BOARD’S QUESTION PAPER & SOLUTION : SEPTEMBER 2021
SUBJECT: PHYSICS | Max. Marks : 70 | Time : 3 Hours
General instructions:
- Section A: Q. No. 1 contains ten multiple choice type questions carrying one mark each. Q. No. 2 contains eight very short answer type questions carrying one mark each.
- Section B: Q. No. 3 to 14 contain twelve short answer type questions carrying two marks each. (Attempt any eight).
- Section C: Q. No. 15 to 26 contain twelve short answer type questions carrying three marks each. (Attempt any eight).
- Section D: Q. No. 27 to 31 contain five long answer type questions carrying four marks each. (Attempt any three).
- Use of logarithmic tables is allowed. Use of a calculator is not allowed.
- Figures to the right indicate full marks.
- For each multiple choice type question, it is mandatory to write the correct answer along with its alphabet, e.g., (a)…/(b)…/(c)…/(d)…. No mark(s) shall be given if only the correct answer or the alphabet of the correct answer is written. Only the first attempt will be considered for evaluation.
Physical constants:
(1) \( h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s} \)
(2) \( \pi = 3.142 \)
(3) \( g = 9.8 \text{ m/s}^2 \)
(4) \( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \)
(5) \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \)
(6) \( 1/4\pi\varepsilon_0 = 9 \times 10^9 \text{ SI unit} \)
(7) \( R = 8.319 \text{ J/mol}\cdot\text{K} \)
SECTION – A
Q. 1. Select and write the correct answer for the following multiple choice type questions: [10]
(i) The SI unit of viscosity is
(a) N·s/m² (b) N·m²/s (c) N²·s²/m (d) m²/N·s.
Answer: (a) N·s/m²
(ii) The colour of a bright shining star is an indication of its
(a) distance from the Earth (b) size (c) temperature (d) mass.
Answer: (c) temperature
(iii) In which thermodynamic process does the volume of the system remain constant?
(a) Isobaric (b) Isothermal (c) Adiabatic (d) Isochoric.
Answer: (d) Isochoric
(iv) If in a resonance tube an oil of density higher than that of water is used, then resonance frequency would
(a) increase (b) decrease (c) slightly increase (d) remain the same.
Answer: (d) remain the same.
(v) In an interference experiment a transparent glass plate with refractive index \(n\) and thickness \(t\) is introduced between one of the slits and the screen, the optical path shifts by
(a) \((n+1)t\) (b) \((n-1)t\) (c) \((n-1)^2t\) (d) \((n-1)t^2\).
Answer: (b) \((n-1)t\)
(vi) For a series LCR circuit at resonance, the impedance of the circuit is equal to
(a) inductive reactance (b) capacitive reactance (c) resistance (d) both inductive and capacitive reactances.
Answer: (c) resistance
(vii) In the Bohr model of an [hydrogen] atom, which of the following is an integral multiple of \(h/2\pi\)?
(a) Kinetic energy (b) Radius of the atom (c) Potential energy (d) Angular momentum
Answer: (d) Angular momentum
(viii) A mass \(m\) attached to a spring oscillates every 2 seconds. If the mass is increased by 2 kg, then the time period increases by 1 second. The initial mass was
(a) 1.6 kg (b) 2.4 kg (c) 3.2 kg (d) 1.4 kg.
Answer: (a) 1.6 kg
(ix) In a biprism experiment, the distance of a point on the screen from the slits is \(1.8 \times 10^{-5}\) m and \(1.23 \times 10^{-5}\) m. If the wavelength of light used is 6000 Å, then the fringe formed at that point is
(a) 10th bright (b) 10th dark (c) 9th dark (d) 9th bright.
Answer: (b) 10th dark
(x) In a common emitter amplifier, the current gain is 80 and the emitter current is 9 mA. The base current is
(a) \(\frac{1}{81}\) mA (b) 8 mA (c) \(\frac{1}{8}\) mA (d) \(\frac{1}{9}\) mA.
Answer: (d) \(\frac{1}{9}\) mA
Q. 2. Answer the following questions: [8]
(i) State the formula for the moment of inertia of a solid sphere about an axis passing through its centre.
The moment of inertia of a solid sphere, of mass \(M\) and radius \(R\), about its diameter is \(I = \frac{2}{5} MR^2\).
(ii) Define angle of contact.
The angle of contact for a liquid-solid pair (a liquid in contact with a solid) is defined as the angle between the surface of the solid and the tangent drawn to the free surface of the liquid at the extreme edge of the liquid, as measured through the liquid.
(iii) What is an isothermal process?
A thermodynamic process that takes place at constant volume is called an isochoric process (or isometric process).
(Note: The solution text in source document describes "Isochoric" but Question asks "Isothermal". However, based on content fidelity rules, the provided solution text says "isochoric process" in response to "constant volume" from Q1(iii). Wait - looking at Q2(iii) in source: "What is an isothermal process?". The solution provided in source image 5 actually says "A thermodynamic process that takes place at constant volume is called an isochoric process". This appears to be a mismatch in the source document itself or a reading error. Let's re-read carefully. Ah, the OCR for Q1(iii) asks about constant volume. The OCR for Q2(iii) asks about Isothermal. The SOLUTION for Q2(iii) in the PDF actually says: "A thermodynamic process that takes place at constant volume is called an isochoric process...". This is strange. Let's look closer. Ah, I see. In the solution image 5, Q2(iii) answer text describes Isochoric. This is likely an error in the source PDF solution key or a misnumbering. However, I must follow the Golden Rule: "Content integrity is the absolute highest priority... strictly verbatim". I will output exactly what the solution text says in the image, even if it seems wrong for the specific question.)
Correction based on careful OCR check: The solution text for Q2(iii) in the image actually corresponds to the definition of Isochoric (constant volume). This is a discrepancy in the original paper/solution key. I will transcribe the solution exactly as printed in the source PDF solution section.
(Note: The solution text in source document describes "Isochoric" but Question asks "Isothermal". However, based on content fidelity rules, the provided solution text says "isochoric process" in response to "constant volume" from Q1(iii). Wait - looking at Q2(iii) in source: "What is an isothermal process?". The solution provided in source image 5 actually says "A thermodynamic process that takes place at constant volume is called an isochoric process". This appears to be a mismatch in the source document itself or a reading error. Let's re-read carefully. Ah, the OCR for Q1(iii) asks about constant volume. The OCR for Q2(iii) asks about Isothermal. The SOLUTION for Q2(iii) in the PDF actually says: "A thermodynamic process that takes place at constant volume is called an isochoric process...". This is strange. Let's look closer. Ah, I see. In the solution image 5, Q2(iii) answer text describes Isochoric. This is likely an error in the source PDF solution key or a misnumbering. However, I must follow the Golden Rule: "Content integrity is the absolute highest priority... strictly verbatim". I will output exactly what the solution text says in the image, even if it seems wrong for the specific question.)
Correction based on careful OCR check: The solution text for Q2(iii) in the image actually corresponds to the definition of Isochoric (constant volume). This is a discrepancy in the original paper/solution key. I will transcribe the solution exactly as printed in the source PDF solution section.
(iv) At which position is the restoring force acting on a particle executing linear SHM maximum?
The magnitude of the restoring force on a particle performing SHM is \(F = kx\), where \(k\) is the force constant and \(x\) is the distance of the particle from the mean position. Hence, the magnitude of the restoring force is maximum when \(x = x_{max} = \pm A\), where \(A\) is the amplitude of the motion.
(v) When is an ac circuit noninductive?
An ac circuit is noninductive when either the inductance is zero (i.e., there are no inductor in the circuit) or the inductive reactance \(X_L\) equals the capacitive reactance \(X_C\).
(vi) What is the phase difference between the input signal voltage and the output signal voltage in a CE amplifier?
The output voltage of a CE amplifier is 180° out of phase with the input signal voltage.
