Maharashtra Board HSC Physics 2016 Solved Paper
Subject: Physics (54) | Date: 2016 | Max Marks: 70
(i) In U. C. M. (Uniform Circular Motion), prove the relation \(\vec{v} = \vec{w} \times \vec{r}\), where symbols have their usual meanings.
Consider a particle performing U.C.M. along a circle of radius \(r\) with angular velocity \(\vec{\omega}\).
The position vector \(\vec{r}\) makes an angle \(\theta\) with the X-axis. In vector form:
$$ \vec{r} = r \cos \theta \hat{i} + r \sin \theta \hat{j} $$
Differentiating with respect to time \(t\):
$$ \vec{v} = \frac{d\vec{r}}{dt} = r(-\sin \theta)\frac{d\theta}{dt}\hat{i} + r(\cos \theta)\frac{d\theta}{dt}\hat{j} $$
Since \(\frac{d\theta}{dt} = \omega\):
$$ \vec{v} = \omega(-r \sin \theta \hat{i} + r \cos \theta \hat{j}) \quad \text{...(1)} $$
Now, consider the cross product \(\vec{\omega} \times \vec{r}\). In U.C.M, \(\vec{\omega}\) is perpendicular to the plane (along Z-axis), so \(\vec{\omega} = \omega \hat{k}\).
$$ \vec{\omega} \times \vec{r} = (\omega \hat{k}) \times (r \cos \theta \hat{i} + r \sin \theta \hat{j}) $$
$$ = \omega r \cos \theta (\hat{k} \times \hat{i}) + \omega r \sin \theta (\hat{k} \times \hat{j}) $$
Using vector rules \(\hat{k} \times \hat{i} = \hat{j}\) and \(\hat{k} \times \hat{j} = -\hat{i}\):
$$ \vec{\omega} \times \vec{r} = \omega r \cos \theta \hat{j} - \omega r \sin \theta \hat{i} $$
Rearranging:
$$ \vec{\omega} \times \vec{r} = \omega (-r \sin \theta \hat{i} + r \cos \theta \hat{j}) \quad \text{...(2)} $$
From (1) and (2), we get: \(\vec{v} = \vec{\omega} \times \vec{r}\).
(ii) Derive an expression for critical velocity of a satellite revolving around the earth in a circular orbit.
Consider a satellite of mass \(m\) revolving around the Earth (mass \(M\)) in a circular orbit of radius \(r\) (where \(r = R + h\)).
The necessary centripetal force is provided by the gravitational force of attraction between the Earth and the satellite.
$$ \text{Centripetal Force} = \text{Gravitational Force} $$
$$ \frac{mv_c^2}{r} = \frac{GMm}{r^2} $$
Where \(v_c\) is the critical velocity.
Canceling \(m\) and one \(r\) from both sides:
$$ v_c^2 = \frac{GM}{r} $$
Taking the square root:
$$ v_c = \sqrt{\frac{GM}{r}} $$
Since \(r = R + h\), the expression becomes:
$$ v_c = \sqrt{\frac{GM}{R+h}} $$
(iii) Obtain an expression for total kinetic energy of a rolling body in the form \(\frac{1}{2}MV^2 \left[1 + \frac{K^2}{R^2}\right]\).
The total kinetic energy (\(E\)) of a rolling body is the sum of its translational kinetic energy (\(E_T\)) and rotational kinetic energy (\(E_R\)).
$$ E = E_T + E_R $$
$$ E = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2 $$
Where \(M\) is mass, \(V\) is linear velocity, \(I\) is moment of inertia, and \(\omega\) is angular velocity.
We know that \(V = R\omega\) \(\Rightarrow \omega = \frac{V}{R}\) and \(I = MK^2\) (where \(K\) is the radius of gyration).
Substituting these values:
$$ E = \frac{1}{2}MV^2 + \frac{1}{2}(MK^2)\left(\frac{V}{R}\right)^2 $$
$$ E = \frac{1}{2}MV^2 + \frac{1}{2}MK^2 \frac{V^2}{R^2} $$
Taking \(\frac{1}{2}MV^2\) common:
$$ E = \frac{1}{2}MV^2 \left[ 1 + \frac{K^2}{R^2} \right] $$
(iv) Define ‘emissive power’ and ‘coefficient of emission of a body’.
