Board Question Paper: March 2014
Physics
Note:
- All questions are compulsory.
- Neat diagrams must be drawn wherever necessary.
- Figures to the right indicate full marks.
- Use of only logarithmic table is allowed.
- All symbols have their usual meaning unless otherwise stated.
HSC Physics Board Papers with Solution
- Physics - March 2025 - English Medium View Answer Key
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- 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 Answer Key
- Physics - March 2014 View Answer Key
- Physics - October 2014 View
- Physics - March 2015 View
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- Physics - March 2016 View
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Section – I
When a capillary tube is dipped in a liquid that wets the glass (like water), the meniscus formed is concave. The pressure on the concave side (inside the liquid) is less than the pressure on the convex side (air).
Let \(P_o\) be the atmospheric pressure. The pressure just above the meniscus is \(P_o\). The pressure just below the meniscus is \(P_o - \frac{2T}{R}\), where \(T\) is surface tension and \(R\) is the radius of the meniscus.
Consider points A (outside capillary, surface level) and B (inside capillary, same level). Pressure at A is \(P_o\). If liquid level were same, pressure at B would be \(P_o - \frac{2T}{R}\). Since \(P_A > P_B\), liquid flows from outside to inside, causing the liquid to rise in the capillary until the hydrostatic pressure \(\rho g h\) compensates for the pressure deficiency.
The graph plots the emissive power (\(E_\lambda\)) on the Y-axis against the wavelength (\(\lambda\)) on the X-axis for different constant temperatures.
Key features:
- Energy is not uniformly distributed.
- At a given temperature, \(E_\lambda\) increases with \(\lambda\), reaches a maximum at \(\lambda_{max}\), and then decreases.
- As Temperature increases, the peak shifts to the left (shorter wavelength).
Given: Escape velocity \(v_e = 11.2\) km/s.
Formula: \(v_e = \sqrt{2gR}\) and critical velocity close to surface \(v_c = \sqrt{gR}\).
Thus, \(v_e = \sqrt{2} \times v_c \implies v_c = \frac{v_e}{\sqrt{2}}\).
Calculation:
\(v_c = \frac{11.2}{1.414} \approx 7.92\) km/s.
Ans: The critical velocity is 7.92 km/s.
Given: Length \(L = 47 \text{ cm} = 0.47 \text{ m}\), Diameter \(d = 5 \text{ cm} = 0.05 \text{ m}\), \(v = 348 \text{ m/s}\).
End correction (e): For one end, \(e = 0.3d = 0.3 \times 0.05 = 0.015 \text{ m}\).
For a pipe open at both ends, corrected length \(L_{eff} = L + 2e\).
\(L_{eff} = 0.47 + 2(0.015) = 0.47 + 0.03 = 0.5 \text{ m}\).
Formula: Fundamental frequency \(n = \frac{v}{2L_{eff}}\).
\(n = \frac{348}{2 \times 0.5} = \frac{348}{1} = 348 \text{ Hz}\).
Ans: The fundamental frequency is 348 Hz.
According to the kinetic theory of gases, the pressure \(P\) exerted by a gas is given by \(P = \frac{1}{3} \rho c^2\), where \(c\) is the RMS velocity.
\(PV = \frac{1}{3} M c^2\) (since \(\rho = M/V\)).
From ideal gas equation, \(PV = nRT\). For 1 mole, \(PV = RT\).
Therefore, \(RT = \frac{1}{3} M c^2 \implies c^2 = \frac{3RT}{M}\).
\(c = \sqrt{\frac{3RT}{M}}\).
Since \(R\) and \(M\) are constants, \(c \propto \sqrt{T}\).
Linear velocity is given by \(\vec{v} = \vec{\omega} \times \vec{r}\).
Differentiating with respect to time to get acceleration \(\vec{a}\):
\(\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(\vec{\omega} \times \vec{r})\)
Using the chain rule:
\(\vec{a} = \left(\frac{d\vec{\omega}}{dt} \times \vec{r}\right) + \left(\vec{\omega} \times \frac{d\vec{r}}{dt}\right)\)
\(\vec{a} = (\vec{\alpha} \times \vec{r}) + (\vec{\omega} \times \vec{v})\)
Here, \(\vec{a_t} = \vec{\alpha} \times \vec{r}\) is the tangential acceleration (due to change in magnitude of velocity).