(vii) Calculate the minimum energy required to take an electron from the ground state to the first excited state in a hydrogen atom.
The energy of the electron in the \(n\)th stationary state in a hydrogen atom (\(Z = 1\)) is
\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]
\(\therefore\) The minimum energy required for the transition of the electron from the ground state (\(n = 1\)) to the first excited state (\(n = 2\)) is
\[ E_2 - E_1 = -13.6 \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = -13.6 \left( \frac{1}{4} - 1 \right) = 13.6 \times \frac{3}{4} = 10.2 \text{ eV} \]
(viii) If a charge of \(50\mu C\) is moving with a speed of \(50 \text{ m/s}\) parallel to the direction of a magnetic field, then what will be the mechanical force acting on the charged particle?
The magnetic force on a charge \(q\) moving with a velocity \(\vec{v}\) in a magnetic field of induction \(\vec{B}\) is \(\vec{F} = q(\vec{v} \times \vec{B})\). Since, in this case, \(\vec{v}\) is parallel to \(\vec{B}\), \(\vec{F} = 0\).
HSC Physics Board Papers with Solution
- Physics - March 2025 - English Medium View Answer Key
- Physics - March 2025 - Marathi Medium View Answer Key
- Physics - March 2025 - Hindi Medium View Answer Key
- Physics - March 2024 - English Medium View Answer Key Answer Key
- Physics - March 2024 - Marathi Medium View Answer Key
- Physics - March 2024 - Hindi Medium View Answer Key
- Physics - March 2023 - English Medium View Answer Key
- Physics - July 2023 - English Medium View Answer Key
- Physics - March 2022 - English Medium View Answer Key
- Physics - July 2022 - English Medium View Answer Key
- Physics - July 2021 - English Medium View Answer Key
- Physics - February 2020 - English Medium View Answer Key
- Physics - March 2013 View
- Physics - October 2013 View
- Physics - March 2014 View
- Physics - October 2014 View
- Physics - March 2015 View
- Physics - July 2015 View
- Physics - March 2016 View
- Physics - July 2016 View
- Physics - March 2017 View
- Physics - July 2017 View
SECTION – B
Attempt any eight of the following questions: [16]
Q. 3. Draw a neat labelled diagram of Fery’s blackbody.
(Refer to standard textbook for detailed diagram showing double-walled hollow sphere, conical projection, aperture, etc.)
Q. 4. Write a note on free expansion in a thermodynamic process.
A free expansion is an irreversible adiabatic process, i.e., a process in which no heat is added to the system or removed from the system. It is characterized by \(Q = W = 0\), implying that the change in the internal energy, \(\Delta U = 0\).
Consider a gas confined by a valve to one half of a double chamber with adiabatic walls while the other half is evacuated. When the gas is in thermal equilibrium, the gas is allowed to expand to fill the entire chamber by opening the valve. No interaction takes place and hence there are no local surroundings. While rushing into a vacuum, the gas does not meet any pressure and hence no work is done by the gas. Hence, \(Q = W = 0\).
Free expansion is an uncontrolled change and the system is not in thermodynamic equilibrium. Free expansion cannot be illustrated with a \(P-V\) diagram as only the initial state and final state are known.
Consider a gas confined by a valve to one half of a double chamber with adiabatic walls while the other half is evacuated. When the gas is in thermal equilibrium, the gas is allowed to expand to fill the entire chamber by opening the valve. No interaction takes place and hence there are no local surroundings. While rushing into a vacuum, the gas does not meet any pressure and hence no work is done by the gas. Hence, \(Q = W = 0\).
Free expansion is an uncontrolled change and the system is not in thermodynamic equilibrium. Free expansion cannot be illustrated with a \(P-V\) diagram as only the initial state and final state are known.
Q. 5. What is magnetization? Write its unit and dimensions.
In an atom, the orbital and spin magnetic moments of its electrons may or may not add up to zero, depending on the electronic configuration. In some materials, their atoms have a net magnetic moment. When such a material is placed in an external magnetic field of induction \(\vec{B_0}\), the magnetic moments tend to align with the applied field. Due to this, the material as a whole acquires a net magnetic moment \(\vec{M}_{net}\) along \(\vec{B_0}\) and the material is said to be magnetized.
The magnetization of a material is defined as the net magnetic moment per unit volume of a material.
SI unit: The ampere per metre (A/m).
Dimensions: \[ [\text{Magnetization}] = \frac{[\text{magnetic moment}]}{[\text{volume}]} = \frac{[L^2I]}{[L^3]} = [L^{-1}I] \]
The magnetization of a material is defined as the net magnetic moment per unit volume of a material.
SI unit: The ampere per metre (A/m).
Dimensions: \[ [\text{Magnetization}] = \frac{[\text{magnetic moment}]}{[\text{volume}]} = \frac{[L^2I]}{[L^3]} = [L^{-1}I] \]
Q. 6. State any two conditions for a steady interference pattern.
Conditions for a steady interference pattern:
(1) The two light sources must be coherent.
(2) The two light sources should be monochromatic.
(1) The two light sources must be coherent.
(2) The two light sources should be monochromatic.
Q. 7. With the help of a suitable diagram, state the expression for Biot-Savart’s law in vector form.
Consider a very short segment of length \(dl\) of a wire carrying a current \(I\). The product \(I\vec{dl}\) is called a current element; the direction of \(\vec{dl}\) is along the wire in the direction of the current.
Biot-Savart law (Laplace law): The magnitude of the incremental magnetic induction \(d\vec{B}\) produced by a current element \(I\vec{dl}\) at a distance \(r\) from it is directly proportional to the magnitude \(I dl\) of the current element, the sine of the angle between the current element \(I\vec{dl}\) and the unit vector \(\hat{r}\) directed from the current element toward the point in question, and inversely proportional to the square of the distance of the point from the current element; the magnetic induction is directed perpendicular to both \(I\vec{dl}\) and \(\hat{r}\) as per the cross product rule. \[ dB \propto \frac{I dl \sin\theta}{r^2} \] \[ \therefore dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2} \quad ... (1) \] In vector form, \[ d\vec{B} = \left( \frac{\mu_0}{4\pi} \right) \frac{I \vec{dl} \times \hat{r}}{r^2} \quad ... (2) \] where \(\hat{r} = \frac{\vec{r}}{r}\) and the constant \(\mu_0\) is the permeability of free space. Equations (1) and (2) are called the Biot-Savart law.
The incremental magnetic induction \(d\vec{B}\) is given by the right-handed screw rule of vector product \(I\vec{dl} \times \hat{r}\). In the figure, the current element \(I\vec{dl}\) and \(\hat{r}\) are in the plane of the page, so that \(d\vec{B}\) points out of the page at point P shown by \(\odot\); at the point Q, \(d\vec{B}'\) points into the page shown by \(\otimes\).
From Eq. (2), \[ \vec{B} = \int d\vec{B} = \frac{\mu_0}{4\pi} \int \frac{I \vec{dl} \times \hat{r}}{r^2} \]
Biot-Savart law (Laplace law): The magnitude of the incremental magnetic induction \(d\vec{B}\) produced by a current element \(I\vec{dl}\) at a distance \(r\) from it is directly proportional to the magnitude \(I dl\) of the current element, the sine of the angle between the current element \(I\vec{dl}\) and the unit vector \(\hat{r}\) directed from the current element toward the point in question, and inversely proportional to the square of the distance of the point from the current element; the magnetic induction is directed perpendicular to both \(I\vec{dl}\) and \(\hat{r}\) as per the cross product rule. \[ dB \propto \frac{I dl \sin\theta}{r^2} \] \[ \therefore dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2} \quad ... (1) \] In vector form, \[ d\vec{B} = \left( \frac{\mu_0}{4\pi} \right) \frac{I \vec{dl} \times \hat{r}}{r^2} \quad ... (2) \] where \(\hat{r} = \frac{\vec{r}}{r}\) and the constant \(\mu_0\) is the permeability of free space. Equations (1) and (2) are called the Biot-Savart law.
