Board Question Paper: October 2013
Physics | Total Marks: 70 | Time: 3 Hours
Derive an expression for the period of motion of a simple pendulum. On which factors does it depend?
A ballet dancer spins about a vertical axis at \(2.5\pi\) rad/s with his both arms outstretched. With the arms folded, the moment of inertia about the same axis of rotation changes by 25%. Calculate the new speed of rotation in r.p.m.
Part 1: Period of Simple Pendulum
Consider a simple pendulum of length \(L\) and mass \(m\). When displaced by a small angle \(\theta\), the restoring force is:
$$F = -mg \sin \theta$$
For small angles, \(\sin \theta \approx \theta = \frac{x}{L}\). Thus,
$$F = -mg \frac{x}{L}$$
Since \(F = ma\), acceleration \(a = -\frac{g}{L}x\). This represents S.H.M where \(\omega^2 = \frac{g}{L}\).
The period \(T = \frac{2\pi}{\omega}\). Substituting \(\omega = \sqrt{\frac{g}{L}}\):
$$T = 2\pi \sqrt{\frac{L}{g}}$$
Factors it depends on:
- Directly proportional to the square root of its length (\(\sqrt{L}\)).
- Inversely proportional to the square root of acceleration due to gravity (\(\sqrt{g}\)).
- Independent of the mass of the bob and the amplitude (for small angles).
Part 2: Numerical
Given:
Initial angular velocity \(\omega_1 = 2.5\pi\) rad/s
Changes in Moment of Inertia: When arms are folded, MI decreases.
\(I_2 = I_1 - 25\% \text{ of } I_1 = 0.75 I_1\)
To Find: New speed in r.p.m (\(N_2\))
Formula: By Law of Conservation of Angular Momentum: \(I_1\omega_1 = I_2\omega_2\)
$$I_1 (2.5\pi) = (0.75 I_1) \omega_2$$
$$\omega_2 = \frac{2.5\pi}{0.75} = \frac{10\pi}{3} \text{ rad/s}$$
Convert to r.p.m. We know \(\omega = \frac{2\pi N}{60}\). So, \(N = \frac{\omega \times 60}{2\pi}\).
$$N_2 = \frac{\frac{10\pi}{3} \times 60}{2\pi}$$
$$N_2 = \frac{10}{3} \times 30 = 100 \text{ r.p.m.}$$
Draw neat labelled diagrams for modes of vibration of an air column in a pipe when it is:
a. open at both ends,
b. closed at one end.
Hence derive an expression for fundamental frequency in each case.
A soap bubble of radius 12 cm is blown. Surface tension of soap solution is 30 dyne/cm. Calculate the work done in blowing the soap bubble.
Part 1: Pipe Open at Both Ends
Length of pipe \(L = \frac{\lambda}{2} \Rightarrow \lambda = 2L\).
Fundamental Frequency \(n_o = \frac{v}{\lambda} = \frac{v}{2L}\).
Part 2: Pipe Closed at One End
Length of pipe \(L = \frac{\lambda}{4} \Rightarrow \lambda = 4L\).
Fundamental Frequency \(n_c = \frac{v}{\lambda} = \frac{v}{4L}\).
Part 3: Numerical
Given:
Radius \(r = 12 \text{ cm}\)
Surface Tension \(T = 30 \text{ dyne/cm}\)
Formula: Work Done \(W = T \times \Delta A\)
Since a soap bubble has two surfaces (inner and outer): \(\Delta A = 2 \times (4\pi r^2 - 0) = 8\pi r^2\).
$$W = T \times 8\pi r^2$$
$$W = 30 \times 8 \times 3.142 \times (12)^2$$
$$W = 240 \times 3.142 \times 144$$
$$W \approx 108590 \text{ ergs}$$
Converting to Joules (\(1 \text{ erg} = 10^{-7} \text{ J}\)): \(W \approx 0.01086 \text{ J}\).
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- In a conical pendulum, a string of length 120 cm is fixed at rigid support and carries a mass of 150 g at its free end. If the mass is revolved in a horizontal circle of radius 0.2 m around a vertical axis, calculate tension in the string (g = 9.8 m/s²).