\(\vec{a_r} = \vec{\omega} \times \vec{v}\) is the radial/centripetal acceleration (due to change in direction).
Total acceleration \(\vec{a} = \vec{a_t} + \vec{a_r}\).
Given: \(m = 1 \text{ kg}\), \(L = 1 \text{ m}\), \(\theta = 30^\circ\), \(g = 9.8 \text{ m/s}^2\).
This is a conical pendulum. The forces are Tension \(T\) and Weight \(mg\).
Resolving Tension:
- Vertical component: \(T \cos \theta = mg\)
- Horizontal component (Centripetal force): \(T \sin \theta = F_{cp}\)
Dividing the two equations: \(\frac{T \sin \theta}{T \cos \theta} = \frac{F_{cp}}{mg} \implies \tan \theta = \frac{F_{cp}}{mg}\).
\(F_{cp} = mg \tan \theta\).
\(F_{cp} = 1 \times 9.8 \times \tan(30^\circ) = 9.8 \times 0.5774\).
\(F_{cp} \approx 5.66 \text{ N}\).
Ans: Centripetal force is 5.66 N.
Given: Mass \(M = 50 \text{ g}\), Radius \(R = 2 \text{ cm}\), Length \(L = 12 \text{ cm}\).
Formula: Moment of inertia of a solid cylinder about a transverse axis passing through the center:
\(I = M \left( \frac{R^2}{4} + \frac{L^2}{12} \right)\).
Calculation:
\(I = 50 \left( \frac{2^2}{4} + \frac{12^2}{12} \right)\)
\(I = 50 \left( \frac{4}{4} + \frac{144}{12} \right) = 50 (1 + 12) = 50 \times 13\).
\(I = 650 \text{ g cm}^2\).
Ans: Moment of inertia is 650 g cm².
Acceleration due to gravity on the surface is \(g = \frac{GM}{R^2}\). Assuming Earth is a uniform sphere of density \(\rho\), \(M = \frac{4}{3}\pi R^3 \rho\). So, \(g = \frac{4}{3}\pi R \rho G\).
At depth \(d\), the body is at distance \((R-d)\) from the center. The effective mass attracting the body is only the inner sphere of radius \((R-d)\).
\(M' = \frac{4}{3}\pi (R-d)^3 \rho\).
Gravity at depth \(d\): \(g_d = \frac{GM'}{(R-d)^2} = \frac{G}{(R-d)^2} \cdot \frac{4}{3}\pi (R-d)^3 \rho\).
\(g_d = \frac{4}{3}\pi (R-d) \rho G\).
Dividing \(g_d\) by \(g\):
\(\frac{g_d}{g} = \frac{R-d}{R} = 1 - \frac{d}{R}\).
Therefore, \(g_d = g \left(1 - \frac{d}{R}\right)\).
Given: Side \(L = 1 \text{ m}\), Area \(A = 1 \times 1 = 1 \text{ m}^2\).
Force \(F = 4.2 \times 10^8 \text{ N}\).
Modulus of Rigidity \(\eta = 14 \times 10^{10} \text{ N/m}^2\).
Formula: \(\eta = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{F/A}{x/L}\), where \(x\) is lateral displacement.
\(x = \frac{FL}{A\eta}\).
Calculation:
\(x = \frac{4.2 \times 10^8 \times 1}{1 \times 14 \times 10^{10}}\)
\(x = \frac{4.2}{14} \times 10^{-2} = 0.3 \times 10^{-2} \text{ m}\).
\(x = 3 \times 10^{-3} \text{ m} = 3 \text{ mm}\).
Ans: Lateral displacement is 3 mm.
Expressions:
P.E. \(= \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2\).
K.E. \(= \frac{1}{2} k (A^2 - x^2) = \frac{1}{2} m \omega^2 (A^2 - x^2)\).
Graph: A plot of Energy vs Displacement shows a parabola opening up for P.E. and a parabola opening down for K.E., intersecting at \(x = \pm A/\sqrt{2}\).
Condition for K.E. = P.E.:
\(\frac{1}{2} k (A^2 - x^2) = \frac{1}{2} k x^2\)
\(A^2 - x^2 = x^2 \implies 2x^2 = A^2 \implies x = \pm \frac{A}{\sqrt{2}}\).
Given Equation: \(y = 0.05 \sin \left( 20\pi t - \frac{\pi x}{6} \right)\).