The incremental magnetic induction \(d\vec{B}\) is given by the right-handed screw rule of vector product \(I\vec{dl} \times \hat{r}\). In the figure, the current element \(I\vec{dl}\) and \(\hat{r}\) are in the plane of the page, so that \(d\vec{B}\) points out of the page at point P shown by \(\odot\); at the point Q, \(d\vec{B}'\) points into the page shown by \(\otimes\).
From Eq. (2), \[ \vec{B} = \int d\vec{B} = \frac{\mu_0}{4\pi} \int \frac{I \vec{dl} \times \hat{r}}{r^2} \]
Q. 8. State Faraday’s laws of electromagnetic induction.
The phenomenon of production of emf in a conductor or circuit by a changing magnetic flux through the circuit is called electromagnetic induction.
Faraday’s laws of electromagnetic induction:
(1) First law: Whenever there is a change in the magnetic flux associated with a circuit, an emf is induced in the circuit.
(2) Second law: The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.
Faraday’s laws of electromagnetic induction:
(1) First law: Whenever there is a change in the magnetic flux associated with a circuit, an emf is induced in the circuit.
(2) Second law: The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.
Q. 9. Explain why the total impedance of a circuit decreases when a capacitor is added in series with an inductor and a resistor.
For an LR circuit, the impedance,
\[ Z_{LR} = \sqrt{R^2 + X_L^2} \]
where \(X_L\) is the reactance of the inductor.
When a capacitor of capacitance \(C\) is added in series with \(L\) and \(R\), the impedance, \[ Z_{LCR} = \sqrt{R^2 + (X_L - X_C)^2} \] where \(X_C\) is the reactance of the capacitor. In the case of an inductor the current lags behind the voltage by a phase angle of \(\pi/2\) rad while in the case of a capacitor the current leads the voltage by a phase angle of \(\pi/2\) rad. The decrease in net reactance decreases the total impedance (\(Z_{LCR} < Z_{LR}\)).
When a capacitor of capacitance \(C\) is added in series with \(L\) and \(R\), the impedance, \[ Z_{LCR} = \sqrt{R^2 + (X_L - X_C)^2} \] where \(X_C\) is the reactance of the capacitor. In the case of an inductor the current lags behind the voltage by a phase angle of \(\pi/2\) rad while in the case of a capacitor the current leads the voltage by a phase angle of \(\pi/2\) rad. The decrease in net reactance decreases the total impedance (\(Z_{LCR} < Z_{LR}\)).
Q. 10. A motorcyclist (treated as a point mass) is to undertake horizontal circles inside the cylindrical wall of a well of inner radius 4 m. The coefficient of static friction between the tyres and the wall is 0.2. Calculate the minimum speed and period necessary to perform this stunt.
Data: \(r = 4 \text{ m}\), \(\mu_s = 0.2\), \(g = 9.8 \text{ m/s}^2\)
\[ v_{min} = \sqrt{\frac{rg}{\mu_s}} = \sqrt{\frac{4 \times 9.8}{0.2}} = \sqrt{4 \times 49} = 14 \text{ m/s} \]
\[ T = \frac{2\pi r}{v_{min}} = \frac{2 \times 3.142 \times 4}{14} = \frac{12.568}{7} = 1.795 \text{ s} \]
Q. 11. Compare the amounts of work done in blowing two soap bubbles of radii in the ratio 4 : 5.
Data: \(\frac{r_1}{r_2} = \frac{4}{5}\)
Work done, \(W = 2TdA\)
\(\therefore W_1 = 2T(4\pi r_1^2)\), \(W_2 = 2T(4\pi r_2^2)\)
\[ \therefore \frac{W_1}{W_2} = \frac{2T(4\pi r_1^2)}{2T(4\pi r_2^2)} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]
Work done, \(W = 2TdA\)
\(\therefore W_1 = 2T(4\pi r_1^2)\), \(W_2 = 2T(4\pi r_2^2)\)
\[ \therefore \frac{W_1}{W_2} = \frac{2T(4\pi r_1^2)}{2T(4\pi r_2^2)} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]
Q. 12. Find the distance between two successive antinodes in a stationary wave on a string vibrating with a frequency of 32 Hz. [Speed of wave = 48 m/s]
Data: \(n = 32 \text{ Hz}\), \(v = 48 \text{ m/s}\)
\[ \lambda = \frac{v}{n} = \frac{48 \text{ m/s}}{32 \text{ Hz}} = 1.5 \text{ m} \] \(\therefore\) Distance between successive antinodes \(= \frac{\lambda}{2} = 0.75 \text{ m}\)
\[ \lambda = \frac{v}{n} = \frac{48 \text{ m/s}}{32 \text{ Hz}} = 1.5 \text{ m} \] \(\therefore\) Distance between successive antinodes \(= \frac{\lambda}{2} = 0.75 \text{ m}\)
Q. 13. The emf of a cell is balanced by a length of 320 cm of the potentiometer wire. When the cell is shunted by a resistance of 50 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.
Data: \(R = 50 \Omega\), \(l_1 = 320 \text{ cm}\), \(l_2 = l_1 - 20 = 320 - 20 = 300 \text{ cm}\)
\(\therefore l_1 - l_2 = 20 \text{ cm}\)
The internal resistance of the cell is \[ r = R \left( \frac{l_1 - l_2}{l_2} \right) = 50 \left( \frac{20}{300} \right) = \frac{10}{3} = 3.333 \Omega \]
\(\therefore l_1 - l_2 = 20 \text{ cm}\)
The internal resistance of the cell is \[ r = R \left( \frac{l_1 - l_2}{l_2} \right) = 50 \left( \frac{20}{300} \right) = \frac{10}{3} = 3.333 \Omega \]
Q. 14. The photoelectric work function for a metal is 5 eV. Calculate the threshold frequency for the metal.
Data: \(\phi_0 = 5 \text{ eV}\), \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\)
\(\phi_0 = h\nu_0\) \[ \therefore \nu_0 = \frac{\phi_0}{h} = \frac{(5 \text{ eV})(1.6 \times 10^{-19} \text{ J/eV})}{6.63 \times 10^{-34} \text{ J}\cdot\text{s}} = \frac{8 \times 10^{15}}{6.63} = 1.207 \times 10^{15} \text{ Hz} \]
\(\phi_0 = h\nu_0\) \[ \therefore \nu_0 = \frac{\phi_0}{h} = \frac{(5 \text{ eV})(1.6 \times 10^{-19} \text{ J/eV})}{6.63 \times 10^{-34} \text{ J}\cdot\text{s}} = \frac{8 \times 10^{15}}{6.63} = 1.207 \times 10^{15} \text{ Hz} \]
log 8 0.9031
log 6.63 -0.8215
______
0.0816
Antilog 0.0816 = 1.207
log 6.63 -0.8215
______
0.0816
Antilog 0.0816 = 1.207
SECTION – C
Attempt any eight of the following questions: [24]
Q. 15. With a neat labelled schematic diagram, explain the experimental setup for photoelectric effect.
Apparatus: A photoelectric cell G consists of the emitting electrode E (emitter) of the material being studied and the collecting electrode C (collector). The electrodes are sealed in an evacuated glass envelope provided with quartz window W that allows the passage of UV radiation and visible light. Monochromatic light of variable frequency from a suitable source S (such as a carbon arc) passes through a pair of polarizers P (permitting a change in the intensity of radiation) and falls on the emitter.
The electric circuit, as shown in the figure, allows the collector potential to be varied from positive through zero to negative with respect to the emitter, and permits the measurement of potential difference and current between the electrodes. When the collector is made negative, the voltmeter is connected in reverse.