Emissive Power (E): The quantity of radiant energy emitted per unit area per unit time by a body at a given temperature is called its emissive power.
Coefficient of Emission (e): The ratio of the emissive power of a body at a given temperature to the emissive power of a perfectly black body at the same temperature is called the coefficient of emission (or emissivity).
$$ e = \frac{E}{E_b} $$
(v) A coin kept at a distance of 5cm from the centre of a turntable of radius 1.5m just begins to slip when the turntable rotates at a speed of 90 r.p.m. Calculate the coefficient of static friction between the coin and the turntable. [ g = 9.8 m/s² ].
Given:
Radius of path \(r = 5 \text{ cm} = 0.05 \text{ m}\)
Angular speed \(N = 90 \text{ r.p.m.}\)
\(\omega = 2\pi N / 60 = 2\pi (90)/60 = 3\pi \text{ rad/s}\)
\(g = 9.8 \text{ m/s}^2\)
To find: Coefficient of static friction \(\mu\).
Formula: The necessary centripetal force is provided by the frictional force.
$$ \mu mg = mr\omega^2 $$
$$ \mu g = r\omega^2 $$
$$ \mu = \frac{r\omega^2}{g} $$
Calculation:
$$ \mu = \frac{0.05 \times (3\pi)^2}{9.8} $$
$$ \mu = \frac{0.05 \times 9 \times 9.87}{9.8} $$ (Using \(\pi^2 \approx 9.87\))
$$ \mu \approx \frac{0.45 \times 9.87}{9.8} \approx 0.453 $$
Answer: The coefficient of static friction is approximately 0.45.
(vi) The fundamental frequency of an air column in a pipe closed at one end is in unison with the third overtone of an open pipe. Calculate the ratio of lengths of their air columns.
Given: Fundamental frequency of closed pipe (\(n_c\)) = Frequency of 3rd overtone of open pipe (\(n'_o\)).
Let \(L_c\) be length of closed pipe and \(L_o\) be length of open pipe.
Formulas:
Fundamental frequency of closed pipe: \( n_c = \frac{v}{4L_c} \)
Frequency of open pipe (p-th overtone): \( n = (p+1) \frac{v}{2L_o} \)
For 3rd overtone of open pipe (\(p=3\)):
$$ n'_o = (3+1)\frac{v}{2L_o} = 4\frac{v}{2L_o} = \frac{2v}{L_o} $$
Calculation:
$$ n_c = n'_o $$
$$ \frac{v}{4L_c} = \frac{2v}{L_o} $$
$$ \frac{1}{4L_c} = \frac{2}{L_o} $$
$$ \frac{L_c}{L_o} = \frac{1}{4 \times 2} = \frac{1}{8} $$
Answer: The ratio of their lengths \(L_c : L_o\) is 1 : 8.
(vii) A particle performing linear S.H.M. has a period of 6.28 seconds and a pathlength of 20cm. What is the velocity when its displacement is 6 cm from mean position?
Given:
Period \(T = 6.28 \text{ s} \approx 2\pi \text{ s}\)
Path length = 20 cm \(\Rightarrow\) Amplitude \(A = 20/2 = 10 \text{ cm}\)
Displacement \(x = 6 \text{ cm}\)
To find: Velocity \(v\).
Formula:
$$ \omega = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 \text{ rad/s} $$
$$ v = \omega \sqrt{A^2 - x^2} $$
Calculation:
$$ v = 1 \times \sqrt{10^2 - 6^2} $$
$$ v = \sqrt{100 - 36} = \sqrt{64} $$
$$ v = 8 \text{ cm/s} $$
Answer: The velocity is 8 cm/s.
(viii) The energy of the free surface of a liquid drop is \(5\pi\) times the surface tension of the liquid. Find the diameter of the drop in C.G.S. system.
Given: Surface Energy \(E = 5\pi T\).