Find \(y\) at \(x = 5\) m and \(t = 0.1\) s:
Substitute values into the phase angle:
\(\phi = 20\pi(0.1) - \frac{\pi(5)}{6} = 2\pi - \frac{5\pi}{6}\).
\(\phi = \frac{12\pi - 5\pi}{6} = \frac{7\pi}{6} \text{ rad}\).
\(y = 0.05 \sin \left( \frac{7\pi}{6} \right)\).
\(\sin(210^\circ) = \sin(180^\circ + 30^\circ) = -\sin(30^\circ) = -0.5\).
\(y = 0.05 \times (-0.5) = -0.025 \text{ m}\).
Ans: The displacement is -0.025 m.
Statement: The moment of inertia of a body about any axis (\(I_o\)) is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass (\(I_c\)) and the product of the mass of the body (\(M\)) and the square of the perpendicular distance (\(h\)) between the two parallel axes.
\(I_o = I_c + Mh^2\).
Proof: [Standard derivation involving integration over mass elements \(dm\) where distance becomes \(r^2 = (x+h)^2 + y^2\)].
Given: \(d = 0.25 \text{ mm} \Rightarrow r = 0.125 \text{ mm} = 1.25 \times 10^{-4} \text{ m}\).
\(T = 0.0245 \text{ N/m}\), \(h = 4 \text{ cm} = 0.04 \text{ m}\), \(\theta = 28^\circ\).
Formula: \(h = \frac{2T \cos \theta}{r \rho g} \implies \rho = \frac{2T \cos \theta}{r h g}\).
Calculation:
\(\rho = \frac{2 \times 0.0245 \times \cos(28^\circ)}{1.25 \times 10^{-4} \times 0.04 \times 9.8}\).
\(\cos(28^\circ) \approx 0.8829\).
\(\rho = \frac{0.049 \times 0.8829}{4.9 \times 10^{-5}}\).
\(\rho = \frac{0.04326}{0.000049} = \frac{43260}{49} \approx 882.8 \text{ kg/m}^3\).
Ans: Density is approx 882.8 kg/m³.
Fund. freq \(n = \frac{1}{2L} \sqrt{\frac{T}{m}}\), where \(m\) is mass per unit length.
\(m = \frac{\text{Mass}}{\text{Length}} = \frac{\text{Vol} \times \rho}{L} = \frac{A L \rho}{L} = A \rho\).
Also, \(Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{l/L} \implies T = \frac{Y A l}{L}\).
Substituting \(T\) and \(m\) into frequency formula:
\(n = \frac{1}{2L} \sqrt{\frac{(Y A l / L)}{A \rho}} = \frac{1}{2L} \sqrt{\frac{Y l}{\rho L}}\). (Proved)
Given: Length changes from \(L\) to \(L' = L - 0.2\). Period changes by 10%.
Since \(T_{period} \propto \sqrt{L}\), if length decreases, period decreases. So \(T'_{period} = 0.9 T_{period}\).
\(\frac{T'}{T} = \sqrt{\frac{L'}{L}} \implies 0.9 = \sqrt{\frac{L-0.2}{L}}\).
Squaring both sides: \(0.81 = \frac{L-0.2}{L}\).
\(0.81L = L - 0.2 \implies 0.19L = 0.2\).
\(L = \frac{0.2}{0.19} \approx 1.053 \text{ m}\).
Ans: Original length is 1.053 m.
- The bulging of earth at the equator and flattening at the poles is due to _______.
- Young’s modulus of material of wire is ‘Y’ and strain energy per unit volume is ‘E’, then the strain is
- The wavelength range of thermal radiation is
- A pipe open at both ends resonates to \(n_1\) and closed at one end to \(n_2\). If joined to form pipe closed at one end, freq is:
- The phase difference between displacement and acceleration of a particle performing S.H.M. is _______.
- Let \(n_1\) and \(n_2\) be two slightly different frequencies... time interval between waxing and immediate next waning is:
- A metal ball cools from 64°C to 50°C in 10 min and to 42°C in next 10 min. Ratio of rates:
Section – II
Current \(I = \frac{e}{T}\), where \(T\) is time period \(T = \frac{2\pi r}{v}\).
\(I = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r}\).
Magnetic moment \(M = I \times A = I \times (\pi r^2)\).