The electric circuit, as shown in the figure, allows the collector potential to be varied from positive through zero to negative with respect to the emitter, and permits the measurement of potential difference and current between the electrodes. When the collector is made negative, the voltmeter is connected in reverse.
Apparatus to study the characteristics of photoelectric effect
S: Source, F: Filter, P: Polarizers, W: Window, E: Emitter, C: Collector, V: Voltmeter, A: Microammeter, Rh: Rheostat, B: Battery.
S: Source, F: Filter, P: Polarizers, W: Window, E: Emitter, C: Collector, V: Voltmeter, A: Microammeter, Rh: Rheostat, B: Battery.
Q. 16. What is a light emitting diode? Explain the working of an LED.
A light-emitting diode (LED) is a forward-biased pn-junction diode formed from compound semiconductor materials such as gallium arsenide (GaAs) in which light emission can take place from direct radiative recombination of excess electron-hole pairs. A photon is emitted when an electron in the conduction band recombines with a hole in the valence band.
In infrared emitting LEDs, the encapsulating plastic lens may be impregnated or coated with phosphorus. Then, phosphorescence of the phosphorus gives off visible light.
Working: An LED is forward-biased with about 1.2 V to 3.6 V at 12 mA to 20 mA. Majority carriers electrons from n-type layer and holes from p-type layer are injected into the active layer. Electrons cross the junction into the p-layer. In the active p-layer, some of these excess minority carriers, electrons, recombine radiatively with majority carriers, holes, thereby emitting photons. The resulting photon has an energy approximately equal to the bandgap of the active layer material. Modifying the bandgap of the active layer creates photons of different energies.
In the energy band diagram this recombination is equivalent to a transition of the electron from a higher energy state in the conduction band to a lower energy state in the valence band. The energy difference is emitted as a photon of energy \(h\nu\).
In infrared emitting LEDs, the encapsulating plastic lens may be impregnated or coated with phosphorus. Then, phosphorescence of the phosphorus gives off visible light.
Working: An LED is forward-biased with about 1.2 V to 3.6 V at 12 mA to 20 mA. Majority carriers electrons from n-type layer and holes from p-type layer are injected into the active layer. Electrons cross the junction into the p-layer. In the active p-layer, some of these excess minority carriers, electrons, recombine radiatively with majority carriers, holes, thereby emitting photons. The resulting photon has an energy approximately equal to the bandgap of the active layer material. Modifying the bandgap of the active layer creates photons of different energies.
In the energy band diagram this recombination is equivalent to a transition of the electron from a higher energy state in the conduction band to a lower energy state in the valence band. The energy difference is emitted as a photon of energy \(h\nu\).
Q. 17. Obtain an expression for the equivalent capacity for the combination of three capacitors connected in series.
In the series arrangement of capacitors, the capacitors are connected end to end and a cell is connected across the combination of the capacitors.
Let \(C_1, C_2, C_3\) be the capacitances of the three capacitors connected in series and \(Q\), the charge on each capacitor. Let \(V_1, V_2, V_3\) be the potential differences across the capacitors.
Now, charge = capacitance \(\times\) potential difference \[ \therefore Q = C_1V_1 = C_2V_2 = C_3V_3 \] \[ \therefore V_1 = \frac{Q}{C_1}, V_2 = \frac{Q}{C_2} \text{ and } V_3 = \frac{Q}{C_3} \] If \(V\) is the potential difference across the combination and \(C_s\) is the equivalent or effective capacitance of the combination, we have, \[ C_s = \frac{Q}{V} \quad \therefore V = \frac{Q}{C_s} \] But, \(V = V_1 + V_2 + V_3\) \[ \therefore \frac{Q}{C_s} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \] \[ \therefore \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
Let \(C_1, C_2, C_3\) be the capacitances of the three capacitors connected in series and \(Q\), the charge on each capacitor. Let \(V_1, V_2, V_3\) be the potential differences across the capacitors.
Now, charge = capacitance \(\times\) potential difference \[ \therefore Q = C_1V_1 = C_2V_2 = C_3V_3 \] \[ \therefore V_1 = \frac{Q}{C_1}, V_2 = \frac{Q}{C_2} \text{ and } V_3 = \frac{Q}{C_3} \] If \(V\) is the potential difference across the combination and \(C_s\) is the equivalent or effective capacitance of the combination, we have, \[ C_s = \frac{Q}{V} \quad \therefore V = \frac{Q}{C_s} \] But, \(V = V_1 + V_2 + V_3\) \[ \therefore \frac{Q}{C_s} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \] \[ \therefore \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
Q. 18. Explain surface tension on the basis of molecular theory.
The phenomenon of surface tension arises due to the cohesive forces between the molecules of a liquid. The net cohesive force on the liquid molecules within the surface film differs from that on molecules deep in the interior of the liquid.
Consider three molecules of a liquid: A molecule A well inside the liquid, and molecules B and C lying within the surface film. The figure also shows their spheres of influence of radius \(R\). (1) The sphere of influence of molecule A is entirely inside the liquid and the molecule is surrounded by its nearest neighbours on all sides. Hence, molecule A is equally attracted from all sides, so that the resultant cohesive force acting on it is zero. Hence, it is free to move anywhere within the liquid.
(2) For molecule B, a part of its sphere of influence is outside the liquid surface. This part contains air molecules whose number is negligible compared to the number of molecules in an equal volume of the liquid. Therefore, molecule B experiences a net cohesive force downward.
(3) For molecule C, the upper half of its sphere of influence is outside the liquid surface. Therefore, the resultant cohesive force on molecule C in the downward direction is maximum.
(4) Thus, all molecules lying within a surface film of thickness equal to \(R\) experience a net cohesive force directed into the liquid.
(5) The surface area is proportional to the number of molecules on the surface. To increase the surface area, molecules must be brought to the surface from within the liquid. For this, work must be done against the cohesive forces. This work is stored in the liquid surface in the form of potential energy. With a tendency to have minimum potential energy, the liquid tries to reduce the number of molecules on the surface so as to have minimum surface area. This is why the surface of a liquid behaves like a stressed elastic membrane.
Consider three molecules of a liquid: A molecule A well inside the liquid, and molecules B and C lying within the surface film. The figure also shows their spheres of influence of radius \(R\). (1) The sphere of influence of molecule A is entirely inside the liquid and the molecule is surrounded by its nearest neighbours on all sides. Hence, molecule A is equally attracted from all sides, so that the resultant cohesive force acting on it is zero. Hence, it is free to move anywhere within the liquid.
(2) For molecule B, a part of its sphere of influence is outside the liquid surface. This part contains air molecules whose number is negligible compared to the number of molecules in an equal volume of the liquid. Therefore, molecule B experiences a net cohesive force downward.
(3) For molecule C, the upper half of its sphere of influence is outside the liquid surface. Therefore, the resultant cohesive force on molecule C in the downward direction is maximum.
(4) Thus, all molecules lying within a surface film of thickness equal to \(R\) experience a net cohesive force directed into the liquid.
(5) The surface area is proportional to the number of molecules on the surface. To increase the surface area, molecules must be brought to the surface from within the liquid. For this, work must be done against the cohesive forces. This work is stored in the liquid surface in the form of potential energy. With a tendency to have minimum potential energy, the liquid tries to reduce the number of molecules on the surface so as to have minimum surface area. This is why the surface of a liquid behaves like a stressed elastic membrane.
Q. 19. Derive an expression for the period of a simple pendulum.
Consider a simple pendulum of length \(L\), suspended from a rigid support O. When displaced from its initial position of rest through a small angle \(\theta\) in a vertical plane and released, it performs oscillations between two extremes, B and C. At B, the forces on the bob are its weight \(m\vec{g}\) and the tension \(\vec{F}_1\) in the string. Resolve \(m\vec{g}\) into two components : \(mg \cos\theta\) in the direction opposite to that of the tension and \(mg \sin\theta\) perpendicular to the string.