Formula: Surface Energy \(E = T \times \text{Surface Area}\).
For a liquid drop, Surface Area \(A = 4\pi r^2\).
$$ E = T(4\pi r^2) $$
Calculation:
Equating the given energy:
$$ T(4\pi r^2) = 5\pi T $$
$$ 4r^2 = 5 $$
$$ r^2 = \frac{5}{4} $$
$$ r = \frac{\sqrt{5}}{2} \text{ cm} $$
Diameter \(d = 2r = 2 \times \frac{\sqrt{5}}{2} = \sqrt{5} \text{ cm}\).
Since \(\sqrt{5} \approx 2.236\).
Answer: The diameter is \(\sqrt{5}\) cm (or approx 2.236 cm).
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(i) A particle rotates in U.C.M. with tangential velocity ‘v’ along a horizontal circle of diameter ‘D’. Total angular displacement of the particle in time ‘t’ is ............
Explanation: Radius \(r = D/2\). Angular velocity \(\omega = v/r = v/(D/2) = 2v/D\). Angular displacement \(\theta = \omega t = \frac{2vt}{D}\).
(ii) Two springs of force constants \(K_1\) and \(K_2\) \((K_1 > K_2)\) are stretched by same force. If \(W_1\) and \(W_2\) be the work done stretching the springs then ............
Explanation: Work done \(W = \frac{F^2}{2K}\). Since \(F\) is same, \(W \propto \frac{1}{K}\). Given \(K_1 > K_2\), therefore \(W_1 < W_2\).
(iii) A and B are two steel wires and the radius of A is twice that of B. If they are stretched by the same load, then the stress on B is ............
Explanation: Stress \(S = \frac{F}{A} = \frac{F}{\pi r^2}\). \(r_A = 2r_B\).
\(S_A \propto \frac{1}{(2r_B)^2} = \frac{1}{4r_B^2}\) and \(S_B \propto \frac{1}{r_B^2}\). Thus \(S_B = 4S_A\).
(iv) If sound waves are reflected from surface of denser medium, there is phase change of ............
(v) A sonometer wire vibrates with frequency \(n_1\) in air under suitable load of specific gravity '\(\sigma\)'. When the load is immersed in water, the frequency of vibration of wire \(n_2\) will be ............
Explanation: Frequency \(n \propto \sqrt{T}\). \(T_{air} = V\sigma g\). \(T_{water} = V\sigma g - V(1)g = Vg(\sigma-1)\). Ratio \(n_2/n_1 = \sqrt{T_{water}/T_{air}} = \sqrt{(\sigma-1)/\sigma}\).
(vi) For polyatomic molecules having ‘\(f\)’ vibrational modes, the ratio of two specific heats, \(\frac{C_P}{C_V}\) is ............
Note: Standard formula for polyatomic gas with f vibrational modes is \(\gamma = \frac{4+f}{3+f}\).
(vii) A body of moment of inertia 5 kgm² rotating with an angular velocity 6 rad/s has the same kinetic energy as a mass of 20 kg moving with a velocity of ............
Explanation: \(E_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(5)(6^2) = 90\) J.
\(E_{trans} = \frac{1}{2}mv^2 = 90 \Rightarrow \frac{1}{2}(20)v^2 = 90 \Rightarrow 10v^2 = 90 \Rightarrow v = 3\) m/s.
Define linear S.H.M. Show that S.H.M. is a projection of U.C.M. on any diameter.
Definition: Linear Simple Harmonic Motion (S.H.M) is defined as the linear periodic motion of a body in which the restoring force (or acceleration) is always directed towards the mean position and its magnitude is directly proportional to the displacement from the mean position.
Proof (Projection of U.C.M.):
- Consider a particle \(P\) moving along a circle of radius \(A\) with constant angular velocity \(\omega\) (U.C.M).
- Let \(M\) be the projection (foot of the perpendicular) of \(P\) on the diameter \(YOY'\) (Y-axis).
- At \(t=0\), let the particle be at \(P_0\) making angle \(\alpha\) with X-axis. At time \(t\), it is at \(P\) such that \(\angle P_0OP = \omega t\). Total angle \(\theta = \omega t + \alpha\).
