BOARD QUESTION PAPER : FEBRUARY 2023
PHYSICS
Time: 3 Hrs. | Max. Marks: 70
The question paper is divided into four sections:
- Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each. Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
- Section B: Q. No. 3 to Q. No. 14 contain Twelve short answer type of questions carrying Two marks each. (Attempt any Eight).
- Section C: Q. No. 15 to Q. No. 26 contain Twelve short answer type of questions carrying Three marks each. (Attempt any Eight).
- Section D: Q. No. 27 to Q. No. 31 contain Five long answer type of questions carrying Four marks each. (Attempt any Three).
- Use of the log table is allowed. Use of calculator is not allowed.
- Figures to the right indicate full marks.
- For each multiple choice type of question, it is mandatory to write the correct answer along with its alphabet. e.g., (a)......./(b)......./(c)......./(d)....... No marks(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.
- (i) \( h = 6.63 \times 10^{-34} \) Js
- (ii) \( c = 3 \times 10^8 \) m/s
- (iii) \( \pi = 3.142 \)
- (iv) \( g = 9.8 \) m/s²
- (v) \( \epsilon_0 = 8.85 \times 10^{-12} \) C² / Nm²
- (vi) \( \mu_0 = 4\pi \times 10^{-7} \) Wb / A–m
SECTION – A
i. If ‘n’ is the number of molecules per unit volume and ‘d’ is the diameter of the molecules, the mean free path ‘λ’ of molecules is
ii. The first law of thermodynamics is consistent with the law of conservation of _______.
iii. \( Y = \overline{A + B} \) is the Boolean expression for _______.
iv. The property of light which remains unchanged when it travels from one medium to another is ________.
v. If a circular coil of 100 turns with a cross-sectional area of 1 m² is kept with its plane perpendicular to the magnetic field of 1 T, the magnetic flux linked with the coil will be ________.
Explanation: \( \phi = NBA \cos \theta \). Here plane is perpendicular to B, so normal is parallel (\( \theta = 0^\circ \)).
\( \phi = 100 \times 1 \times 1 \times \cos 0 = 100 \) Wb.
vi. If ‘θ’ represents the angle of contact made by a liquid which completely wets the surface of the container then ________.
vii. The LED emits visible light when its ________.
viii. Soft iron is used to make the core of transformer because of its _______.
ix. If the maximum kinetic energy of emitted electrons in photoelectric effect is 2eV, the stopping potential will be _______.
Explanation: \( K_{max} = eV_s \). If \( K_{max} = 2 \) eV, then \( V_s = 2 \) V.
x. The radius of eighth orbit of electron in H-atom will be more than that of fourth orbit by a factor of _______.
Explanation: \( r_n \propto n^2 \). \( \frac{r_8}{r_4} = \frac{8^2}{4^2} = \frac{64}{16} = 4 \).
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 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
i. What is the value of resistance for an ideal voltmeter?
ii. What is the value of force on a closed circuit in a magnetic field?
iii. What is the average value of alternating current over a complete cycle?
iv. An electron is accelerated through a potential difference of 100 volt. Calculate de-Broglie wavelength in nm.
Given: \( V = 100 \) V
Calculation: \( \lambda = \frac{1.228}{\sqrt{100}} = \frac{1.228}{10} = 0.1228 \) nm.
v. If friction is made zero for a road, can a vehicle move safely on this road?
vi. State the formula giving relation between electric field intensity and potential gradient.
vii. Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by \( x = 5 \sin \left(\frac{\pi t}{3}\right) \) m.
Velocity \( v = \frac{dx}{dt} = \omega A \cos(\omega t) \).
\( v = \frac{\pi}{3} \times 5 \times \cos\left(\frac{\pi t}{3}\right) \).
At \( t = 1 \) s: \( v = \frac{5\pi}{3} \cos\left(\frac{\pi}{3}\right) = \frac{5\pi}{3} \cos(60^\circ) \).
