹 = 100 𝑁 Elongation ∆𝑙 = 1.5 mm = 1.5 × 10−3 m To Determine: (i) Strain =? (ii) Young′s modulus Y =? (iii) Energy stored in the wire =? −3 Calculations: (i) Strain = ∆𝑙 = 1.5×10= 6 × 10−4 𝑙2.5 (F)( 100 ) (ii) Young′s modulus Y = Stress =A= 10−5 =100= 1.66 × 1010 𝑃𝑎 StrainStrain6×10−410−5×6×10−4 2 (iii) Energy stored in the wire = 1 F𝐴(∆𝑙)2 = 1 1.66×1010×10−5×(1.5×10−3) = 0.075 𝐽 = 7.5 × 10−2𝐽 2𝑙22.5 17.4 What stress would cause a wire to increase in length by 0.01% if the Young's modulus of the wire is 12 x 1010 Pa. What force would produce this stress if the diameter of the wire is 0.56 mm? Given Data: % 𝐸𝑙𝑜𝑛𝑔𝑎𝑡i𝑜𝑛 = 0.01 %, Young Modulus Y = 12 × 1010 Pa, Diameter 𝑑 = 0.56 mm = 0.56 × 10−3 𝑚 To Determine: Force 𝐹 =? Calculations: % Elongation = 0.01 % ⟹ ∆𝑙 × 100 = 0.01 ⟹ ∆𝑙 = 0.01 = 10−4 𝑙𝑙100 F Stress(A)∆𝑙F∆𝑙 Young Modulus Y = Strain ⟹ 𝑌 = ∆𝑙 ⟹ 𝑌 × ( 𝑙 ) = 𝐴 ⟹ 𝐹 = 𝐴 [𝑌 × ( 𝑙 )] ( 𝑙 ) 𝜋𝑑2∆𝑙3.14 × (0.56 × 10−3)2 ⟹ 𝐹 = () [𝑌 × ()] ⟹ 𝐹 = [] [12 × 1010 × 10−4] = 2.95 𝑁 4𝑙4 17.5 The length of a steel wire is 1.0 m and its cross-sectional area is 0. 03 × 10−4 𝑚2. Calculate the work done in stretching the wire when a force of 100 N is applied within the elastic region. Young's modulus of steel is 3. 0 × 1011 𝑁𝑚−2. Given Data: Length of wire 𝑙 = 1 𝑚, Cross-Sectional Area 𝐴 = 0.03 × 10−4 𝑚2, Force 𝐹 = 100 𝑁 Young Modulus Y = 3.0 × 1011 𝑁𝑚−2 To Determine: W𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 = 1 (𝐹)(∆𝑙) =?, Here ∆𝑙 =? 2 F Calculations: Young Modulus Y = Stress ⟹ 𝑌 = (A) ⟹ 𝑌 = F×l ⟹ ∆𝑙 = F×l =100×1 Strain(∆𝑙)∆𝑙×𝐴F×𝐴3.0×1011×0.03×10−4 𝑙 ⟹ ∆𝑙 = 1.11 × 10−4 𝑚 Now W𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 = 1 (𝐹)(∆𝑙) = 1 (100)(1.11 × 10−4) = 5.55 × 10−3𝐽 22 17.6 A cylindrical copper wire and a cylindrical steel wire each of length 1.5 m and diameter 2.0 mm are joined at one end to form a composite wire 3.0 m long. The wire is loaded until its length becomes 3.003 m. Calculate the strain in copper and steel wires and the force applied to the wire. (Young's modulus of copper is 1. 2 × 1011 𝑃𝑎 and for steel is 2. 0 × 1011 𝑃𝑎). Given Data: Case 1 (Copper wire): Length 𝑙𝑐 = 1.5 𝑚, Diameter 𝑑𝑐 = 2 𝑚𝑚 = 0.002 𝑚 Case 2 (Steel wire): Length 𝑙𝑠 = 1.5 𝑚, Diameter 𝑑𝑠 = 2 𝑚𝑚 = 0.002 𝑚 Length of Composite Wire 𝑙 = 𝑙𝑐 + 𝑙𝑐 = 3 𝑚, Final Length 𝑙′ = 3.003 𝑚, Elongation ∆𝑙 = 0.003 𝑚 Young Modulus for Copper 𝑌𝑐 = 1.2 × 1011 𝑃𝑎 , Young Modulus for Steel 𝑌𝑠 = 2 × 1011 𝑃𝑎 To Determine: (a) Strain in Copper Wire s𝑐 =?, (b) Strain in Steel Wire s𝑠 =?, (c) Force 𝐹 =? Calculations: As Young Modulus Y = Stress 𝜎, Strain 𝗌 For Copper wire 𝑌 = 𝜎𝑐 ⟹ 𝜎 = 𝑌 s − − − (1), For Steel wire 𝑌 = 𝜎𝑠 ⟹ 𝜎 = 𝑌 s − − − (2) 𝑐𝗌𝑐𝑐𝑐 𝑐𝑠𝗌𝑠𝑠𝑠 𝑠 ∆𝑙𝑐∆𝑙𝑠 As both wire experience equal stress: 𝜎𝑐 = 𝜎𝑠 ⟹ 𝑌𝑐s𝑐 = 𝑌𝑠s𝑠 ⟹ 𝑌𝑐 ( 𝑙 ) = 𝑌𝑠 ( 𝑙 ) 𝑐𝑠 ⟹ 𝑌𝑐∆𝑙𝑐 = 𝑌𝑠∆𝑙𝑠 − − − (3)∴ 𝑙𝑐 = 𝑙𝑠 = 1.5 𝑚 Let ∆𝑙𝑐 = 𝑥 − − − (4), then ∆𝑙𝑠 = 0.003 − 𝑥 − − − − (5) Equation (3) Becomes: 1.2 × 1011 × 𝑥 = 2 × 1011 × (0.003 − 𝑥) ⟹ 1.2 𝑥 = 0.006 − 2𝑥 ⟹ 3.2 𝑥 = 0.006 ⟹ 𝑥 = 0.006 = 1.872 × 10−3 𝑚. 3.2 Therefore, by equation (4) and (5): ∆𝑙𝑐 = 1.872 × 10−3 𝑚 & ∆𝑙𝑠 = 0.003 − 1.872 × 10−3 −3 (a) Strain in Copper Wire s = ∆𝑙𝑐 = 1.872 ×10= 1.25 × 10−3 & 𝑐𝑙𝑐1.5 −3 (b) Strain in Steel Wire s = ∆𝑙𝑠 = 0.003−1.872 ×10= 0.75 × 10−3 𝑠𝑙𝑠1.5 (c) As 𝑆𝑟𝑒𝑠𝑠 (𝜎) = 𝐹𝑜𝑟𝑐𝑒 (𝐹) ⟹ 𝐹 = 𝜎𝐴 = 𝜎 𝐴∴ 𝑆𝑟𝑒𝑠𝑠 𝑜𝑛 𝐵𝑜𝑡ℎ 𝑤i𝑟𝑒𝑠 i𝑠 𝑠𝑎𝑚𝑒 i. 𝑒. , 𝜎 = 𝜎 𝐴𝑟𝑒𝑎 (𝐴)𝑐𝑐 𝜋𝑑23.14 × (0.002)2 ⟹ 𝐹 = 𝑌𝑐s𝑐 () = 1.2 × 1011 × 1.25 × 10−3 ×= 471.2 𝑁∴ 𝐵𝑦 (1) 44 F.Sc. Physics, (1st Year), Multiple Choice Questions (MCQs) CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-1st-year-multiple- choice.html F.Sc. Physics, (2nd Year), Multiple Choice Questions (MCQs) CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-2nd-year-multiple- choice.html NUMERICAL PROBLEMS F.Sc. Physics, Chapter # 18: ELECTRONICS 18.1 The current flowing into the base of a transistor is 100 µA. Find its collector current 𝐼𝑐, its emitter current 𝐼𝐸 and the ratio 𝐼𝑐, if the value of current gain β is 100. 𝐼𝐸 Given Data: Base Current 𝐼𝐵 = 100 µA = 100 × 10−6𝐴 = 10−4𝐴, Current Gain 𝛽 = 100 To Determine: (a) Collector Current 𝐼 =?, (b) Emitter Current 𝐼 =?, (c) 𝐼𝑐 =? 𝑐𝐸𝐼𝐸 Calculations: (a) Collector Current 𝐼𝑐 = 𝛽𝐼𝐵 = 100 × 100 × 10−6 = 10−2𝐴 = 10 𝑚𝐴 (b) Emitter Current 𝐼𝐸 = 𝐼𝑐 + 𝐼𝐵 = 10−2𝐴 + 10−4𝐴 = 0.0101 𝐴 = 10.1 𝑚𝐴 (c) 𝐼𝑐 = 10 𝑚𝐴 = 0.99 𝐼𝐸10.1 𝑚𝐴 18.2 Fig. shows a transistor which operates a relay as the switch S is closed. The relay is energized by a current of 10 mA. Calculate the value 𝑅𝑆 which will just make the relay operate. The current gain β of the transistor is 200. When the transistor conducts, its 𝑉𝐵𝐸 can be assumed to be 0.6 V. Given Data: Collector Current 𝐼𝑐 = 10 mA = 10−2𝐴, Current Gain 𝛽 = 200, 𝑉𝐵𝐸 = 0.6 𝑉, 𝑉𝑐𝑐 = 9 𝑉 To Determine: Base Resistance 𝑅𝑆 =? Calculations: As 𝛽 = 𝐼𝑐 ⟹ 𝐼 = 𝐼𝑐 = 10−2 = 0.5 × 10−4 𝐴 𝐼𝐵𝐵𝛽200 Applying KVR on input circuit: 𝑉𝑐𝑐 − 𝐼𝐵𝑅𝑆 − 𝑉𝐵𝐸 = 0 ⟹ 𝐼𝐵𝑅𝑆 = 𝑉𝑐𝑐 − 𝑉𝐵𝐸 𝑉𝑐𝑐 − 𝑉𝐵𝐸9 − 0.6 ⟹ 𝑅𝑆 =𝐼= 0.5 × 10−4 = 168000 Ω = 168 𝑘Ω 𝐵 18.3 In circuit (Fig.), there is negligible potential drop between B and E, if β is 100. Calculate i. base current ii. collector current iii. potential drop across 𝑅𝐶 iv.𝑉𝐶𝐸 Given Data: 𝑉𝑐𝑐 = 9 𝑉, 𝑉𝐵𝐸 ≈ 0, 𝑅𝐵 = 800 𝑘Ω = 800000 Ω, 𝑅𝐶 = 1 𝑘Ω = 1000 Ω, 𝛽 = 100 To Determine: (a) base current 𝐼𝐵 =?, (b) collector current 𝐼𝑐 =?, (c) potential drop across 𝑅𝐶 =?, (d) 𝑉𝐶𝐸 =? Calculations: (a) Applying KVR on Input Circuit: 𝑉𝑐𝑐 − 𝐼𝐵𝑅𝐵 − 𝑉𝐵𝐸 = 0 ⟹ 𝐼𝐵𝑅𝐵 = 𝑉𝑐𝑐 − 𝑉𝐵𝐸 𝑉𝑐𝑐 − 𝑉𝐵𝐸9 − 0 ⟹ 𝐼𝐵 === 11.25 × 10−6𝐴 𝑅𝐵800000 (b) Collector Current 𝐼𝑐 = 𝛽𝐼𝐵 = 100 × 11.25 × 10−6 = 11.25 × 10−4𝐴 (c) Potential Drop Across 𝑅𝐶 = 𝑉𝐶 = 𝐼𝐶 𝑅𝐶 = 11.25 × 10−4 × 1000 = 1.125 𝑉 (d) Applying KVR on Output Circuit: 𝑉𝑐𝑐 − 𝐼𝐶𝑅𝐶 − 𝑉𝐶𝐸 = 0 ⟹ 𝑉𝐶𝐸 = 𝑉𝑐𝑐 − 𝐼𝐶𝑅𝐶 ⟹ 𝑉𝐶𝐸 = 11.25 × 10−4 × 1000 = 9 − 1.125 = 7.875 𝑉 18.4 Calculate the output of the op-amp circuit shown in Fig. Given Data: Let 𝑅1 = 10 𝑘Ω = 10000 Ω, 𝑅2 = 4 𝑘Ω = 4000 Ω, 𝑅3 = 20 𝑘Ω = 20000 Ω To Determine: Output Voltage 𝑉𝑜 =? Calculations: Let I1 = Current Through R1, I2 = Current Through R2, I3 = Current Through R3 By Kirchhoff’s Current Rule: I + I = I ⟹ (5−0) + (−2−0) = (0−𝑉𝑜- ⟹ 0.5 × 10−3 − 0.5 × 10−3 = −𝑉𝑜 ⟹ −𝑉𝑜 = 0 ⟹ 𝑉 =-𝑜 18.5 Calculate the gain of non-inverting amplifier shown in Fig. Given Data: Let 𝑅1 = 10 𝑘Ω = 10000 Ω, 𝑅2 = 40 𝑘Ω = 40000 Ω To Determine: Gain 𝐺 =? Calculations: For Non-Inverting Op-Amp: 𝐺 = 1 + 𝑅2 = 1 + 40000 = 1 + 4 = 5 𝑅110000 F.Sc. Physics, (1st Year), Complete Physics Notes CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-complete-physics- notes.html F.Sc. Physics, (1st Year), Multiple Choice Questions (MCQs) CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-1st-year-multiple- choice.html F.Sc. Physics, (1st Year), Exercise Short Questions CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/exercise-short-questions-fsc- physics.html NUMERICAL PROBLEM F.Sc. Physics, Chapter # 19: DAWN OF MODERN PHYSICS 19.1 A particle called the pion lives on the average only about 2. 6 × 10−8 𝑠 when at rest in the laboratory. It then changes to another form. How long would such a particle live when shooting through the space at 0.95 c? Given Data: Proper Time 𝑡𝑜 = 2.6 × 10−8 𝑠, Relative Speed 𝑣 = 0.95 c To Determine: Relativistic Time 𝑡 =? Calculations: 𝑡 = 𝑡𝑜 = 2.6×10−8 = 2.6×10−8 = 2.6×10−8 = 8.3 × 10−8 𝑠 √1−𝑣2√1−(0.95 c)2√1−(0.95 )2𝑐2√1−(0.95 )2 𝑐2𝑐2𝑐2 19.2 What is the mass of a 70 kg man in a space rocket traveling at 0.8 c from us as measured from Earth? Given Data: Proper Mass 𝑚𝑜 = 70 kg , Relative Speed 𝑣 = 0.8 c To Determine: Relativistic Time 𝑚 =? Calculations: 𝑚 = 𝑚𝑜 =70=70=70= 116.7 𝑘𝑔 √1−𝑣2√1−(0.8 c)2√1−(0.8 )2𝑐2√1−(0.8)2 𝑐2𝑐2𝑐2 19.3 Find the energy of photon in a) Radiowave of wavelength 100 m b) Green light of wavelength 550 nm c) X-ray with wavelength 0.2 nm Given Data: (a) Wavelength of Radiowave 𝜆1 = 100 m, (b) Wavelength of Green Light 𝜆2 = 550 nm = 550 × 10−9𝑚, (c) Wavelength of X-ray 𝜆3 = 0.2 nm = 0.2 × 10−9𝑚 To Determine: (a) Energy of Radiowave 𝐸1 =?, (b) Energy of Green Light 𝐸2 =?, (c) Energy of X-ray 𝐸3 =? −348−28 Calculations: (a) Energy of Radiowave 𝐸 = hc = 6.63×10×3×10 = 19.89 × 10−28 𝐽 = 19.89×10= 1.24 × 10−8 𝑒𝑉-×10−19 −348−19 (b) Energy of Green Light 𝐸 = hc = 6.63×10×3×10 = 3.6 × 10−19 𝐽 = 3.6×10= 2.25 𝑒𝑉 22550×10−91.6×10−19 −348−16 (c) Energy of X-ray 𝐸 = hc = 6.63×10×3×10 = 9.945 × 10−16 𝐽 = 9.945×10= 6215 𝑒𝑉 330.2×10−91.6×10−19 19.4 Yellow light of 577 nm wavelength is incident on a cesium surface. The stopping voltage is found to be 0.25 V. Find a) the Maximum K. E. of the photoelectrons b) the work function of cesium Given Data: Wavelength of Yellow Light 𝜆 = 577 nm = 577 × 10−9𝑚, Stopping Potential 𝑉0 = 0.25 V To Determine: (a) Maximum K. E. 