\(mg \cos\theta\) is balanced by the tension in the string. \(mg \sin\theta\) restores the bob to the equilibrium position.
Restoring force, \(F = -mg \sin\theta\)
If \(\theta\) is small and expressed in radian, \[ \sin\theta \approx \theta = \frac{\text{arc}}{\text{radius}} = \frac{AB}{OB} = \frac{x}{L} \] \[ \therefore F = -mg\theta = -mg\frac{x}{L} \quad ... (1) \] Since \(m, g\) and \(L\) are constant, \[ F \propto (-x) \quad ... (2) \] Thus, the net force on the bob is in the direction opposite to that of displacement \(x\) of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, \(a = \frac{F}{m} = -\frac{g}{L} x \quad ... (3)\)
Therefore, acceleration per unit displacement \(= \left| \frac{a}{x} \right| = \frac{g}{L} \quad ... (4)\)
Period of SHM, \[ T = \frac{2\pi}{\sqrt{\text{acceleration per unit displacement}}} = \frac{2\pi}{\sqrt{g/L}} \] \[ \therefore T = 2\pi \sqrt{\frac{L}{g}} \quad ... (5) \] This gives the expression for the period of a simple pendulum.
\(mg \cos\theta\) is balanced by the tension in the string. \(mg \sin\theta\) restores the bob to the equilibrium position.
Restoring force, \(F = -mg \sin\theta\)
If \(\theta\) is small and expressed in radian, \[ \sin\theta \approx \theta = \frac{\text{arc}}{\text{radius}} = \frac{AB}{OB} = \frac{x}{L} \] \[ \therefore F = -mg\theta = -mg\frac{x}{L} \quad ... (1) \] Since \(m, g\) and \(L\) are constant, \[ F \propto (-x) \quad ... (2) \] Thus, the net force on the bob is in the direction opposite to that of displacement \(x\) of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, \(a = \frac{F}{m} = -\frac{g}{L} x \quad ... (3)\)
Therefore, acceleration per unit displacement \(= \left| \frac{a}{x} \right| = \frac{g}{L} \quad ... (4)\)
Period of SHM, \[ T = \frac{2\pi}{\sqrt{\text{acceleration per unit displacement}}} = \frac{2\pi}{\sqrt{g/L}} \] \[ \therefore T = 2\pi \sqrt{\frac{L}{g}} \quad ... (5) \] This gives the expression for the period of a simple pendulum.
Q. 20. State Huygens’ principle. Explain geometrical construction of a plane wavefront.
Huygens’ principle: Every point on a wavefront acts as a secondary source of light and sends out secondary wavelets in all directions. The secondary wavelets travel with the speed of light in the medium. These wavelets are effective only in the forward direction and not in the backward direction. At any instant, the forward-going envelope or the surface of tangency to these wavelets gives the position of the new wavefront at that instant.
Huygens’ construction of a plane wavefront: A plane wavefront may be treated as a part of a spherical or cylindrical wave at a very great distance from a point source or an extended source, such that the wavefront has a negligible curvature. Let A, B, C, D, ..., be points on a plane wavefront in a homogeneous isotropic medium in which the speed of light, taken to be monochromatic, is \(v\).
In a time, \(t=T\), secondary wavelets with points A, B, C, D, ... as secondary sources travel a distance \(vT\). To find the position of the wavefront after a time \(t=T\), we draw spheres of radii \(vT\) with A, B, C, ... as centres. The envelope or the surface of tangency to these spheres is a plane A'B'C'. This plane, the new wavefront, is at a perpendicular distance \(vT\) from the original wavefront in the direction of propagation of the wave. Thus, in an isotropic medium, plane wavefronts are propagated as planes.
Huygens’ construction of a plane wavefront: A plane wavefront may be treated as a part of a spherical or cylindrical wave at a very great distance from a point source or an extended source, such that the wavefront has a negligible curvature. Let A, B, C, D, ..., be points on a plane wavefront in a homogeneous isotropic medium in which the speed of light, taken to be monochromatic, is \(v\).
In a time, \(t=T\), secondary wavelets with points A, B, C, D, ... as secondary sources travel a distance \(vT\). To find the position of the wavefront after a time \(t=T\), we draw spheres of radii \(vT\) with A, B, C, ... as centres. The envelope or the surface of tangency to these spheres is a plane A'B'C'. This plane, the new wavefront, is at a perpendicular distance \(vT\) from the original wavefront in the direction of propagation of the wave. Thus, in an isotropic medium, plane wavefronts are propagated as planes.
Q. 21. Obtain the expression for Bohr magneton.
In the Bohr model of a hydrogen atom, the electron of charge \(-e\) performs a uniform circular motion around the positively charged nucleus. Let \(r\), \(v\) and \(T\) be the orbital radius, speed and period of motion of the electron. Then,
\[ T = \frac{2\pi r}{v} \quad ... (1) \]
The orbital magnetic moment associated with this orbital current loop has a magnitude,
\[ I = \frac{e}{T} = \frac{ev}{2\pi r} \quad ... (2) \]
Therefore, the magnetic dipole moment associated with this electronic current loop has a magnitude
\[ M_o = \text{current} \times \text{area of the loop} \]
\[ = I(\pi r^2) = \frac{ev}{2\pi r} \times \pi r^2 = \frac{1}{2}evr \quad ... (3) \]
Multiplying and dividing the right hand side of the above expression by the electron mass \(m_e\),
\[ M_o = \frac{e}{2m_e}(m_evr) = \frac{e}{2m_e}L_o \quad ... (4) \]
According to Bohr’s second postulate of stationary orbits in his theory of hydrogen atom, the angular momentum of the electron in the \(n\)th stationary orbit is equal to \(n\frac{h}{2\pi}\), where \(h\) is the Planck constant and \(n\) is a positive integer. Thus, for an orbital electron,
\[ L_o = m_evr = \frac{nh}{2\pi} \quad ... (5) \]
Substituting for \(L_o\) in Eq. (4),
\[ M_o = \frac{enh}{4\pi m_e} \]
For \(n=1\), \(M_o = \frac{eh}{4\pi m_e}\).
The quantity \(\frac{eh}{4\pi m_e}\) is a fundamental constant called the Bohr magneton, \(\mu_B\).
\(\mu_B = 9.274 \times 10^{-24} \text{ J/T (or A}\cdot\text{m}^2) = 5.788 \times 10^{-5} \text{ eV/T}\).
The quantity \(\frac{eh}{4\pi m_e}\) is a fundamental constant called the Bohr magneton, \(\mu_B\).
\(\mu_B = 9.274 \times 10^{-24} \text{ J/T (or A}\cdot\text{m}^2) = 5.788 \times 10^{-5} \text{ eV/T}\).
Q. 22. The wavelength of two sound waves in air are \(\frac{82}{173}\) m and \(\frac{82}{171}\) m. They produce 9 beats per second. Calculate the velocity of sound in air.
Data: \(\lambda_1 = \frac{82}{173} \text{ m}\), \(\lambda_2 = \frac{82}{171} \text{ m}\)
Let \(n_1\) and \(n_2\) be the corresponding frequencies.
\(\therefore v = n_1\lambda_1 = n_2\lambda_2\), where \(v\) is the speed of sound in air.
But \(\lambda_1 < \lambda_2\)
\(\therefore n_1 > n_2\)
\(\therefore n_1 - n_2 = 9\)
\[ \therefore v \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = 9 \] \[ \therefore v \left( \frac{173}{82} - \frac{171}{82} \right) = 9 \] \[ \therefore v = \frac{9 \times 82}{2} = 9 \times 41 = 369 \text{ m/s} \]
Let \(n_1\) and \(n_2\) be the corresponding frequencies.
\(\therefore v = n_1\lambda_1 = n_2\lambda_2\), where \(v\) is the speed of sound in air.