- The displacement of projection \(M\) from origin \(O\) is \(y\).
- In \(\Delta OMP\), \(\sin(\omega t + \alpha) = \frac{OM}{OP} = \frac{y}{A}\).
- Therefore, \(y = A \sin(\omega t + \alpha)\).
- Differentiating twice w.r.t \(t\), we get acceleration \(a = -\omega^2 y\).
- This equation \(a = -\omega^2 y\) satisfies the condition of S.H.M. Thus, the projection of U.C.M. on any diameter is S.H.M.
A metal sphere cools at the rate of 4°C / min. when its temperature is 50°C. Find its rate of cooling at 45°C if the temperature of surroundings is 25°C.
Given:
Surrounding temp \(\theta_0 = 25^\circ C\).
Case 1: \(\frac{d\theta_1}{dt} = 4^\circ C/\text{min}\) at \(\theta_1 = 50^\circ C\).
Case 2: Find \(\frac{d\theta_2}{dt}\) at \(\theta_2 = 45^\circ C\).
Formula: Newton's Law of Cooling: \(\frac{d\theta}{dt} = K(\theta - \theta_0)\).
Calculation:
From Case 1:
$$ 4 = K(50 - 25) $$
$$ 4 = K(25) \Rightarrow K = \frac{4}{25} $$
For Case 2:
$$ \frac{d\theta_2}{dt} = K(45 - 25) $$
$$ \frac{d\theta_2}{dt} = \frac{4}{25} (20) $$
$$ \frac{d\theta_2}{dt} = \frac{4 \times 4}{5} = \frac{16}{5} = 3.2 $$
Answer: The rate of cooling at 45°C is 3.2°C / min.
Explain analytically how the stationary waves are formed. Hence show that the distance between node and adjacent antinode is \(\frac{\lambda}{4}\).
Consider two identical progressive waves traveling along the same path in opposite directions.
$$ y_1 = A \sin 2\pi(nt - x/\lambda) $$
$$ y_2 = A \sin 2\pi(nt + x/\lambda) $$
According to the principle of superposition, resultant displacement \(y = y_1 + y_2\).
$$ y = A [\sin 2\pi(nt - x/\lambda) + \sin 2\pi(nt + x/\lambda)] $$
Using \(\sin C + \sin D = 2 \sin((C+D)/2) \cos((C-D)/2)\):
$$ y = 2A \sin(2\pi nt) \cos(2\pi x/\lambda) $$
$$ y = R \sin(2\pi nt) $$ where Resultant Amplitude \(R = 2A \cos(2\pi x/\lambda)\).
Distance between Node and Antinode:
1. Node: \(R=0 \Rightarrow \cos(2\pi x/\lambda) = 0 \Rightarrow 2\pi x/\lambda = \pi/2 \Rightarrow x = \lambda/4\).
2. Antinode: \(R=\pm 2A \Rightarrow \cos(2\pi x/\lambda) = \pm 1 \Rightarrow 2\pi x/\lambda = 0 \Rightarrow x = 0\).
Distance = \(x_{node} - x_{antinode} = \lambda/4 - 0 = \lambda/4\).
A set of 48 tuning forks is arranged in a series of descending frequencies such that each fork gives 4 beats per second with preceding one. The frequency of first fork is 1.5 times the frequency of the last fork, find the frequency of the first and 42nd tuning fork.
Given:
Total forks \(n = 48\).
Descending order, Beats \(d = 4\). (Since descending, \(n_{i+1} = n_i - 4\)).
Relation: \(n_1 = 1.5 n_{48}\).