𝐾. 𝐸𝑚𝑎𝑥 =?, (b) Work Function 𝜑 =? Calculations: (a) Maximum K. E. 𝐾. 𝐸𝑚𝑎𝑥 = 𝑒𝑉0 = 1.6 × 10−19 × 0.25 = 4 × 10−20 𝐽 (b) By Quantum Theory of Photoelectric Effect 𝜑 = ℎƒ − 𝐾. 𝐸⟹ 𝜑 = ℎ𝑐 − 𝐾. 𝐸∴ ƒ = 𝑐 𝑚𝑎𝑥 𝑚𝑎𝑥 −348−19 ⟹ 𝜑 = 6.63×10×3×10 − 4 × 10−20 = 3.147 × 10−19 𝐽 = 3.047×10𝑒𝑉 = 1.9 𝑒𝑉 577×10−91.6×10−19 19.5 X-rays of wavelength 22 pm are scattered from a carbon target. The scattered radiation being viewed at 85° to the incident beam. What is Compton shift? Given Data: Wavelength 𝜆 = 22 pm = 22 × 10−12𝑚, Scattering Angle 𝜃 = 85° To Determine: Compton Shift ∆𝜆 =? −34 Calculations: ∆𝜆 = ℎ (1 − cos 𝜃) =6.63×10(1 − cos 85°) = 2.428 × 10−12 × 0.913 𝑚𝑜𝑐9.1×10−31×3×108 ⟹ ∆𝜆 = 2.2 × 10−12 𝑚 = 2.2 𝑝𝑚 19.6 A 90 keV X-ray photon is fired at a carbon target and Compton scattering occurs. Find the wavelength of the incident photon and the wavelength of the scattered photon for scattering angle of (a) 30° (b) 60° Given Data: Energy of Photon 𝐸 = 90 keV = 90 × 103 × 1.6 × 10−19𝐽 = 1.44 × 10−14 𝐽 To Determine: (a) Wavelength of Incident Photon 𝜆 =?, (b) Wavelength of Scattered Photon at 30°: 𝜆′ =?, (c) Wavelength of Scattered Photon at 60°: 𝜆′′ =? −348 Calculations: (a) As 𝐸 = hc ⟹ 𝜆 = hc = 6.63×10×3×10 = 13.8 × 10−12 𝑚 = 13.8 𝑝𝑚 𝐸1.44×10−14 (b) ∆𝜆 = ℎ (1 − cos 𝜃) ⟹ 𝜆′ − 𝜆 = ℎ (1 − cos 𝜃) ⟹ 𝜆′ = 𝜆 + ℎ (1 − cos 𝜃) 𝑚𝑜𝑐𝑚𝑜𝑐𝑚𝑜𝑐 6.63 × 10−34 ⟹ 𝜆′ = 13.8 × 10−12 +(1 − cos 30°) = 13.8 × 10−12 + 0.324 × 10−12 9.1 × 10−31 × 3 × 108 ⟹ 𝜆′ = 14.12 × 10−12 𝑚 = 14.12 𝑝𝑚 −34 (c) 𝜆′′ = 𝜆 + ℎ (1 − cos 𝜃) = 13.8 × 10−12 +6.63×10(1 − cos 60°) = 13.8 × 10−12 + 1.21 × 10−12 𝑚𝑜𝑐9.1×10−31×3×108 ⟹ 𝜆′′ = 15 × 10−12 𝑚 = 15 𝑝𝑚 19.7 What is the maximum wavelength of the two photons produced when a positron annihilates an electron? The rest mass energy of each is 0.51 MeV. Given Data: Energy 𝐸 = 0.51 MeV = 0.51 × 106 × 1.6 × 10−19𝐽 = 8.16 × 10−14 𝐽 To Determine: Wavelength 𝜆 =? −348 Calculations: As 𝐸 = hc ⟹ 𝜆 = hc = 6.63×10×3×10 = 2.44 × 10−12 𝑚 = 2.44 𝑝𝑚 𝐸8.16×10−14 19.8 Calculate the wavelength of a) a 140 g ball moving at 40 𝑚𝑠−1 b) a proton moving at the same speed c) an electron moving at the same speed Given Data: Speed 𝑣 = 40 𝑚𝑠−1 (a) Mass of Ball 𝑚1 = 140 g = 0.14 kg, (b) Mass of Proton 𝑚2 = 1.67 × 10−27 kg, (c) Mass of Electron 𝑚3 = 9.1 × 10−31 kg To Determine: (a) Wavelength associated with Ball 𝜆1 =?, (b) Wavelength associated with Proton 𝜆2 =?, (c) Wavelength associated with Electron 𝜆3 =? −34 Calculations: (a) 𝜆 = h = 6.63×10= 1.18 × 10−34 𝑚 1𝑚1𝑣0.14×40 −34 (b) 𝜆 = h =6.63×10= 9.92 × 10−9 𝑚 = 9.92 𝑛𝑚 2𝑚2𝑣1.67×10−27×40 −34 (c) 𝜆 = h = 6.63×10= 1.82 × 10−5 𝑚 3𝑚3𝑣9.1×10−31×40 19.9 What is the de Broglie wavelength of an electron whose kinetic energy is 120 eV? Given Data: 𝐾. 