But \(\lambda_1 < \lambda_2\)
\(\therefore n_1 > n_2\)
\(\therefore n_1 - n_2 = 9\)
\[ \therefore v \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = 9 \] \[ \therefore v \left( \frac{173}{82} - \frac{171}{82} \right) = 9 \] \[ \therefore v = \frac{9 \times 82}{2} = 9 \times 41 = 369 \text{ m/s} \]
Q. 23. A circular loop of radius 9.2 cm carries a current of 2.3 A. Obtain the magnitude of magnetic field at the centre of the loop.
Data: \(R = z = 9.2 \times 10^{-2} \text{ m}\), \(I = 2.3 \text{ A}\), \(N = 1\)
The magnitude of the magnetic induction, \[ B = \frac{\mu_0 NI}{2R} \] \[ = \frac{(4\pi \times 10^{-7})(1)(2.3)}{2(9.2 \times 10^{-2})} \] \[ = \frac{(9.2\pi \times 10^{-5})}{2 \times 9.2} = \frac{3.142}{2} \times 10^{-5} = 1.571 \times 10^{-5} \text{ T} \]
The magnitude of the magnetic induction, \[ B = \frac{\mu_0 NI}{2R} \] \[ = \frac{(4\pi \times 10^{-7})(1)(2.3)}{2(9.2 \times 10^{-2})} \] \[ = \frac{(9.2\pi \times 10^{-5})}{2 \times 9.2} = \frac{3.142}{2} \times 10^{-5} = 1.571 \times 10^{-5} \text{ T} \]
Q. 24. 0.5 mole of a gas at a temperature of 450 K expands isothermally from an initial volume of 3 L to a final volume of 9 L. (a) What is the work done by the gas? (b) How much heat is supplied to the gas?
Data: \(n = 0.5 \text{ mol}\), \(T = 450 \text{ K}\), \(V_i = 3 \text{ L}\), \(V_f = 9 \text{ L}\), \(R = 8.319 \text{ J/mol}\cdot\text{K}\)
(a) The work done by a gas in an isothermal expansion is \[ W = nRT \ln \left(\frac{V_f}{V_i}\right) = nRT \left[ 2.303 \times \log \left( \frac{V_f}{V_i} \right) \right] \] \[ \therefore W = (0.5)(8.319)(450)(2.303) \times \log \left(\frac{9}{3}\right) \] \[ = (8.319)(225)(2.303) \times \log 3 \] \[ = (8.319)(225)(2.303)(0.4771) \] \[ = 2057 \text{ J} = 2.057 \text{ kJ} \] (b) At constant temperature, change in internal energy, \(\Delta U = 0\).
\(\therefore\) The heat supplied to the gas, \[ Q = \Delta U + W = 0 + 2.057 = 2.057 \text{ kJ} \]
(a) The work done by a gas in an isothermal expansion is \[ W = nRT \ln \left(\frac{V_f}{V_i}\right) = nRT \left[ 2.303 \times \log \left( \frac{V_f}{V_i} \right) \right] \] \[ \therefore W = (0.5)(8.319)(450)(2.303) \times \log \left(\frac{9}{3}\right) \] \[ = (8.319)(225)(2.303) \times \log 3 \] \[ = (8.319)(225)(2.303)(0.4771) \] \[ = 2057 \text{ J} = 2.057 \text{ kJ} \] (b) At constant temperature, change in internal energy, \(\Delta U = 0\).
\(\therefore\) The heat supplied to the gas, \[ Q = \Delta U + W = 0 + 2.057 = 2.057 \text{ kJ} \]
log 8.319 0.9201
log 225 +2.3522
log 2.303 +0.3623
log 0.4771 +1.6786
______
3.3132
Antilog 3.3132 = 2057
log 225 +2.3522
log 2.303 +0.3623
log 0.4771 +1.6786
______
3.3132
Antilog 3.3132 = 2057
Q. 25. An alternating emf \(e = 200 \sin 314.2t\) volt is applied between the terminals of an electric bulb whose filament has a resistance of 100 Ω. Calculate (a) the rms current (b) the frequency of the ac signal (c) the period of the ac signal.
Data: \(e = 200 \sin 314.2t \text{ V}\), \(R = 100 \Omega\)
Comparing the given equation with \(e = e_0 \sin \omega t\), we get the peak value of the applied emf, \(e_0 = 200 \text{ V}\) and \(\omega = 314.2 \text{ rad/s}\).
\(\therefore\) The peak current is \[ i_0 = \frac{e_0}{R} = \frac{200}{100} = 2 \text{ A} \] \(\therefore\) The rms current is \[ i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ A} \] The frequency of the applied emf, \[ f = \frac{\omega}{2\pi} = \frac{314.2}{2 \times 3.142} = \frac{100}{2} = 50 \text{ Hz} \] The period of the applied emf, \[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \text{ s} \]
Comparing the given equation with \(e = e_0 \sin \omega t\), we get the peak value of the applied emf, \(e_0 = 200 \text{ V}\) and \(\omega = 314.2 \text{ rad/s}\).
\(\therefore\) The peak current is \[ i_0 = \frac{e_0}{R} = \frac{200}{100} = 2 \text{ A} \] \(\therefore\) The rms current is \[ i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ A} \] The frequency of the applied emf, \[ f = \frac{\omega}{2\pi} = \frac{314.2}{2 \times 3.142} = \frac{100}{2} = 50 \text{ Hz} \] The period of the applied emf, \[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \text{ s} \]
Q. 26. Two charges of magnitudes 5 nC and -2 nC are placed at points (2 cm, 0, 0) and (20 cm, 0, 0) in a region of space where there is no external field. Find the electrostatic potential energy of the system.
Data: \(q_1 = 5 \text{ nC} = 5 \times 10^{-9} \text{ C}\), \(q_2 = -2 \text{ nC} = -2 \times 10^{-9} \text{ C}\)
\(r = (20 - 2) \text{ cm} = 18 \text{ cm} = 18 \times 10^{-2} \text{ m}\), \(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2\)
The electrostatic potential energy of the system, \[ U = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r} \] \[ = (9 \times 10^9) \frac{(5 \times 10^{-9})(-2 \times 10^{-9})}{18 \times 10^{-2}} \] \[ = -5 \times 10^{-7} = -0.5 \times 10^{-6} \text{ J} = -0.5 \mu\text{J} \]
\(r = (20 - 2) \text{ cm} = 18 \text{ cm} = 18 \times 10^{-2} \text{ m}\), \(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2\)
The electrostatic potential energy of the system, \[ U = \frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r} \] \[ = (9 \times 10^9) \frac{(5 \times 10^{-9})(-2 \times 10^{-9})}{18 \times 10^{-2}} \] \[ = -5 \times 10^{-7} = -0.5 \times 10^{-6} \text{ J} = -0.5 \mu\text{J} \]
SECTION – D
Attempt any three of the following questions: [12]
Q. 27. Using the principle of energy conservation, derive the expression for the minimum speeds at different locations along a vertical circular motion controlled by gravity.
Consider a particle of mass \(m\) attached to a string and revolved in a vertical circle of radius \(r\). At every instant of its motion, the particle is acted upon by its weight \(m\vec{g}\) and the tension \(\vec{T}\) in the string. The particle may not complete the circle if the string slackens before the particle reaches the top. This requires that the particle must have some minimum speed.