\( v = \frac{5\pi}{3} \times \frac{1}{2} = \frac{5\pi}{6} \) m/s.
viii. Write the mathematical formula for Bohr magneton for an electron revolving in nth orbit.
\( M_o = n \left( \frac{eh}{4\pi m_e} \right) \)
Where \( n \) is the principal quantum number. (The value \( \frac{eh}{4\pi m_e} \) is the Bohr Magneton constant \( \mu_B \)).
SECTION – B
Attempt any EIGHT questions of the following: [16]
Q.3. Define coefficient of viscosity. State its formula and S.I. units.
Formula: \( \eta = \frac{F}{A(\frac{dv}{dx})} \)
Where \( F \) is viscous force, \( A \) is area of layer, and \( \frac{dv}{dx} \) is velocity gradient.
S.I. Unit: N s / m² (Newton-second per square meter) or Pa s (Pascal-second) or Poiseuille (Pl).
Q.4. Obtain an expression for magnetic induction of a toroid of ‘N’ turns about an axis passing through its centre and perpendicular to its plane.
\( \oint \vec{B} \cdot \vec{dl} = \mu_0 I_{enclosed} \)
For a toroid of mean radius \( r \), the path is a circle of circumference \( 2\pi r \). By symmetry, B is constant and tangent to the path.
\( B (2\pi r) = \mu_0 (NI) \)
Where \( N \) is the total number of turns and \( I \) is the current.
\( B = \frac{\mu_0 N I}{2\pi r} \)
If \( n \) is the number of turns per unit length (\( n = \frac{N}{2\pi r} \)), then:
\( B = \mu_0 n I \).
Q.5. State and prove principle of conservation of angular momentum.
Proof:
The rate of change of angular momentum is equal to the external torque:
\( \vec{\tau}_{ext} = \frac{d\vec{L}}{dt} \)
If \( \vec{\tau}_{ext} = 0 \), then \( \frac{d\vec{L}}{dt} = 0 \).
Therefore, \( \vec{L} = \text{constant} \).
\( I_1 \omega_1 = I_2 \omega_2 \) (for a rigid body).
Q.6. Obtain an expression for equivalent capacitance of two capacitors C1 and C2 connected in series.
Total potential \( V = V_1 + V_2 \).
We know \( V = \frac{Q}{C} \).
\( \frac{Q}{C_{eq}} = \frac{Q}{C_1} + \frac{Q}{C_2} \)
Dividing by \( Q \):
\( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \)
Q.7. Explain, why the equivalent inductance of two coils connected in parallel is less than the inductance of either of the coils.
\( \frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} \)
Mathematically, the sum of reciprocals is always greater than the reciprocal of the smallest individual value. Therefore, the reciprocal of the sum (which is \( L_{eq} \)) must be smaller than the smallest individual inductance.
Physically, connecting coils in parallel provides multiple paths for magnetic flux development for a given voltage change rate, reducing the overall opposition (back EMF) for the total current change, effectively lowering the inductance.
Q.8. How will you convert a moving coil galvanometer into an ammeter?
1. Connect a low resistance wire, called a shunt resistance (S), in parallel with the galvanometer coil.
2. The value of the shunt resistance is calculated using the formula: \( S = \left( \frac{I_g}{I - I_g} \right) G \), where \( I \) is the total current to be measured, \( I_g \) is the full-scale deflection current of the galvanometer, and \( G \) is the galvanometer resistance.
Q.9. A 100 Ω resistor is connected to a 220 V, 50 Hz supply. Calculate: i. r.m.s. value of current and ii. net power consumed over the full cycle
i. r.m.s. value of current:
\( I_{rms} = \frac{V_{rms}}{R} = \frac{220}{100} = 2.2 \) A.
ii. Net power consumed:
For a pure resistor, \( P = V_{rms} I_{rms} \).
\( P = 220 \times 2.2 = 484 \) W.