𝐸. = 120 eV = 120 × 1.6 × 10−19𝐽 = 192 × 10−12 𝐽, Mass 𝑚 = 9.1 × 10−31 kg To Determine: De Broglie Wavelength 𝜆 =? Calculations: As De Broglie Wavelength 𝜆 = h− − − − − (1) 𝐻𝑒𝑟𝑒 𝑣 =? 𝑚𝑣 −12−12 As 𝐾. 𝐸. = 192 × 10−12 𝐽 ⟹ 1 𝑚𝑣2 = 192 × 10−12 𝐽 ⟹ 𝑣2 = 2×192×10⟹ 𝑣 = √2×192×10 2𝑚𝑚 ⟹ 𝑣 = √2×192×10−12 = 6.5 × 106 𝑚𝑠−1 9.1×10−31 −34 Putting values in (1): 𝜆 =6.63×10= 1.12 × 10−10 𝑚 9.1×10−31×6.5×- An electron is placed in a box about the size of an atom that is about 1. 0 × 10−10 𝑚. What is the velocity of the electron? Given Data: Size of Box ∆𝑥 = 1.0 × 10−10 𝑚, Mass 𝑚 = 9.1 × 10−31 kg To Determine: Velocity ∆𝑣 =? Calculations: By Uncertainty Principle: ∆𝑥. ∆𝑝 = ℎ ⟹ ∆𝑥. 𝑚∆𝑣 = ℎ ⟹ ∆𝑣 = ℎ ∆𝑥.𝑚 −34 ⟹ ∆𝑣 =6.63×10= 7.9 × 106 𝑚𝑠−1 1.0×10−10×9.1×10−31 NUMERICAL PROBLEM F.Sc. Physics, Chapter # 20: ATOMIC SPECTRA 20.1 A hydrogen atoms is in its ground state (n = 1). Using Bohr's theory, calculate (a) the radius of the orbit (b) the linear momentum of the electron (c) the angular momentum of the electron (d) the kinetic energy (e) the potential energy and (f) the total energy. Given Data: 𝑛 = 1, Mass of electron 𝑚 = 9.1 × 10−31 𝑘𝑔 To Determine: (a) Radius 𝑟 =?, (b) Linear Momentum 𝑝 =? (c) Angular Momentum 𝐿 =? (d) Kinetic Energy 𝐾 =? (e) Potential Energy 𝑈 =? (f) Total Energy 𝑇 =? 2 2 Calculations: (a) As 𝑟 =𝑛 ℎ. For 1st orbit, 𝑛 = 1. Therefore, 𝑛4𝜋2𝐾𝑒2𝑚 2 Radius of 1st orbit: 𝑟 =(1)2(6.63×10−34)= 5.39 × 10−11 𝑚 14×(3.14)2×9×109×(1.6×10−19)2×9.1×10−31 2𝜋𝐾𝑒2 (b) Linear Momentum 𝑝𝑛 = 𝑚𝑣𝑛 = 𝑚 (). For 1st orbit, 𝑛 = 1. Therefore, 𝑛ℎ 2 Linear Momentum 𝑝 = 𝑚 (2𝜋𝐾𝑒2) = 9.1 × 10−31 × (2×3.14×9×109×(1.6×10−19) ) = 1.99 × 10−24 Ns 11×ℎ1×6.63×10−34 (c) Angular Momentum 𝐿1 = 𝑚𝑣1𝑟1 = 𝑝1𝑟1 = 1.99 × 10−24 × 5.39 × 10−11 = 1.05 × 10−34 Js 2 2 (d) Kinetic Energy 𝐾 = 1 𝑚𝑣2 = 1 𝑚 2𝜋𝐾𝑒 ) . For 1st Orbit, 𝑛 = 1. Therefore, 𝑛2𝑛2( 𝑛ℎ 2 2 12𝜋𝐾𝑒2 212×3.14×9×109×(1.6×10−19) 𝐾1 =× 𝑚 × () =× 9.1 × 10−31 × () = 2.17 × 10−18 𝐽 21×ℎ21×6.63×10−34 −18 ⟹ 𝐾 = 2.17×10𝑒𝑉 = 13.6 𝑒𝑉 11.6×10−19 2 (e) Potential Energy 𝑈 = − 𝐾𝑒2 = − 𝐾𝑒2 = − 9×109×(1.6×10−19) = −4.35 × 10−18 𝐽 𝑟𝑛𝑟15.39×10−11 −18 ⟹ 𝑈 = −4.35×10𝑒𝑉 = −27.2 𝑒𝑉 1.6×10−19 (f) Total Energy 𝑇 = 𝐾 + 𝑈 = 13.6 𝑒𝑉 + (−27.2 𝑒𝑉) = −13.6 𝑒𝑉 20.2 What are the energies in eV of quanta of wavelength? λ = 400, 500 and 700 nm. Given Data: Wavelengths of Quanta: (a) 𝜆1 = 400 𝑛𝑚 = 400 × 10−9 𝑚, (b) 𝜆2 = 500 𝑛𝑚 = 500 × 10−9 𝑚, (c) 𝜆3 = 700 𝑛𝑚 = 700 × 10−9 𝑚 To Determine: Energies of Quanta: (a) 𝐸1 =?, (b) 𝐸2 =?, (c) 𝐸3 =? −348−19 Calculations: (a) 𝐸 = hc = 6.63×10×3×10 = 4.97 × 10−19 𝐽 = 4.97×10= 3.10 𝑒𝑉 11400×10−91.6×10−19 −348−19 (b) 𝐸 = hc = 6.63×10×3×10 = 3.98 × 10−19 𝐽 = 3.98×10= 2.49 𝑒𝑉 22500×10−91.6×10−19 −348−19 (c) 𝐸 = hc = 6.63×10×3×10 = 2.84 × 10−19 𝐽 = 2.84×10= 1.77 𝑒𝑉 33700×10−91.6×10−19 20.3 An electron jumps from a level 𝐸i = −3. 