\(M = \frac{ev}{2\pi r} \times \pi r^2 = \frac{evr}{2}\).
A photoelectric cell consists of an evacuated glass or quartz bulb containing two electrodes:
- Emitter (Cathode): A semi-cylindrical metal plate coated with photosensitive material (like Cesium) to emit electrons when light falls on it.
- Collector (Anode): A wire loop or rod placed at the center of the cathode to collect the emitted electrons.
The assembly is connected to an external circuit with a battery and galvanometer.
Given: \(\mu = 1.633\).
Formula: Brewster's Law \(\mu = \tan i_p\).
\(i_p = \tan^{-1}(1.633)\).
From log tables, \(\tan(58^\circ 30') \approx 1.632\).
Ans: Angle of incidence is approx 58° 30'.
Given: \(T_1 = 300 \text{ K}\), \(\chi_1 = 2.4 \times 10^{-5}\), \(\chi_2 = 3.6 \times 10^{-5}\).
Formula: Curie's Law \(\chi \propto \frac{1}{T} \implies \chi_1 T_1 = \chi_2 T_2\).
\(2.4 \times 10^{-5} \times 300 = 3.6 \times 10^{-5} \times T_2\).
\(T_2 = \frac{2.4 \times 300}{3.6} = \frac{2}{3} \times 300 = 200 \text{ K}\).
Ans: Temperature is 200 K.
Transmitter: A device that processes the information signal (message), modulates it with a carrier wave, amplifies it, and transmits it over the communication channel (via antenna).
Receiver: A device that captures the transmitted signal from the channel, demodulates it to extract the original information, amplifies it, and presents it in a usable form (audio/video).
Given: \(L = 1/\sqrt{\pi} \text{ m}\), \(B = 4 \times 10^{-3} \text{ T}\), \(e = 16 \text{ mV} = 16 \times 10^{-3} \text{ V}\).
Formula: \(e = \frac{1}{2} B \omega L^2 = \frac{1}{2} B (2\pi n) L^2 = B \pi n L^2\).
Substitute values:
\(16 \times 10^{-3} = 4 \times 10^{-3} \times \pi \times n \times \left( \frac{1}{\sqrt{\pi}} \right)^2\).
\(16 = 4 \pi n \left( \frac{1}{\pi} \right)\).
\(16 = 4n \implies n = 4 \text{ rps}\).
Ans: 4 revolutions per second.
Given: \(E = 2.072 \text{ eV} = 2.072 \times 1.6 \times 10^{-19} \text{ J}\).
Formula: \(E = h c \bar{\nu} \implies \bar{\nu} = \frac{E}{hc}\).
\(\bar{\nu} = \frac{2.072 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34} \times 3 \times 10^8}\).
\(\bar{\nu} = \frac{3.3152}{19.89} \times 10^7\).
\(\bar{\nu} \approx 0.1666 \times 10^7 = 1.67 \times 10^6 \text{ m}^{-1}\).
Ans: Wave number is \(1.67 \times 10^6 \text{ m}^{-1}\).
Law: The line integral of magnetic field \(\vec{B}\) around any closed path in free space is equal to \(\mu_0\) times the net current \(I\) enclosed by the path. \(\oint \vec{B} \cdot \vec{dl} = \mu_0 I\).
Toroid derivation: Consider a toroid with \(N\) turns carrying current \(I\). Choose an Amperian loop as a circle of radius \(r\) inside the toroid.
By symmetry, \(B\) is constant along the loop and parallel to \(dl\). So \(\oint B \cdot dl = B(2\pi r)\).
Current enclosed \(I_{enclosed} = N \times I\).
\(B(2\pi r) = \mu_0 N I \implies B = \frac{\mu_0 N I}{2\pi r}\).
Or \(B = \mu_0 n I\) where \(n\) is turns per unit length.
Given: \(n=2\), constants provided.
Formula: \(r_n = \frac{\epsilon_0 n^2 h^2}{\pi m e^2}\).
Substituting values: \(r_2 = \frac{8.85 \times 10^{-12} \times 2^2 \times (6.63 \times 10^{-34})^2}{\pi \times 9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^2}\).
Alternatively, knowing \(r_1 \approx 0.53\) Å, \(r_n = n^2 r_1\).
\(r_2 = 2^2 \times 0.53 \text{ Å} = 4 \times 0.53 = 2.12 \text{ Å}\).