Q.10. A bar magnet of mass 120 g in the form of a rectangular parallelepiped, has dimensions l = 40 mm, b = 100 mm and h = 80 mm, with its dimension ‘h’ vertical, the magnet performs angular oscillations in the plane of the magnetic field with period π seconds. If the magnetic moment is 3.4 Am², determine the influencing magnetic field.
Mass \( m = 120 \) g = 0.12 kg.
Dimensions: \( l = 40 \) mm = 0.04 m, \( b = 100 \) mm = 0.1 m, \( h = 80 \) mm (vertical).
Since \( h \) is vertical, the magnet oscillates about the vertical axis passing through its center. The relevant moment of inertia \( I \) involves the horizontal dimensions \( l \) and \( b \).
\( I = \frac{Mass(l^2 + b^2)}{12} \)
\( I = \frac{0.12((0.04)^2 + (0.1)^2)}{12} = 0.01 (0.0016 + 0.01) \)
\( I = 0.01 (0.0116) = 1.16 \times 10^{-4} \) kg m².
Period \( T = \pi \) s. Magnetic Moment \( \mu = 3.4 \) Am².
Formula: \( T = 2\pi \sqrt{\frac{I}{\mu B}} \)
Squaring both sides: \( T^2 = 4\pi^2 \frac{I}{\mu B} \)
\( \pi^2 = 4\pi^2 \frac{1.16 \times 10^{-4}}{3.4 \times B} \)
\( 1 = \frac{4 \times 1.16 \times 10^{-4}}{3.4 \times B} \)
\( B = \frac{4.64 \times 10^{-4}}{3.4} \approx 1.365 \times 10^{-4} \) T.
Q.11. Distinguish between free vibrations and forced vibrations (Two points).
- Cause: Free vibrations occur when a body is disturbed and left to oscillate under its own restoring force. Forced vibrations occur when a body oscillates under the influence of an external periodic force.
- Frequency: In free vibrations, the body oscillates with its natural frequency. In forced vibrations, the body oscillates with the frequency of the external periodic force (driver frequency).
- Amplitude: Amplitude of free vibrations decreases over time due to damping. Amplitude of forced vibrations remains constant as long as the external force is applied.
Q.12. Compare the rate of loss of heat from a metal sphere at 827 °C with rate of loss of heat from the same at 427 °C, if the temperature of surrounding is 27 °C.
Temperatures in Kelvin:
\( T_1 = 827 + 273 = 1100 \) K
\( T_2 = 427 + 273 = 700 \) K
\( T_0 = 27 + 273 = 300 \) K
Ratio \( \frac{R_1}{R_2} = \frac{1100^4 - 300^4}{700^4 - 300^4} \)
\( \frac{R_1}{R_2} = \frac{(11)^4 \times 10^8 - 3^4 \times 10^8}{(7)^4 \times 10^8 - 3^4 \times 10^8} = \frac{14641 - 81}{2401 - 81} \)
\( \frac{R_1}{R_2} = \frac{14560}{2320} = \frac{1456}{232} \approx 6.276 \)
Ratio is approx 182:29 or 6.276:1
Q.13. An ideal mono-atomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. Calculate the ratio of final pressure to its initial pressure.
For mono-atomic gas, \( \gamma = 5/3 \).
\( P_1^{1-5/3} T_1^{5/3} = P_2^{1-5/3} T_2^{5/3} \)
\( P_1^{-2/3} T_1^{5/3} = P_2^{-2/3} (2T_1)^{5/3} \)
\( \left(\frac{P_2}{P_1}\right)^{-2/3} = \left(\frac{T_1}{2T_1}\right)^{5/3} = \left(\frac{1}{2}\right)^{5/3} \)
\( \frac{P_2}{P_1} = \left( \left(\frac{1}{2}\right)^{5/3} \right)^{-3/2} = \left(\frac{1}{2}\right)^{-2.5} = 2^{2.5} \)
\( 2^{2.5} = 2^2 \times 2^{0.5} = 4\sqrt{2} \approx 5.657 \).
Q.14. Disintegration rate of a radio-active sample is 1010 per hour at 20 hours from the start. It reduces to 5 × 109 per hour after 30 hours. Calculate the decay constant.