Calculation:
Using AP formula for the last term: \(n_{last} = n_1 + (N-1)(-d)\)
$$ n_{48} = n_1 + (48 - 1)(-4) $$
$$ n_{48} = n_1 - 188 $$
Substitute \(n_{48} = \frac{n_1}{1.5} = \frac{2n_1}{3}\):
$$ \frac{2n_1}{3} = n_1 - 188 $$
$$ 188 = n_1 - \frac{2n_1}{3} = \frac{n_1}{3} $$
$$ n_1 = 188 \times 3 = 564 \text{ Hz} $$
Now, find frequency of 42nd fork (\(n_{42}\)):
$$ n_{42} = n_1 + (42 - 1)(-4) $$
$$ n_{42} = 564 + (41)(-4) $$
$$ n_{42} = 564 - 164 $$
$$ n_{42} = 400 \text{ Hz} $$
Answer: Frequency of 1st fork is 564 Hz and 42nd fork is 400 Hz.
(i) What is the decrease in weight of a body of mass 600 kg when it is taken in a mine of depth 5000 m? [ Radius of earth = 6400 km, g = 9.8 m/s² ]
Given:
Mass \(m = 600 \text{ kg}\).
Depth \(d = 5000 \text{ m} = 5 \text{ km}\).
Radius \(R = 6400 \text{ km} = 6400000 \text{ m}\).
Formula:
Gravity at depth \(d\): \(g_d = g(1 - \frac{d}{R})\).
Weight at surface \(W = mg\).
Weight at depth \(W_d = m g_d = mg(1 - \frac{d}{R}) = W - W\frac{d}{R}\).
Decrease in weight \(\Delta W = W - W_d = mg \frac{d}{R}\).
Calculation:
$$ \Delta W = 600 \times 9.8 \times \frac{5000}{6400000} $$
$$ \Delta W = 5880 \times \frac{5}{6400} $$
$$ \Delta W = \frac{29400}{6400} = \frac{294}{64} $$
$$ \Delta W \approx 4.59 \text{ N} $$
Answer: Decrease in weight is 4.59 N.
(ii) State and prove theorem of parallel axes about moment of inertia.
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 distance (\(h\)) between the two parallel axes.
$$ I_o = I_c + Mh^2 $$
Proof:
Let an axis pass through \(O\). Let parallel axis pass through Center of Mass \(C\). Distance \(OC = h\). Take element of mass \(dm\) at point \(P\).
\(I_o = \int OP^2 dm\) and \(I_c = \int CP^2 dm\).
By geometry (drawing perpendicular from P to extended OC at D), \(OP^2 = CP^2 + h^2 + 2h(CD)\).
Integrating: \(\int OP^2 dm = \int CP^2 dm + h^2 \int dm + 2h \int CD dm\).
Since \(C\) is center of mass, \(\int CD dm = 0\).
Thus, \(I_o = I_c + M h^2\).
(iii) Derive Laplace’s law for spherical membrane of bubble due to surface tension.
Consider a soap bubble of radius \(r\) and surface tension \(T\). Let excess pressure inside be \(P = P_i - P_o\).
Work done to increase radius by \(dr\) against excess pressure:
$$ dW = \text{Force} \times \text{distance} = (P \times 4\pi r^2) \times dr $$
Increase in surface area (bubble has 2 surfaces):
$$ dA = 2 \times [4\pi(r+dr)^2 - 4\pi r^2] \approx 2 \times 8\pi r dr = 16\pi r dr $$
Work done as Surface Energy:
$$ dW = T \times dA = T(16\pi r dr) $$
Equating both works:
$$ P(4\pi r^2) dr = T(16\pi r dr) $$
$$ Pr = 4T $$
$$ P = \frac{4T}{r} $$
(iv) A steel wire having cross sectional area \(1.5 \text{ mm}^2\) when stretched by a load produces a lateral strain \(1.5 \times 10^{-5}\). Calculate the mass attached to the wire.