5 × 10−19 𝐽 to 𝐸𝑓 = −1. 20 × 10−18 𝐽. What is the wavelength of the emitted light? Given Data: Energy in ground state 𝐸i = −3.5 × 10−19 𝐽,Energy in Excited State 𝐸ƒ = −1.20 × 10−18 𝐽 To Determine: Wavelength 𝜆 =? −348−348 Calculations: ∆𝐸 = hc ⟹ 𝜆 = hc =hc=6.63×10×3×10= 6.63×10×3×10 = 234 × 10−9 𝑚 ∆𝐸𝐸ƒ−𝐸i−1.20×10−18−(−3.5 ×10−19)8.5 ×10−19 20.4 Find the wavelength of the spectral line corresponding to the transition in hydrogen from n = 6 state to n = 3 state? Given Data: 𝑝 = 3, 𝑛 = 6 To Determine: Wavelength 𝜆 =? Calculations: 1 = 𝑅 ( 1 − 1 ) = 1.0974 × 107 × ( 1 − 1 ) = 1.0974 × 107 × 1 − 1 ) 𝐻 𝑝2𝑛- − 1336 ⟹= 1.0974 × 107 × () = 1.0974 × 107 × () ⟹ 𝜆 == 1093 𝑛𝑚 𝜆- × 107 × 3 20.5 Compute the shortest wavelength radiation in the Balmer series? What value of n must be used? Given Data: For Balmer series 𝑝 = 2, For Shortest Wavelength 𝑛 = ∞ To Determine: Shortest Wavelength 𝜆𝑠 =? Calculations: 1 = 𝑅 ( 1 − 1 ) = 1.0974 × 107 × ( 1 − 1 ) = 1.0974 × 107 × 1 − 0) 𝐻 𝑝2𝑛222∞2( 𝑠4 ⟹ 1 = 1.0974 × 107 × 1) ⟹ 𝜆 =4= 365.5 𝑛𝑚 (𝑠7 𝑠41.0974×10 20.6 Calculate the longest wavelength of radiation for the Paschen series. Given Data: For Paschen Series 𝑝 = 3, For Shortest Wavelength 𝑛 = 4 To Determine: Longest Wavelength 𝜆𝐿 =? Calculations: 1 = 𝑅 ( 1 − 1 ) = 1.0974 × 107 × ( 1 − 1 ) = 1.0974 × 107 × 1 − 1 ) 𝐻 𝑝2𝑛- 𝐿 ⟹ 1 = 1.0974 × 107 × (16−9) = 1.0974 × 107 × ( 7 ) ⟹ 𝜆 =16= 1875 𝑛𝑚 𝐿144144𝑠1.0974×107×7 20.7 Electrons in an X-ray tube are accelerated through a potential difference of 3000 V. If these electrons were slowed down in a target, what will be the minimum wavelength of X—rays produced? Given Data: Potential Difference 𝑉0 = 3000 𝑉 To Determine: Minimum Wavelength 𝜆𝑚i𝑛 =? −348 Calculations: 𝐸 = hc ⟹ 𝑉 𝑒 = hc ⟹ 𝜆= hc = 6.63×10×3×10 = 4.14 × 10−10 𝑚 𝑚i𝑛0𝑚i𝑛𝑚i𝑛𝑉0𝑒3000×1.6×10−19 20.8 The wavelength of K X-ray from copper is 1. 377 × 10−10 𝑚. What is the energy difference between the two levels from which this transition results? Given Data: Wavelength 𝜆 = 1.377 × 10−10 𝑚 To Determine: Energy Difference ∆𝐸 =? −348−16 Calculations: ∆𝐸 = hc = 6.63×10×3×10 = 14.14 × 10−16 𝐽 = 14.14×10𝑒𝑉 = 9.025 × 103 𝑒𝑉 1.377 × 10−101.6×10−19 20.9 A tungsten target is struck by electrons that have been accelerated from rest through 40 kV potential difference. Find the shortest wavelength of the bremsstrahlung radiation emitted. Given Data: Potential Difference 𝑉0 = 40 𝑘𝑉 = 40000 𝑉 To Determine: Minimum Wavelength 𝜆𝑚i𝑛 =? −348 Calculations: 𝐸 = hc ⟹ 𝑉 𝑒 = hc ⟹ 𝜆= hc = 6.63×10×3×10 = 0.31 × 10−10 𝑚 𝑚i𝑛0𝑚i𝑛𝑚i𝑛𝑉0𝑒40000×1.6×10−19 20.10 The orbital electron of a hydrogen atom moves with a speed of 5. 