At \( t_1 = 20 \) h, \( R_1 = 10^{10} \).
At \( t_2 = 30 \) h, \( R_2 = 5 \times 10^9 \).
\( \frac{R_2}{R_1} = \frac{e^{-\lambda(30)}}{e^{-\lambda(20)}} = e^{-10\lambda} \)
\( \frac{5 \times 10^9}{10^{10}} = 0.5 \)
\( 0.5 = e^{-10\lambda} \Rightarrow \ln(0.5) = -10\lambda \)
\( -0.693 = -10\lambda \)
\( \lambda = 0.0693 \) hr\(^{-1}\).
SECTION – C
Attempt any EIGHT questions of the following: [24]
Q.15. Derive laws of reflection of light using Huygens’ principle.
1. Consider a plane wavefront AB incident obliquely on a plane reflecting surface MN.
2. At t=0, point A touches the surface. Point B is at distance 'ct' from C (where C is the point on surface where secondary wavelet from B will strike).
3. During time 't', a secondary wavelet from A grows into a hemisphere of radius 'ct'.
4. Draw a tangent plane from C to the hemisphere at A (point E). This tangent plane CE represents the reflected wavefront.
5. Triangles AEC and ABC are congruent (Common hypotenuse AC, AE=BC=ct, 90° angles).
6. Therefore, angle \( \angle BAC \) (angle of incidence i) equals angle \( \angle ECA \) (angle of reflection r).
7. Hence, \( i = r \). This proves the law of reflection.
Q.16. State postulates of Bohr’s atomic model.
2. Electrons can revolve only in those orbits where the angular momentum is an integral multiple of \( \frac{h}{2\pi} \). (\( L = mvr = \frac{nh}{2\pi} \)).
3. An electron radiates energy only when it jumps from a higher energy orbit to a lower energy orbit. The energy difference corresponds to the photon emitted: \( E_2 - E_1 = h\nu \).
Q.17. Define and state unit and dimensions of : i. Magnetization ii. Magnetic susceptibility
Definition: Net magnetic moment per unit volume of a material.
Unit: Ampere/meter (A/m).
Dimensions: \( [L^{-1} M^0 T^0 I^1] \).
ii. Magnetic Susceptibility (\( \chi \)):
Definition: The ratio of magnetization (Mz) to the magnetic intensity (H). \( \chi = \frac{Mz}{H} \).
Unit: No unit (It is a pure number).
Dimensions: Dimensionless \( [M^0 L^0 T^0 I^0] \).
Q.18. With neat labelled circuit diagram, describe an experiment to study the characteristics of photoelectric effect.
1. Monochromatic light from a source falls on the photosensitive plate (Emitter) through a quartz window.
2. Electrons are emitted and accelerated towards the Collector plate by a potential difference.
3. The photoelectric current is measured by a microammeter.
4. By varying the potential (using a commutator) and intensity/frequency of light, characteristics like saturation current and stopping potential are studied.
Q.19. Explain the use of potentiometer to determine internal resistance of a cell.
2. Step 1 (Open Circuit): Key in the shunt is open. The balancing length \( l_1 \) is found where the galvanometer shows zero deflection. Here, \( E \propto l_1 \).
3. Step 2 (Closed Circuit): Key in the shunt is closed with a resistance \( R \). Current flows from the cell. The terminal potential difference \( V \) is balanced at length \( l_2 \). Here, \( V \propto l_2 \).
4. Formula: Internal resistance \( r = R \left( \frac{E}{V} - 1 \right) = R \left( \frac{l_1}{l_2} - 1 \right) \).
Q.20. Explain the working of n-p-n transistor in common base configuration.
2. Emitter-Base junction is forward biased (Emitter negative, Base positive). Collector-Base junction is reverse biased (Collector positive, Base negative).
3. Electrons from the n-type emitter are injected into the p-type base. Being thin and lightly doped, most electrons (approx 95%) cross the base and reach the collector junction.