(i) At the top (point A): Let \(v_1\) be the speed of the particle and \(T_1\) the tension in the string at the top. Here, both \(\vec{T}_1\) and weight \(m\vec{g}\) are vertically downward. Hence, the net force on the particle towards the centre O is \(T_1 + mg\), which is the necessary centripetal force. \[ \therefore T_1 + mg = \frac{mv_1^2}{r} \quad ... (1) \] To find the minimum value of \(v_1\) that the particle must have at the top, we consider the limiting case when the tension \(T_1\) just becomes zero. \[ \therefore \frac{mv_1^2}{r} = mg \] that is, the particle's weight alone is the necessary centripetal force at the point A. \[ \therefore v_1^2 = gr \quad \therefore v_1 = \sqrt{gr} \quad ... (2) \] (ii) At the bottom (point B): Let \(v_2\) be the speed at the bottom. Taking the reference level for zero potential energy to be the bottom of the circle, the particle has only kinetic energy \(\frac{1}{2}mv_2^2\) at the lowest point. \[ \text{Total energy at the bottom} = \text{KE} + \text{PE} = \frac{1}{2}mv_2^2 + 0 = \frac{1}{2}mv_2^2 \quad ... (3) \] As the particle goes from the bottom to the top of the circle, it rises through a height \(h = 2r\). Therefore, its potential energy at the top is \[ mgh = mg(2r) \] and, from Eq. (2), its minimum kinetic energy there is \[ \frac{1}{2}mv_1^2 = \frac{1}{2}mgr \quad ... (4) \] \[ \text{Minimum total energy at the top} = \text{KE} + \text{PE} = \frac{1}{2}mgr + 2mgr = \frac{5}{2}mgr \quad ... (5) \] Assuming that the total energy of the particle is conserved, total energy at the bottom = total energy at the top. Then, from Eqs. (4) and (5), \[ \frac{1}{2}mv_2^2 = \frac{5}{2}mgr \] The minimum speed the particle must have at the lowest position is \[ v_2 = \sqrt{5gr} \quad ... (6) \] (iii) At the midway (point C): Let \(v_3\) be the speed at point C, so that its kinetic energy is \(\frac{1}{2}mv_3^2\).
At C, the particle is at a height \(r\) from the bottom of the circle. Therefore, its potential energy at C is \(mgr\). \[ \text{Total energy at C} = \frac{1}{2}mv_3^2 + mgr \quad ... (7) \] From the law of conservation of energy, total energy at C = total energy at B \[ \therefore \frac{1}{2}mv_3^2 + mgr = \frac{5}{2}mgr \] \[ \therefore v_3^2 = 5gr - 2gr = 3gr \] \(\therefore\) The minimum speed the particle must have midway up is \[ v_3 = \sqrt{3gr} \quad ... (8) \]
(i) At the top (point A): Let \(v_1\) be the speed of the particle and \(T_1\) the tension in the string at the top. Here, both \(\vec{T}_1\) and weight \(m\vec{g}\) are vertically downward. Hence, the net force on the particle towards the centre O is \(T_1 + mg\), which is the necessary centripetal force. \[ \therefore T_1 + mg = \frac{mv_1^2}{r} \quad ... (1) \] To find the minimum value of \(v_1\) that the particle must have at the top, we consider the limiting case when the tension \(T_1\) just becomes zero. \[ \therefore \frac{mv_1^2}{r} = mg \] that is, the particle's weight alone is the necessary centripetal force at the point A. \[ \therefore v_1^2 = gr \quad \therefore v_1 = \sqrt{gr} \quad ... (2) \] (ii) At the bottom (point B): Let \(v_2\) be the speed at the bottom. Taking the reference level for zero potential energy to be the bottom of the circle, the particle has only kinetic energy \(\frac{1}{2}mv_2^2\) at the lowest point. \[ \text{Total energy at the bottom} = \text{KE} + \text{PE} = \frac{1}{2}mv_2^2 + 0 = \frac{1}{2}mv_2^2 \quad ... (3) \] As the particle goes from the bottom to the top of the circle, it rises through a height \(h = 2r\). Therefore, its potential energy at the top is \[ mgh = mg(2r) \] and, from Eq. (2), its minimum kinetic energy there is \[ \frac{1}{2}mv_1^2 = \frac{1}{2}mgr \quad ... (4) \] \[ \text{Minimum total energy at the top} = \text{KE} + \text{PE} = \frac{1}{2}mgr + 2mgr = \frac{5}{2}mgr \quad ... (5) \] Assuming that the total energy of the particle is conserved, total energy at the bottom = total energy at the top. Then, from Eqs. (4) and (5), \[ \frac{1}{2}mv_2^2 = \frac{5}{2}mgr \] The minimum speed the particle must have at the lowest position is \[ v_2 = \sqrt{5gr} \quad ... (6) \] (iii) At the midway (point C): Let \(v_3\) be the speed at point C, so that its kinetic energy is \(\frac{1}{2}mv_3^2\).
At C, the particle is at a height \(r\) from the bottom of the circle. Therefore, its potential energy at C is \(mgr\). \[ \text{Total energy at C} = \frac{1}{2}mv_3^2 + mgr \quad ... (7) \] From the law of conservation of energy, total energy at C = total energy at B \[ \therefore \frac{1}{2}mv_3^2 + mgr = \frac{5}{2}mgr \] \[ \therefore v_3^2 = 5gr - 2gr = 3gr \] \(\therefore\) The minimum speed the particle must have midway up is \[ v_3 = \sqrt{3gr} \quad ... (8) \]
Q. 28. Explain the conversion of a moving-coil galvanometer (MCG) into an ammeter. Obtain the necessary formula. State any two advantages of a potentiometer over a voltmeter.
A moving-coil galvanometer is converted into an ammeter by reducing its effective resistance by connecting a low resistance \(S\) across the coil. Such a parallel low resistance is called a shunt since it shunts a part of the current around the coil. That makes it possible to increase the range of currents over which the meter is useful.
Let \(I\) be the maximum line current to be measured and \(I_g\) the current for which the galvanometer of resistance \(G\) shows a full-scale deflection. Then, the shunt resistance \(S\) should be such that the remaining current \(I - I_g = I_s\) is shunted through it.
In the parallel combination, the p.d. across the galvanometer = the p.d. across the shunt \[ \therefore I_g G = I_s S = (I - I_g) S \] \[ \therefore S = \left( \frac{I_g}{I - I_g} \right) G \quad ... (1) \] This is the required resistance of the shunt. The scale of the galvanometer is then calibrated so as to read the current in ampere or its submultiples (mA or \(\mu\)A) directly.
Advantages of a potentiometer over a voltmeter:
(1) The cell, whose emf is being measured, draws no current from the circuit at the null point. Thus, the potentiometer measures the open-circuit potential difference across its terminals, or the emf \(E\). A voltmeter will measure the terminal potential difference, \(V\), of the cell in a closed circuit. This is because the resistance of a voltmeter is high but not infinite and hence the voltmeter is not ideal.
(2) By setting up a suitably small potential gradient along a long potentiometer wire, any small voltage can be measured. Increasing the length of the wire effectively decreases the potential gradient, and increases both the precision and accuracy of measurement.
(3) The adjustment of a potentiometer is a ‘null’ method which does not, in any way, depend on the calibration of the galvanometer. The galvanometer is used only to detect the current, not to measure it. The accuracy of a voltmeter is limited by its calibration.
(4) Since a potentiometer can measure both the emf and terminal potential difference of a cell, the internal resistance of the cell can be found.
Let \(I\) be the maximum line current to be measured and \(I_g\) the current for which the galvanometer of resistance \(G\) shows a full-scale deflection. Then, the shunt resistance \(S\) should be such that the remaining current \(I - I_g = I_s\) is shunted through it.
In the parallel combination, the p.d. across the galvanometer = the p.d. across the shunt \[ \therefore I_g G = I_s S = (I - I_g) S \] \[ \therefore S = \left( \frac{I_g}{I - I_g} \right) G \quad ... (1) \] This is the required resistance of the shunt. The scale of the galvanometer is then calibrated so as to read the current in ampere or its submultiples (mA or \(\mu\)A) directly.
Advantages of a potentiometer over a voltmeter:
(1) The cell, whose emf is being measured, draws no current from the circuit at the null point. Thus, the potentiometer measures the open-circuit potential difference across its terminals, or the emf \(E\). A voltmeter will measure the terminal potential difference, \(V\), of the cell in a closed circuit. This is because the resistance of a voltmeter is high but not infinite and hence the voltmeter is not ideal.