456 × 105 𝑚𝑠−1 (a) Find the value of the quantum number n associated with this electron. (b) Calculate the radius of this orbit. (c) Find the energy of the electron in this orbit. Given Data: Speed of Electron 𝑣𝑛 = 5.456 × 105 𝑚𝑠−1 To Determine: (a) Quantum Number 𝑛 =?, (b) Orbital Radii 𝑟𝑛 =?, (c) Energy of Electron 𝐸𝑛 =?, 2 Calculations: (a) As 𝑣 = 2𝜋𝑘𝑒2 ⟹ 𝑛 = 2𝜋𝑘𝑒2 = 2×3.14×9×109×(1.6×10−19) = 4 𝑛𝑛ℎℎ𝑣𝑛6.63×10−34×5.456 ×105 2 (b) As Orbital Radii 𝑟 =𝑛2ℎ2=(4)2(6.63×10−34)= 0.846 × 10−9 𝑚 𝑛4𝜋2𝐾𝑒2𝑚4×(3.14)2×9×109×(1.6×10−19)2×9.1×10−31 24 (c) Energy of Electron 𝐸 = 1 4𝜋2𝑘2𝑒4𝑚) = 1 (4×(3.14)2×(9×109) ×(1.6×10−19) ×9.1×10−31) 𝑛𝑛2 (ℎ2(4)2(6.63×10−34 2 ) 1.35 × 10−19 ⟹ 𝐸𝑛 = −1.35 × 10−19 𝐽 = −𝑒𝑉 = −0.85 𝑒𝑉 1.6 × 10−19 F.Sc. Physics, (2nd Year), Complete Physics Notes CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-2nd-year-complete- physics.html F.Sc. Physics, (2nd Year), Multiple Choice Questions (MCQs) CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/fsc-physics-2nd-year-multiple- choice.html F.Sc. Physics, (2nd Year), Exercise Short Questions CLICK THE LINK TO DOWNLOAD https://houseofphy.blogspot.com/2017/12/exercise-short-questions-fsc- physics_10.html NUMERICAL PROBLEM F.Sc. Physics, Chapter # 21: NUCLEAR PHYSICS 21 .1 Find the mass defect and the binding energy for tritium, if the atomic mass of tritium is- u. Given Data: Mass of Tritium 𝑚𝑇 =- µ, Mass of Proton 𝑚𝑃 =- µ, Mass of Neutron 𝑚𝑁 =- µ, For Tritium 𝑍 = 1, 𝐴 = 3 To Determine: (a) Mass Defect ∆𝑚 =?, (b) Binding Energy 𝐸𝐵 =? Calculations: (a) ∆𝑚 = [𝑍 𝑚𝑃 + (𝐴 − 𝑍)𝑚𝑁] = [1 ×- + 2 ×- ] =- µ (b) 𝐸𝐵 = ∆𝑚𝑐2 =- × 931 𝑀𝑒𝑉 = 7.97 𝑀𝑒𝑉 21.2 The half-life of 91𝑆𝑟 is 9.70 hours. Find its decay constant. Given Data: Half Life 𝑇1/2 = 9.7 ℎ𝑜𝑢𝑟𝑠 = 34920 𝑠 To Determine: Decay Constant 𝜆 =? Calculations: As 𝑇= 0.693 ⟹ 𝜆 = 0.693 = 0.693 = 1.98 × 10−5 𝑠−1 1/2 𝑇- 21.3 The element 234𝑃𝑎 is unstable and decays by β-emission with a half-life 6.66 s. State the nuclear reaction and the daughter nuclei. Given Data: Parent Nucleus: 𝐴X = 234𝑃𝑎, Half Life 𝑇1/2 = 6.66 s Z91 To Determine: Nuclear Reaction, Daughter Nucleus for 𝛽 − 𝑑𝑒𝑐𝑎𝑦: Z 𝐴𝑌 =? +1 Calculations: As for 𝛽 − 𝑑𝑒𝑐𝑎𝑦: 𝐴X →𝐴𝑌 + 0𝛽. ZZ+1−1 For Present Case: 234𝑃𝑎 → 234𝑌 + 0𝛽, where 234𝑌 = 234𝑈 9192−19292 21 .4 Find the energy associated with the following reaction: (Mass of 1𝐻 = 1. 00784 µ) 14𝑁 + 4𝐻𝑒 → 17𝑂 + 1𝐻 What does negative sign indicate? 7281 Given Data: Mass of 14𝑁 = 14.0031 µ, Mass of 4𝐻𝑒 = 4.00264 µ, Mass of 17𝑂 = 16.991 µ, 728 Mass of 1𝐻 = 1.00784 µ 1 To Determine: Associated Energy 𝑄 =? Calculations: Mass Defect ∆𝑚 = (Mass of 14𝑁 + Mass of 4𝐻𝑒) − (𝑀𝑎𝑠𝑠 𝑜ƒ 17𝑂 + 𝑀𝑎𝑠𝑠 𝑜ƒ 1𝐻) 7281 ⟹ ∆𝑚 = (14.0031 + 4.00264) − (16.991 + 1.00784) = −0.0012 µ Now Associated Energy 𝑄 = ∆𝑚𝑐2 = −0.0012 × 931 𝑀𝑒𝑉 = 1.12 𝑀𝑒𝑉 The negative sign indicate that energy is needed to start nuclear reaction. 