4. These electrons are swept into the collector by the strong reverse bias field, constituting collector current \( I_C \).
5. A small fraction recombines in the base, forming base current \( I_B \).
6. Relation: \( I_E = I_B + I_C \).
Q.21. State the differential equation of linear S.H.M. Hence, obtain expression for : i. acceleration ii. velocity
i. Acceleration:
From equation, \( \frac{d^2x}{dt^2} = -\omega^2 x \).
Since \( a = \frac{d^2x}{dt^2} \), \( a = -\omega^2 x \).
ii. Velocity:
\( a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx} = -\omega^2 x \).
Integrate both sides: \( \int v \, dv = -\omega^2 \int x \, dx \).
\( \frac{v^2}{2} = -\frac{\omega^2 x^2}{2} + C \).
At amplitude (extremes), \( x = \pm A, v = 0 \). Thus \( C = \frac{\omega^2 A^2}{2} \).
\( v^2 = \omega^2 (A^2 - x^2) \Rightarrow v = \pm \omega \sqrt{A^2 - x^2} \).
Q.22. Two tuning forks of frequencies 320 Hz and 340 Hz are sounded together to produce sound wave. The velocity of sound in air is 326.4 m/s. Calculate the difference in wavelengths of these waves.
Wavelength \( \lambda = \frac{v}{n} \).
\( \lambda_1 = \frac{326.4}{320} = 1.02 \) m.
\( \lambda_2 = \frac{326.4}{340} = 0.96 \) m.
Difference \( \Delta \lambda = \lambda_1 - \lambda_2 = 1.02 - 0.96 = 0.06 \) m.
Q.23. In a biprism experiment, the fringes are observed in the focal plane of the eye-piece at a distance of 1.2 m from the slit. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eye-piece, 90 cm from the eye-piece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.
\( x_{20} = 20 \beta \Rightarrow \beta = \frac{4 \times 10^{-3}}{20} = 0.2 \times 10^{-3} \) m.
2. Determining distance between sources (d) using displacement method:
The lens is 90 cm from eyepiece, so \( v = 90 \) cm, \( u = 120 - 90 = 30 \) cm (assuming total D is slit to eyepiece).
Magnification \( m = \frac{v}{u} = \frac{90}{30} = 3 \).
We are given the distance between magnified images \( d_1 = 0.9 \) cm \( = 9 \times 10^{-3} \) m.
Since \( m = \frac{d_1}{d} \), \( d = \frac{d_1}{m} = \frac{0.9}{3} = 0.3 \) cm \( = 3 \times 10^{-3} \) m.
3. Wavelength calculation:
\( \beta = \frac{\lambda D}{d} \Rightarrow \lambda = \frac{\beta d}{D} \)
\( \lambda = \frac{(0.2 \times 10^{-3}) (3 \times 10^{-3})}{1.2} \)
\( \lambda = \frac{0.6 \times 10^{-6}}{1.2} = 0.5 \times 10^{-6} \) m.
\( \lambda = 5000 \) Å.
Q.24. Calculate the current flowing through two long parallel wires carrying equal currents and separated by a distance of 1.35 cm experiencing a force per unit length of 4.76 × 10–2 N/m.
Given: \( I_1 = I_2 = I \), \( d = 1.35 \) cm \( = 1.35 \times 10^{-2} \) m, \( F/L = 4.76 \times 10^{-2} \) N/m.
\( 4.76 \times 10^{-2} = \frac{4\pi \times 10^{-7} \times I^2}{2\pi \times 1.35 \times 10^{-2}} \)
\( 4.76 \times 10^{-2} = \frac{2 \times 10^{-7} I^2}{1.35 \times 10^{-2}} \)
\( I^2 = \frac{4.76 \times 10^{-2} \times 1.35 \times 10^{-2}}{2 \times 10^{-7}} \)
\( I^2 = \frac{6.426 \times 10^{-4}}{2 \times 10^{-7}} = 3.213 \times 10^3 = 3213 \)
\( I = \sqrt{3213} \approx 56.68 \) A.