- Explain the behaviour of metal wire under increasing load.
- Derive an expression for one dimensional simple harmonic progressive wave travelling in the direction of positive X-axis. Express it in ‘two’ different forms.
- The kinetic energy of nitrogen per unit mass at 300 K is \(2.5 \times 10^6\) J/kg. Find the kinetic energy of 4 kg oxygen at 600 K. (Molecular weight of nitrogen = 28, Molecular weight of oxygen = 32).
i. Conical Pendulum Tension
Given: \(L = 120 \text{ cm} = 1.2 \text{ m}\), \(m = 150 \text{ g} = 0.15 \text{ kg}\), \(r = 0.2 \text{ m}\).
Let \(\theta\) be the semi-vertical angle. From geometry: \(\sin \theta = \frac{r}{L} = \frac{0.2}{1.2} = \frac{1}{6}\).
Using identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\cos \theta = \sqrt{1 - (1/6)^2} = \sqrt{35/36} \approx \frac{5.916}{6} \approx 0.986\).
For a conical pendulum, the vertical component of tension balances weight: \(T \cos \theta = mg\).
$$T = \frac{mg}{\cos \theta} = \frac{0.15 \times 9.8}{0.986}$$
$$T = \frac{1.47}{0.986} \approx 1.49 \text{ N}$$
ii. Behaviour of metal wire under increasing load
When a wire is subjected to gradually increasing load, the stress-strain graph shows:
- Proportionality Limit: Stress is directly proportional to strain (Hooke's Law valid).
- Elastic Limit: Upon removing load, the wire regains original shape.
- Yield Point: Beyond elastic limit, plastic deformation begins. The wire extends permanently.
- Breaking Stress: The maximum stress the wire can sustain.
- Fracture Point: The wire breaks.
iii. Progressive Wave Equation
Consider a particle at origin executing SHM: \(y = A \sin(\omega t)\).
A particle at distance \(x\) to the right lags in phase by \(\delta = \frac{2\pi x}{\lambda}\).
Equation: \(y = A \sin(\omega t - kx)\)
Different Forms:
1. Using \(\omega = \frac{2\pi}{T}\) and \(k = \frac{2\pi}{\lambda}\):
$$y = A \sin 2\pi \left(\frac{t}{T} - \frac{x}{\lambda}\right)$$
2. Using \(v = n\lambda\):
$$y = A \sin \frac{2\pi}{\lambda} (vt - x)$$
iv. Kinetic Energy of Gas
Given: \(KE_{N_2}/kg = E_1 = 2.5 \times 10^6\), \(T_1 = 300 K\), \(M_{N_2} = 28\).
Target: \(KE_{O_2}\) for mass \(m=4 \text{ kg}\), \(T_2 = 600 K\), \(M_{O_2} = 32\).
KE per unit mass \(E \propto \frac{T}{M_0}\).
$$\frac{E_2 (per kg)}{E_1} = \frac{T_2}{T_1} \times \frac{M_{N_2}}{M_{O_2}}$$
$$\frac{E_2}{2.5 \times 10^6} = \frac{600}{300} \times \frac{28}{32} = 2 \times \frac{7}{8} = 1.75$$
\(E_2 (per kg) = 1.75 \times 2.5 \times 10^6 = 4.375 \times 10^6 \text{ J/kg}\).
Total KE for 4 kg = \(4 \times 4.375 \times 10^6 = 17.5 \times 10^6 \text{ J}\).
- A racing car completes 5 rounds of a circular track in 2 minutes. Find the radius of the track if the car has uniform centripetal acceleration of \(\pi^2\) m/s².
- A body weighs 4.0 kg-wt on the surface of the Earth. What will be its weight on the surface of a planet whose mass is \(\frac{1}{8}\)th of the mass of the Earth and radius half \(\left(\frac{1}{2}\right)\) of that of the Earth?
- Define radius of gyration. Explain its physical significance.
- A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate: a. Angular frequency, b. frequency of vibration.
- Show that the surface tension of a liquid is numerically equal to the surface energy per unit area.
- Explain black body radiation spectrum in terms of wavelength.