\(2.12 \times 10^{-10} \text{ m}\).
Ans: Radius is \(2.12 \times 10^{-10}\) m.
Forward Bias: P-side connected to positive, N-side to negative. Depletion region width decreases. Holes from P and electrons from N move towards junction, cross it, and constitute current. Resistance is low.
Reverse Bias: P-side connected to negative, N-side to positive. Majority carriers move away from junction. Depletion region widens. Potential barrier increases. No current flows (except negligible leakage current). Resistance is very high.
The diagram represents a Wheatstone bridge of capacitors. There are 5 capacitors in total (4 forming the arms, 1 in the middle), all of 6 µF.
Since \(C_1/C_2 = C_3/C_4\) (all are equal), the bridge is balanced. No charge flows/stores in the central capacitor.
Equivalent Capacitance: The circuit becomes two parallel branches. Each branch has two 6 µF capacitors in series.
Series capacitance \(C_s = (6 \times 6)/(6 + 6) = 3 \text{ µF}\).
Total capacitance (parallel) \(C_{eq} = 3 + 3 = 6 \text{ µF}\).
Charge:
Total charge \(Q = C_{eq} V = 6 \text{ µF} \times 240 \text{ V} = 1440 \text{ µC}\).
Charge divides equally between the two branches (symmetry): 720 µC per branch.
In series, charge is same. So charge on \(C_1, C_2, C_3, C_4\) is 720 µC each.
Ans: Charge on outer four capacitors is 720 µC each. Charge on middle capacitor is 0.
A Fresnel biprism creates two coherent virtual sources \(S_1\) and \(S_2\) from a single slit \(S\). Interference fringes are observed on a screen. By measuring bandwidth \(X\), distance \(D\) (source to screen), and distance \(d\) (between virtual sources), wavelength is \(\lambda = X d / D\).
Given: \(\mu_g = 1.5 = 3/2\), \(\mu_w = 1.333 \approx 4/3\).
\(v_w - v_g = 2.7 \times 10^7\).
Since \(v = c/\mu\):
\(\frac{c}{\mu_w} - \frac{c}{\mu_g} = 2.7 \times 10^7\).
\(c \left( \frac{3}{4} - \frac{2}{3} \right) = 2.7 \times 10^7\).
\(c \left( \frac{9 - 8}{12} \right) = 2.7 \times 10^7\).
\(c \times \frac{1}{12} = 2.7 \times 10^7\).
\(c = 12 \times 2.7 \times 10^7 = 32.4 \times 10^7 = 3.24 \times 10^8 \text{ m/s}\).
Ans: Velocity of light in air is \(3.24 \times 10^8 \text{ m/s}\).
Principle: Mutual induction. Changing current in primary coil induces EMF in secondary coil.
Derivation: \(e_p = -N_p \frac{d\phi}{dt}\), \(e_s = -N_s \frac{d\phi}{dt}\). Ratio \(\frac{e_s}{e_p} = \frac{N_s}{N_p}\).
Setup: Ring connected at diametrically opposite points in left gap. This forms two semi-circles in parallel.
If \(R_{ring}\) is total resistance, each half is \(R_{ring}/2\). Parallel combo is \(R_{ring}/4\).
Left gap resistance \(X = R_{ring}/4\). Right gap \(R = 11 \Omega\).
Null point \(l_x = 45 \text{ cm}\), \(l_R = 55 \text{ cm}\).
Wheatstone condition: \(\frac{X}{R} = \frac{l_x}{l_R}\).
\(\frac{X}{11} = \frac{45}{55} = \frac{9}{11}\).
\(X = 9 \Omega\).
\(R_{ring}/4 = 9 \implies R_{ring} = 36 \Omega\).
Ans: Resistance of metal ring is 36 Ω.
- Intensity of electric field at a point close to and outside a charged conducting cylinder is proportional to _______.
- When a hole is produced in P-type semiconductor, there is _______.
- The outermost layer of the earth’s atmosphere is _______.
- Accuracy of potentiometer can be easily increased by _______.
- When electron in hydrogen atom jumps from second orbit to first, wavelength is \(\lambda\). From third to first, wavelength is:
- An ideal voltmeter has _______.
- The resolving power of telescope of aperture 100 cm for light of wavelength \(5.5 \times 10^{-7}\) m is _______.