Q.25. An alternating voltage given by e = 140 sin (314.2 t) is connected across a pure resistor of 50 Ω. Calculate : i. the frequency of the source ii. the r.m.s current through the resistor
Given: \( e = 140 \sin(314.2 t) \), \( R = 50 \, \Omega \).
Comparing, \( E_0 = 140 \) V, \( \omega = 314.2 \) rad/s.
i. Frequency:
\( \omega = 2\pi f \Rightarrow f = \frac{314.2}{2 \times 3.142} = \frac{314.2}{6.284} \approx 50 \) Hz.
ii. r.m.s current:
\( E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{140}{1.414} \approx 99.0 \) V.
\( I_{rms} = \frac{E_{rms}}{R} = \frac{99.0}{50} = 1.98 \) A.
(Approximation: \( E_{rms} \approx 100 \) V, \( I_{rms} \approx 2 \) A).
Q.26. An electric dipole consists of two opposite charges each of magnitude 1 μC, separated by 2 cm. The dipole is placed in an external electric field of 105 N/C. Calculate the : i. maximum torque experienced by the dipole and ii. work done by the external field to turn the dipole through 180 °.
Dipole moment \( p = q(2a) = 10^{-6} \times 0.02 = 2 \times 10^{-8} \) Cm.
i. Maximum Torque:
\( \tau_{max} = pE \sin(90^\circ) = pE \)
\( \tau_{max} = 2 \times 10^{-8} \times 10^5 = 2 \times 10^{-3} \) Nm.
ii. Work done (turning through 180°):
Assuming starting from stable equilibrium (\( \theta_1 = 0^\circ \)) to (\( \theta_2 = 180^\circ \)).
\( W = pE(\cos \theta_1 - \cos \theta_2) \)
\( W = 2 \times 10^{-3} (\cos 0 - \cos 180) \)
\( W = 2 \times 10^{-3} (1 - (-1)) = 4 \times 10^{-3} \) J.
SECTION – D
Attempt any THREE questions of the following: [12]
Q.27. On the basis of kinetic theory of gases obtain an expression for pressure exerted by gas molecules enclosed in a container on its walls.
1. Consider a cube of side \( L \) containing \( N \) molecules of mass \( m \).
2. A molecule moving with velocity \( v_x \) collides elastically with the wall perpendicular to x-axis.
3. Change in momentum per collision: \( \Delta p = -mv_x - (mv_x) = -2mv_x \). Momentum imparted to wall = \( +2mv_x \).
4. Time between collisions: \( \Delta t = \frac{2L}{v_x} \).
5. Force exerted by one molecule: \( f_x = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L} \).
6. Total force by all molecules: \( F_x = \frac{m}{L} \sum v_x^2 \).
7. Pressure \( P = \frac{F}{L^2} = \frac{m}{L^3} \sum v_x^2 = \frac{m}{V} \sum v_x^2 \).
8. Due to randomness, \( \sum v_x^2 = \frac{1}{3} \sum v^2 \).
9. Mean square velocity \( \overline{v^2} = \frac{\sum v^2}{N} \).
10. Final Expression: \( P = \frac{1}{3} \frac{N m \overline{v^2}}{V} \) or \( P = \frac{1}{3} \rho \overline{v^2} \).
Q.28. i. Derive an expression for energy stored in the magnetic field in terms of induced current.
ii. A wire 5 m long is supported horizontally at a height of 15 m along east-west direction. When it is about to hit the ground, calculate the average e.m.f. induced in it. (g = 10 m/s²)
Work done against induced EMF to maintain current is stored as energy.
Power \( P = |e| I = L \frac{dI}{dt} I \).
Work \( dW = P dt = L I \, dI \).
Total Energy \( U = \int_0^I L I \, dI = \frac{1}{2} L I^2 \).
ii. Numerical:
Given: \( l = 5 \) m, \( h = 15 \) m, \( g = 10 \) m/s².