- ‘g’ is the acceleration due to gravity on the surface of the Earth and ‘R’ is the radius of the Earth. Show that acceleration due to gravity at height ‘h’ above the surface of the Earth is \(g_h = g \left(\frac{R}{R+h}\right)^2\).
- In Melde’s experiment, the number of loops on a string changes from 7 to 5 by the addition of 0.015 kgwt. Find the initial tension applied to the string.
i. Radius of Track
\(\omega = \frac{2\pi N}{t} = \frac{2\pi(5)}{120} = \frac{\pi}{12} \text{ rad/s}\).
\(a_c = r\omega^2 = \pi^2\).
\(r \left(\frac{\pi}{12}\right)^2 = \pi^2 \Rightarrow r \frac{\pi^2}{144} = \pi^2\).
\(r = 144 \text{ m}\).
ii. Weight on Planet
\(g = \frac{GM}{R^2}\). \(g' = \frac{G(M/8)}{(R/2)^2} = \frac{G(M/8)}{R^2/4} = \frac{1}{2} \frac{GM}{R^2} = \frac{g}{2}\).
Weight \(W' = m g' = m (g/2) = \frac{W_{earth}}{2} = \frac{4.0}{2} = 2.0 \text{ kg-wt}\).
iii. Radius of Gyration (K)
Definition: The distance from the axis of rotation to a point where the entire mass of the body is supposed to be concentrated, such that the moment of inertia remains the same.
Significance: It measures the distribution of mass about the axis of rotation. \(I = MK^2\).
iv. Spring Oscillation
\(m=1\), \(k=16\).
a. Angular frequency \(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{16}{1}} = 4 \text{ rad/s}\).
b. Frequency \(n = \frac{\omega}{2\pi} = \frac{4}{2\pi} = \frac{2}{\pi} \text{ Hz} \approx 0.637 \text{ Hz}\).
v. Surface Tension and Energy
Consider a rectangular frame with a movable wire. Work done to pull wire by \(dx\) against surface tension \(T\) (for 2 surfaces) is \(dW = T(2L)dx\). Increase in area \(dA = 2Ldx\).
Surface Energy \(E = dW = T dA\).
Therefore, \(T = \frac{dW}{dA} = \text{Surface Energy per unit area}\).
vi. Black Body Spectrum
1. Energy is not distributed uniformly.
2. At a given temp, there is a wavelength \(\lambda_{max}\) with maximum energy.
3. As temperature increases, \(\lambda_{max}\) shifts to shorter wavelengths (Wien's Law).
4. Area under curve represents total energy \(\propto T^4\).
vii. Gravity at Altitude
\(g = \frac{GM}{R^2}\).
At height \(h\), distance is \(R+h\): \(g_h = \frac{GM}{(R+h)^2}\).
Dividing both equations: \(\frac{g_h}{g} = \frac{R^2}{(R+h)^2}\).
Hence, \(g_h = g \left(\frac{R}{R+h}\right)^2\).
viii. Melde's Experiment
Law: \(Tp^2 = \text{constant}\).
\(T_1 (7)^2 = T_2 (5)^2\). Given \(T_2 = T_1 + 0.015\).
\(49 T_1 = 25 (T_1 + 0.015)\).
\(49 T_1 = 25 T_1 + 0.375\).
\(24 T_1 = 0.375 \Rightarrow T_1 = 0.015625 \text{ kg-wt}\).
i. A planet is revolving around a star in a circular orbit of radius R with a period T. If the gravitational force between the planet and the star is proportional to \(R^{-5/2}\), then:
(A) \(T^2 \propto R^{5/2}\) (B) \(T^2 \propto R^{-7/2}\) (C) \(T^2 \propto R^{3/2}\) (D) \(T^2 \propto R^4\)
Note: Calculating theoretically for \(F \propto R^{-5/2}\) yields \(T^2 \propto R^{7/2}\). If \(F \propto R^{-3/2}\), then \(T^2 \propto R^{5/2}\). Assuming option typo or specific context.
ii. If ‘L’ is the angular momentum and ‘I’ is the moment of inertia of a rotating body, then \(\frac{L^2}{2I}\) represents its:
(A) rotational P.E. (B) total energy (C) rotational K.E. (D) translational K.E.