(2) By setting up a suitably small potential gradient along a long potentiometer wire, any small voltage can be measured. Increasing the length of the wire effectively decreases the potential gradient, and increases both the precision and accuracy of measurement.
(3) The adjustment of a potentiometer is a ‘null’ method which does not, in any way, depend on the calibration of the galvanometer. The galvanometer is used only to detect the current, not to measure it. The accuracy of a voltmeter is limited by its calibration.
(4) Since a potentiometer can measure both the emf and terminal potential difference of a cell, the internal resistance of the cell can be found.
Q. 29. State (a) Stefan-Boltzmann law of radiation (b) Wien’s displacement law.
The difference between the two molar specific heats of a gas is 6000 J/kg·K. If the ratio of the specific heats is 1.4, calculate the molar specific heat at constant volume.
The difference between the two molar specific heats of a gas is 6000 J/kg·K. If the ratio of the specific heats is 1.4, calculate the molar specific heat at constant volume.
The Stefan-Boltzmann law: The rate of emission of radiant energy per unit area, i.e., the power radiated per unit area of a perfect blackbody, is directly proportional to the fourth power of its absolute temperature. OR The quantity of radiant energy emitted by a perfect blackbody per unit time per unit surface area of the body is directly proportional to the fourth power of its absolute temperature.
Wien’s displacement law: The wavelength for which the emissive power of a blackbody is maximum, is inversely proportional to the absolute temperature of the blackbody. OR For a blackbody at an absolute temperature \(T\), the product of \(T\) and the wavelength \(\lambda_m\) corresponding to the maximum radiation of energy is a constant.
\(\lambda_m T = b\), a constant.
Data: \(S_P - S_V = 6000 \text{ J/kg}\cdot\text{K}\), \(S_P/S_V = 1.4\)
The unit J/kg·K and the magnitude of the difference indicate that the quantities are the principal specific heat capacities \(S_P\) and \(S_V\), and not molar specific heats.
\[ S_P/S_V = 1.4 \quad \therefore S_P = 1.4 S_V \] \[ S_P - S_V = 6000 \] \[ \therefore 1.4 S_V - S_V = 6000 \] \[ \therefore S_V = \frac{6000}{0.4} = \frac{60}{4} \times 10^3 = 15 \text{ kJ/kg}\cdot\text{K} \]
Wien’s displacement law: The wavelength for which the emissive power of a blackbody is maximum, is inversely proportional to the absolute temperature of the blackbody. OR For a blackbody at an absolute temperature \(T\), the product of \(T\) and the wavelength \(\lambda_m\) corresponding to the maximum radiation of energy is a constant.
\(\lambda_m T = b\), a constant.
Data: \(S_P - S_V = 6000 \text{ J/kg}\cdot\text{K}\), \(S_P/S_V = 1.4\)
The unit J/kg·K and the magnitude of the difference indicate that the quantities are the principal specific heat capacities \(S_P\) and \(S_V\), and not molar specific heats.
\[ S_P/S_V = 1.4 \quad \therefore S_P = 1.4 S_V \] \[ S_P - S_V = 6000 \] \[ \therefore 1.4 S_V - S_V = 6000 \] \[ \therefore S_V = \frac{6000}{0.4} = \frac{60}{4} \times 10^3 = 15 \text{ kJ/kg}\cdot\text{K} \]
Q. 30. Define: (a) self inductance (b) mutual inductance.
A straight conductor is moving with a velocity of 3 m/s at right angles to a magnetic field of magnitude \(4.5 \times 10^{-5} \text{ Wb/m}^2\). If an emf developed between its ends is \(1.35 \times 10^{-4}\) volt, calculate the length of the straight conductor.
A straight conductor is moving with a velocity of 3 m/s at right angles to a magnetic field of magnitude \(4.5 \times 10^{-5} \text{ Wb/m}^2\). If an emf developed between its ends is \(1.35 \times 10^{-4}\) volt, calculate the length of the straight conductor.
Definition (Self Inductance): The self inductance or the coefficient of self induction of a coil is defined as the emf induced in the coil per unit time rate of change of current in the same coil. OR the self inductance of a coil is the ratio of magnetic flux linked with the coil to the current in it.
Definition (Mutual Inductance): The mutual inductance or the coefficient of mutual induction of two magnetically linked coils is equal to the flux linkage of one coil per unit current in the neighbouring coil. OR The mutual inductance or the coefficient of mutual induction of two magnetically linked coils is numerically equal to the emf induced in one coil (secondary) per unit time rate of change of current in the neighbouring coil (primary).
Data: \(v = 3 \text{ m/s}\), \(B = 4.5 \times 10^{-5} \text{ T}\), \(e = 1.35 \times 10^{-4} \text{ V}\)
\(e = Blv\) \(\therefore\) The length of the conductor, \[ l = \frac{e}{Bv} = \frac{1.35 \times 10^{-4}}{(4.5 \times 10^{-5})(3)} = \frac{1.35 \times 10^{-4}}{13.5 \times 10^{-5}} = 1 \text{ m} \]
Definition (Mutual Inductance): The mutual inductance or the coefficient of mutual induction of two magnetically linked coils is equal to the flux linkage of one coil per unit current in the neighbouring coil. OR The mutual inductance or the coefficient of mutual induction of two magnetically linked coils is numerically equal to the emf induced in one coil (secondary) per unit time rate of change of current in the neighbouring coil (primary).
Data: \(v = 3 \text{ m/s}\), \(B = 4.5 \times 10^{-5} \text{ T}\), \(e = 1.35 \times 10^{-4} \text{ V}\)
\(e = Blv\) \(\therefore\) The length of the conductor, \[ l = \frac{e}{Bv} = \frac{1.35 \times 10^{-4}}{(4.5 \times 10^{-5})(3)} = \frac{1.35 \times 10^{-4}}{13.5 \times 10^{-5}} = 1 \text{ m} \]
Q. 31. State any two limitations of Bohr’s atomic model.
The half-life of a radioactive species is 3.2 days. Calculate the decay constant (per day).
The half-life of a radioactive species is 3.2 days. Calculate the decay constant (per day).
Limitations of Bohr’s atomic model:
(1) The model cannot explain the relative intensities of spectral lines even in the hydrogen spectrum.
(2) The model cannot explain the atomic spectra of many electron atoms of higher elements.
(3) The model cannot account for the Zeeman effect and Stark effect (fine structure of spectral lines as revealed in the presence of strong magnetic field and electric field, respectively).
Data: \(T_{1/2} = 3.2 \text{ d}\)
The half-life period of the radioactive substance, \[ T_{1/2} = \frac{0.693}{\lambda} \] \(\therefore\) The decay constant of the radioactive substance, \[ \lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{3.2} = 0.2165 \text{ day}^{-1} \]
(1) The model cannot explain the relative intensities of spectral lines even in the hydrogen spectrum.
(2) The model cannot explain the atomic spectra of many electron atoms of higher elements.
(3) The model cannot account for the Zeeman effect and Stark effect (fine structure of spectral lines as revealed in the presence of strong magnetic field and electric field, respectively).
Data: \(T_{1/2} = 3.2 \text{ d}\)
The half-life period of the radioactive substance, \[ T_{1/2} = \frac{0.693}{\lambda} \] \(\therefore\) The decay constant of the radioactive substance, \[ \lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{3.2} = 0.2165 \text{ day}^{-1} \]
log 0.693 \(\bar{1}.8407\)
log 3.2 \(-0.5051\)
______
\(\bar{1}.3356\)
Antilog \(\bar{1}.3356 = 0.2166\)
log 3.2 \(-0.5051\)
______
\(\bar{1}.3356\)
Antilog \(\bar{1}.3356 = 0.2166\)