Wire is East-West. As it falls, it cuts the horizontal component of Earth's magnetic field (\( B_H \)) which points North.
Velocity calculation: \( v^2 = u^2 + 2gh = 0 + 2(10)(15) = 300 \).
Final velocity \( v = \sqrt{300} = 17.32 \) m/s.
Average velocity \( v_{avg} = \frac{0 + 17.32}{2} = 8.66 \) m/s.
Average EMF \( e = B_H l v_{avg} \).
Note: Value of \( B_H \) is not provided in question data. Assuming standard value \( B_H \approx 3.6 \times 10^{-5} \) T.
\( e = (3.6 \times 10^{-5}) \times 5 \times 8.66 \)
\( e \approx 1.56 \times 10^{-3} \) V or 1.56 mV.
Q.29. i. Derive an expression for the work done during an isothermal process.
ii. 104 J of work is done on certain volume of a gas. If the gas releases 125 kJ of heat, calculate the change in internal energy of the gas.
Work \( W = \int_{V_1}^{V_2} P \, dV \).
For isothermal, \( PV = nRT \Rightarrow P = \frac{nRT}{V} \).
\( W = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT [\ln V]_{V_1}^{V_2} \).
\( W = nRT \ln\left(\frac{V_2}{V_1}\right) \).
ii. Numerical:
Given: Work done on the gas \( \Delta W = -104 \) J.
Heat released \( \Delta Q = -125 \) kJ = \( -125000 \) J.
First Law: \( \Delta Q = \Delta U + \Delta W \).
\( -125000 = \Delta U - 104 \).
\( \Delta U = -125000 + 104 = -124896 \) J.
Internal energy decreases by 124896 J.
Q.30. i. Obtain the relation between surface energy and surface tension.
ii. Calculate the work done in blowing a soap bubble to a radius of 1 cm. The surface tension of soap solution is 2.5 × 10–2 N/m.
Consider a rectangular frame with a movable side. Work done to increase surface area by \( dA \) against surface tension \( T \) is \( dW = F dx = 2Tl dx \) (2 surfaces).
\( dA = 2l dx \).
\( dW = T (2l dx) = T dA \).
Surface Energy per unit area = Surface Tension (\( E = T \)).
ii. Numerical:
Soap bubble has 2 free surfaces.
\( r = 1 \) cm = \( 10^{-2} \) m. \( T = 2.5 \times 10^{-2} \) N/m.
Increase in area \( \Delta A = 2 \times (4\pi r^2 - 0) = 8\pi r^2 \).
Work \( W = T \Delta A = T (8\pi r^2) \).
\( W = (2.5 \times 10^{-2}) \times 8 \times 3.142 \times (10^{-2})^2 \).
\( W = 0.2 \times 3.142 \times 10^{-4} \)
\( W \approx 6.284 \times 10^{-5} \) J.
Q.31. Derive expressions for linear velocity at lowest position, mid-way position and the top-most position for a particle revolving in a vertical circle, if it has to just complete circular motion without string slackening at top.
Condition for just completing circle: Tension \( T_C = 0 \).
\( mg + T_C = \frac{mv_C^2}{r} \Rightarrow mg = \frac{mv_C^2}{r} \).
\( v_{top} = \sqrt{gr} \).
2. Lowest Position (A):
By conservation of energy between Top and Bottom (Height diff \( 2r \)):
\( \frac{1}{2}mv_A^2 = \frac{1}{2}mv_C^2 + mg(2r) \).
\( v_A^2 = v_C^2 + 4gr = gr + 4gr = 5gr \).
\( v_{bottom} = \sqrt{5gr} \).
3. Mid-way Position (B) (Horizontal):
By conservation of energy between Bottom and Mid (Height diff \( r \)):
\( \frac{1}{2}mv_A^2 = \frac{1}{2}mv_B^2 + mg(r) \).
\( v_B^2 = v_A^2 - 2gr = 5gr - 2gr = 3gr \).
\( v_{mid} = \sqrt{3gr} \).