iii. A particle executing linear S.H.M. has velocities \(v_1\) and \(v_2\) at distances \(x_1\) and \(x_2\) respectively from the mean position. The angular velocity of the particle is:
(A) \(\sqrt{\frac{x_1^2 - x_2^2}{v_2^2 - v_1^2}}\) (B) \(\sqrt{\frac{v_2^2 - v_1^2}{x_1^2 - x_2^2}}\)
iv. A metal rod having coefficient of linear expansion (\(\alpha\)) and Young’s modulus (Y) is heated to raise the temperature by \(\Delta \theta\). The stress exerted by the rod is:
(A) \(\frac{Y\alpha}{\Delta \theta}\) (B) \(\frac{Y\Delta \theta}{\alpha}\) (C) \(Y\alpha\Delta \theta\) (D) \(\frac{\alpha\Delta \theta}{Y}\)
v. A big drop of radius R is formed from 1000 droplets of water. The radius of a droplet will be:
(A) 10 R (B) R/10 (C) R/100 (D) R/1000
vi. Apparent frequency of the sound heard by a listener is less than the actual frequency of sound emitted by source. In this case:
(C) listener moves away from the source.
vii. The substance which allows heat radiations to pass through is:
(A) iron (B) water vapour (C) wood (D) dry air
i. (A) or (C) [Context dependent]
Calculation: \(F = \frac{mv^2}{R} \propto R^{-5/2} \Rightarrow v^2 \propto R^{-1.5} \Rightarrow v \propto R^{-0.75}\).
\(T = \frac{2\pi R}{v} \propto \frac{R}{R^{-0.75}} = R^{1.75} = R^{7/2}\).
(Most likely intended answer is \(R^{7/2}\) or a typo in question exponent).
ii. (C) rotational K.E.
iii. (B) \(\sqrt{\frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}}\) (matches form B adjusted for signs)
iv. (C) \(Y\alpha\Delta \theta\)
v. (B) R/10
vi. (C) listener moves away from the source.
vii. (D) dry air
Obtain an expression for the induced e.m.f. in a coil rotating with uniform angular velocity in uniform magnetic field. Plot a graph of variation of induced e.m.f. against phase \((\theta = \omega t)\) over one cycle.
The energy density at a point in a medium of dielectric constant 6 is \(26.55 \times 10^6 \text{ J/m}^3\). Calculate electric field intensity at that point. (\(\epsilon_0 = 8.85 \times 10^{-12}\) SI units).
Part 1: Induced EMF derivation
Consider a coil of \(N\) turns and area \(A\) rotating in magnetic field \(B\) with angular velocity \(\omega\).
Magnetic flux \(\phi = N \vec{B} \cdot \vec{A} = NBA \cos(\omega t)\).
Induced EMF \(e = -\frac{d\phi}{dt} = -NBA \frac{d}{dt}(\cos \omega t)\).
$$e = NBA \omega \sin(\omega t)$$
Peak EMF \(e_0 = NBA\omega\). Thus, \(e = e_0 \sin(\omega t)\).
Part 2: Numerical
Given: \(K = 6\), \(u = 26.55 \times 10^6\), \(\epsilon_0 = 8.85 \times 10^{-12}\).
Formula: Energy density \(u = \frac{1}{2} K \epsilon_0 E^2\).
$$E^2 = \frac{2u}{K \epsilon_0} = \frac{2 \times 26.55 \times 10^6}{6 \times 8.85 \times 10^{-12}}$$
$$E^2 = \frac{53.1 \times 10^6}{53.1 \times 10^{-12}} = 10^{18}$$
$$E = \sqrt{10^{18}} = 10^9 \text{ V/m}$$
Write notes on –
a. Nuclear fission b. Nuclear fusion
A galvanometer has a resistance of 16 \(\Omega\). It shows full scale deflection, when a current of 20 mA is passed through it. The only shunt resistance available is 0.06 \(\Omega\) which is not appropriate to convert a galvanometer into an ammeter. How much resistance should be connected in series with the coil of galvanometer, so that the range of ammeter is 8 A?