21 .5 Determine the energy associated with the following reaction: 14𝐶 → 14𝑁 +0 Given Data: Mass of 14𝐶 =- µ, Mass of 14𝑁 =- µ, Mass of 0𝑒 =- µ 67−1 To Determine: Associated Energy 𝑄 =? Calculations: Mass Defect ∆𝑚 = (Mass of 14𝐶) − (𝑀𝑎𝑠𝑠 𝑜ƒ 14𝑁 + 𝑀𝑎𝑠𝑠 𝑜ƒ 0𝑒) 67−1 ⟹ ∆𝑚 = -) − - -) =- µ Now Associated Energy 𝑄 = ∆𝑚𝑐2 =- × 931 𝑀𝑒𝑉 = 3.77 𝑀𝑒𝑉 21 .6 lf 233𝑈 decays twice by α-emission, what is the resulting isotope? Given Data: Parent Nucli: : 𝐴X = 233𝑈 Z92 To Determine: As for 𝛼 − 𝑑𝑒𝑐𝑎𝑦: 𝐴X → 𝐴−4𝑌 + 4𝐻𝑒 ZZ−22 Calculations: 1st Decay: 233𝑈 → 229𝑌 + 4𝐻𝑒, Here 229𝑌 = 229𝑇ℎ-nd Decay: 229𝑇ℎ → 225𝑌′ + 4𝐻𝑒, Here 225𝑌′ = 225𝑅𝑎- So resulting isotope is radium 21 .7 Calculate the energy (in MeV) released in the following fusion reaction; 2𝐻 + 3𝐻 → 4𝐻𝑒 + 1𝑛 1120 Given Data: Mass of 2𝐻 =- µ, Mass of 3𝐻 = 3.01605 µ, Mass of 4𝐻𝑒 = 4.00263 µ, 112 Mass of 1𝑛 =- µ 0 To Determine: Associated Energy 𝑄 =? Calculations: Mass Defect ∆𝑚 = (Mass of 2𝐻 + Mass of 3𝐻) − (𝑀𝑎𝑠𝑠 𝑜ƒ 4𝐻𝑒 + 𝑀𝑎𝑠𝑠 𝑜ƒ 1𝑛) 1120 ⟹ ∆𝑚 = - + 3.01605) − (4.00263 -) =- µ Now Associated Energy 𝑄 = ∆𝑚𝑐2 =- µ × 931 𝑀𝑒𝑉 = 17.6 𝑀𝑒𝑉 21 .8 A sheet of lead 5.0 mm thick reduces the intensity of a beam of γ-rays by a factor 0.4. Find half value thickness of lead sheet which will reduce the intensity to half of its initial value. Given Data: Case 1 Thickness 𝑥1 = 5 𝑚𝑚 = 5 × 10−3𝑚, Intensity 𝐼1 = 0.4 𝐼0 Case 2 Intensity 𝐼2 = 0.5 𝐼0 To Determine: Thickness of Lead Sheet for case 2: 𝑥2 =? Calculations: As 𝐼 = 𝐼0𝑒𝜇𝑥 For Case 1: 𝐼1 = 𝐼0𝑒𝜇𝑥1 ⟹ 0.4 𝐼0 = 𝐼0𝑒𝜇𝑥1 ⟹ 0.4 = 𝑒𝜇𝑥1 ⟹ ln(0.4) = 𝜇𝑥1 − − − − (1) For Case 2: 𝐼2 = 𝐼0𝑒𝜇𝑥2 ⟹ 0.5 𝐼0 = 𝐼0𝑒𝜇𝑥2 ⟹ 0.5 = 𝑒𝜇𝑥2 ⟹ ln(0.5) = 𝜇𝑥2 − − − − (2) Dividing (1) and (2): 𝜇𝑥2 = ln(0.5) ⟹ 𝑥2 = −0.693 ⟹ 𝑥 = 0.693 𝑥 = 0.693 × 5 × 10−3 = 3.78 × 10−3𝑚 = 3.78 𝑚𝑚 𝜇𝑥1ln(0.4)𝑥1−- 10.916 21 .9 Radiation from a point source obeys the inverse square law. If the count rate at a distance of 1.0 m from Geiger counter is 360 counts per minute, what will be its count rate at 3.0 m from the source? Given Data: 𝐼𝘢 1Case 1 Count Rate 𝐼 = 360 𝑐𝑜𝑢𝑛𝑡𝑠 , Distance 𝑟 = 1 𝑚 𝑟21𝑚i𝑛𝑢𝑡𝑒𝑠1 Case 1, Distance 𝑟3 = 3 𝑚 To Determine: Count Rate for case 2: 𝐼2 =? Calculations: Given 𝐼𝘢 1 ⟹ 𝐼 = 𝑘 𝑟2𝑟2 For Case 1: 𝐼 = 𝑘 − − − − (1),For Case 2: 𝐼 = 𝑘− − − − (2) 1𝑟22𝑟2 12 ( 𝑘 )( 1 ) 22222 Dividing (1) and (2): 𝐼2 = r2 ⟹ 𝐼2 = r2 ⟹ 𝐼2 = 𝑟1 ⟹ 𝐼 = 𝑟1 𝐼 = (1) × 360 = 1 × 360 = 40 𝑐𝑜𝑢𝑛𝑡𝑠 𝐼1𝑘𝐼11𝐼1𝑟22𝑟2 1(3)29𝑚i𝑛𝑢𝑡𝑒𝑠 ( 2)( 2)22 r1r1 21.10 A 75 kg person receives a whole body radiation dose of 24 m-rad, delivered by α-particles for which RBE factor is 12. Calculate (a) the absorbed energy in joules, and (b) the equivalent dose in rem. Given Data: Mass of Person 𝑚 = 75 𝑘𝑔, RBE factor = 12 Radiation Dose 𝐷 = 24 m rad = 24 × 10−3 rad = 24 × 10−3 × 0.01 𝐺𝑦 = 24 × 10−5 𝐺𝑦 To Determine: (a) Absorbed Energy in joules 𝐸 =?, (b) Equivalent Dose in rem 𝐷𝑒 =? Calculations: (a) As 𝐸 = 𝐷 × 𝑚 = 24 × 10−5 × 75 = 18 × 10−3𝐽 (b) 𝐷𝑒 = 𝐷 × 𝑅𝐵𝐸 = 24 × 10−5 × 12 = 2.88 × 10−3𝑆𝑣 = 0.288 𝑟𝑒𝑚∴ 1 𝑆𝑣 = 100 𝑟𝑒𝑚 B.Sc. 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