Part 1: Notes
- Nuclear Fission: The process of splitting a heavy nucleus (like Uranium-235) into two lighter nuclei of comparable masses, releasing a large amount of energy and neutrons. Used in nuclear reactors.
- Nuclear Fusion: The process where light nuclei (like Hydrogen) combine to form a heavier nucleus (like Helium), releasing immense energy. This is the source of energy in stars. Requires very high temperature.
Part 2: Numerical
Given: \(G = 16 \Omega\), \(I_g = 20 \text{ mA} = 0.02 \text{ A}\), Range \(I = 8 \text{ A}\). Shunt \(S = 0.06 \Omega\).
Logic: We need to connect a resistance \(R_x\) in series with the galvanometer coil such that this combination \((G + R_x)\), when shunted by \(S\), gives the required range.
Voltage across Shunt = Voltage across (Galvanometer + Series R)
$$(I - I_g) S = I_g (G + R_x)$$
$$(8 - 0.02) \times 0.06 = 0.02 (16 + R_x)$$
$$7.98 \times 0.06 = 0.02 (16 + R_x)$$
$$0.4788 = 0.32 + 0.02 R_x$$
$$0.02 R_x = 0.1588$$
$$R_x = 7.94 \Omega$$
- Draw a well labelled diagram of photoelectric cell. Explain the observations made by Hertz and Lenard about the phenomenon of photoelectric emission.
- Explain the working of transistor as a switch.
- The refractive indices of water for red and violet colours are 1.325 and 1.334 respectively. Find the difference between the velocities of rays for these two colours in water. (c = \(3 \times 10^8\) m/s)
- In Young’s experiment, the ratio of intensity at the maxima and minima in an interference pattern is 36 : 9. What will be the ratio of the intensities of two interfering waves?
i. Photoelectric Effect
Observations:
1. Emission is instantaneous.
2. There exists a minimum frequency (Threshold frequency) below which no emission occurs.
3. Photocurrent is directly proportional to intensity of incident light.
ii. Transistor as Switch
A transistor works as a switch in Cut-off and Saturation regions.
OFF State (Cut-off): Input voltage is low, base current is zero, collector current is zero. Switch is open.
ON State (Saturation): Input voltage is high, collector current is maximum (\(V_{CE} \approx 0\)). Switch is closed.
iii. Velocity Difference
\(v = c / \mu\).
\(v_R = \frac{3 \times 10^8}{1.325} \approx 2.2641 \times 10^8 \text{ m/s}\).
\(v_V = \frac{3 \times 10^8}{1.334} \approx 2.2488 \times 10^8 \text{ m/s}\).
Difference \(\Delta v = v_R - v_V \approx 0.0153 \times 10^8 = 1.53 \times 10^6 \text{ m/s}\).
iv. Intensity Ratio
\(\frac{I_{max}}{I_{min}} = \frac{36}{9} = \frac{4}{1}\).
\(\frac{A_{max}}{A_{min}} = \sqrt{\frac{4}{1}} = \frac{2}{1}\).
\(\frac{a_1+a_2}{a_1-a_2} = 2 \Rightarrow a_1+a_2 = 2a_1 - 2a_2 \Rightarrow a_1 = 3a_2\).
Ratio of Intensities \(\frac{I_1}{I_2} = \left(\frac{a_1}{a_2}\right)^2 = (3)^2 = 9:1\).
- Explain the principle of potentiometer.
- What is − a. magnetization and b. magnetic intensity?
- What do you mean by polar molecules and non-polar molecules? Give ‘one’ example each.
- The minimum angular separation between two stars is \(4 \times 10^{-6}\) rad, if telescope is used to observe them with an objective of aperture 16 cm. Find the wavelength of light used.
- Explain the need for modulation related to the size of antenna (aerial).
- Four resistances 4 \(\Omega\), 8 \(\Omega\), X \(\Omega\) and 6 \(\Omega\) are connected in a series so as to form Wheatstone’s network. If the network is balanced, find the value of ‘X’.
- The magnetic susceptibility of annealed iron at saturation is 4224. Find the permeability of annealed iron at saturation. (\(\mu_0 = 4\pi \times 10^{-7}\) SI unit)
i. Principle of Potentiometer: The potential drop across any portion of a wire of uniform cross-section and composition is directly proportional to the length of that portion, provided a constant current flows through it. \(V \propto L\).
ii. Definitions:
a. Magnetization (M): Net magnetic dipole moment per unit volume.
b. Magnetic Intensity (H): The capability of a magnetic field to magnetize a material. \(B = \mu H\).
iii. Polar vs Non-polar:
Polar: Center of positive charge does not coincide with center of negative charge. Have permanent dipole moment. Ex: \(H_2O\), \(HCl\).
Non-polar: Centers coincide. No permanent dipole moment. Ex: \(H_2\), \(O_2\).
iv. Telescope Resolution
\(d\theta = \frac{1.22 \lambda}{D}\).
Given \(d\theta = 4 \times 10^{-6}\), \(D = 16 \text{ cm} = 0.16 \text{ m}\).
\(\lambda = \frac{d\theta \times D}{1.22} = \frac{4 \times 10^{-6} \times 0.16}{1.22} \approx 5.24 \times 10^{-7} \text{ m} = 5240 \mathring{A}\).
v. Antenna Size: For effective transmission, antenna height must be at least \(\lambda/4\). For audio frequencies (low f, large \(\lambda\)), the antenna size becomes impractical (kilometers). Modulation increases frequency, reducing \(\lambda\) and making antenna size practical.
vi. Wheatstone's Network
Assuming cyclic order P, Q, S, R (or standard arms):
\(\frac{4}{8} = \frac{X}{6}\) (or similar arrangement).
If \(4/8 = X/6 \Rightarrow 1/2 = X/6 \Rightarrow X=3 \Omega\).
(Note: Order matters. If 4, 8, X, 6 is cyclic \(P, Q, R, S\), then \(P/Q = S/R \Rightarrow 4/8 = 6/X \Rightarrow 0.5 = 6/X \Rightarrow X=12\)). Standard assumption often leads to \(X=3\) or \(X=12\). Assuming standard bridge balance \(R_1/R_2 = R_3/R_4 \Rightarrow 4/8 = X/6 \rightarrow X=3\Omega\).
vii. Permeability
\(\chi = 4224\).
\(\mu_r = 1 + \chi = 4225\).
\(\mu = \mu_0 \mu_r = 4\pi \times 10^{-7} \times 4225\).
\(\mu \approx 5.3 \times 10^{-3} \text{ H/m}\).
i. A ray of light passes from a vacuum to a medium of refractive index (\(\mu\)). The angle of incidence is found to be twice the angle of refraction. The angle of incidence is:
(C) \(2 \cos^{-1}\left(\frac{\mu}{2}\right)\)
ii. The fringes produced in diffraction pattern are of:
(B) unequal width with varying intensity
iii. If ‘R’ is the radius of dees and ‘B’ be the magnetic field of induction in which positive charges (q) of mass (m) escape from the cyclotron, then its maximum speed (vmax) is:
(C) \(\frac{qBR}{m}\)
iv. The number of photoelectrons emitted:
(D) varies directly with intensity
v. The width of depletion region of p-n junction diode is:
(C) 50 nm to 500 nm (Typically \(10^{-7}\)m to \(10^{-6}\)m range)
vi. A device that converts one form of energy into another form is termed as:
(A) transducer
vii. A transformer converts 240 V AC to 60 V AC. The secondary has 75 turns. The number of turns in primary are:
(D) 300
i. (C) Hint: \(i=2r\). \(\mu = \frac{\sin 2r}{\sin r} = 2\cos r\). \(r = \cos^{-1}(\mu/2)\).
ii. (B) Diffraction fringes have central max width \(2\lambda D/a\), others \(\lambda D/a\).
iii. (C) \(qvB = mv^2/R \Rightarrow v = qBR/m\).
iv. (D) Intensity determines photon count, hence electron count.
v. (C) Standard width is approx \(0.1\) to \(1\) micron.
vi. (A) Transducer.
vii. (D) \(\frac{N_p}{N_s} = \frac{V_p}{V_s} \Rightarrow \frac{N_p}{75} = \frac{240}{60} = 4 \Rightarrow N_p = 300\).