\((Y_{steel} = 2 \times 10^{11} \text{ N/m}^2 \text{, Poisson’s ratio } \sigma = 0.291 \text{, } g = 9.8 \text{ m/s}^2)\)
Given:
Area \(A = 1.5 \text{ mm}^2 = 1.5 \times 10^{-6} \text{ m}^2\)
Lateral strain = \(1.5 \times 10^{-5}\)
\(\sigma = 0.291\)
Step 1: Find Longitudinal Strain
$$ \sigma = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} $$
$$ \text{Long. Strain} = \frac{1.5 \times 10^{-5}}{0.291} \approx 5.1546 \times 10^{-5} $$
Step 2: Find Stress
$$ Y = \frac{\text{Stress}}{\text{Long. Strain}} $$
$$ \text{Stress} = Y \times \text{Strain} = 2 \times 10^{11} \times 5.1546 \times 10^{-5} $$
$$ \text{Stress} = 10.309 \times 10^6 \text{ N/m}^2 $$
Step 3: Find Force (Load)
$$ F = \text{Stress} \times A = 10.309 \times 10^6 \times 1.5 \times 10^{-6} $$
$$ F = 15.46 \text{ N} $$
Step 4: Find Mass
$$ m = F/g = 15.46 / 9.8 \approx 1.58 \text{ kg} $$
Answer: Mass attached is 1.58 kg.
(i) What is ‘diffraction of light’? Explain its two types.
Definition: The phenomenon of bending of light around the sharp corners of an obstacle or slit and spreading into the region of geometrical shadow is called diffraction of light.
Types:
- Fresnel Diffraction: The source and the screen are at finite distances from the diffracting element. Incident wavefronts are spherical or cylindrical. No lenses are required.
- Fraunhofer Diffraction: The source and the screen are effectively at infinite distances from the aperture. Incident wavefronts are plane. Convex lenses are used to focus the light.
(ii) Draw a neat labelled diagram for the construction of ‘cyclotron’.
(Please refer to standard textbook diagrams for the exact visual. Key components described below)
- Dees (D1 and D2): Two hollow semicircular metal chambers.
- Electromagnet: Strong North and South poles providing perpendicular magnetic field.
- High Frequency Oscillator: Connected to Dees to provide alternating electric field.
- Source: Ion source at the center.
- Target/Deflector plate: At the exit.
(iii) Distinguish between ‘paramagnetic’ and ‘ferromagnetic’ substances.
| Paramagnetic | Ferromagnetic |
|---|---|
| Weakly attracted by magnets. | Strongly attracted by magnets. |
| Susceptibility (\(\chi\)) is small and positive. | Susceptibility (\(\chi\)) is very large and positive. |
| Examples: Aluminum, Platinum. | Examples: Iron, Nickel, Cobalt. |
(iv) Write a short note on surface wave propagation of electromagnetic waves.
Also known as Ground Wave propagation. In this mode, radio waves travel along the surface of the earth. The ground induces current in the wave, causing attenuation. This is suitable only for low frequencies (up to 2 MHz) like AM radio broadcasts, as attenuation increases with frequency.
(v) The combined resistance of a galvanometer of resistance 500\(\Omega\) and its shunt is 21\(\Omega\). Calculate the value of shunt.
Given:
Galvanometer Resistance \(G = 500 \Omega\).
Combined (Parallel) Resistance \(R_{eq} = 21 \Omega\).
Formula:
$$ \frac{1}{R_{eq}} = \frac{1}{G} + \frac{1}{S} $$
$$ \frac{1}{21} = \frac{1}{500} + \frac{1}{S} $$
$$ \frac{1}{S} = \frac{1}{21} - \frac{1}{500} = \frac{500 - 21}{10500} = \frac{479}{10500} $$
$$ S = \frac{10500}{479} \approx 21.92 \Omega $$
Answer: Shunt resistance is approximately 21.92 \(\Omega\).
(vi) The susceptibility of magnesium at 200 K is \(1.8 \times 10^{-5}\). At what temperature will the susceptibility decrease by \(6 \times 10^{-6}\)?
Given:
\(T_1 = 200\) K, \(\chi_1 = 1.8 \times 10^{-5}\).
Decrease = \(6 \times 10^{-6}\). So \(\chi_2 = 1.8 \times 10^{-5} - 0.6 \times 10^{-5} = 1.2 \times 10^{-5}\).
Formula: Curie's Law: \(\chi \propto \frac{1}{T} \Rightarrow \chi_1 T_1 = \chi_2 T_2\).
$$ (1.8 \times 10^{-5}) \times 200 = (1.2 \times 10^{-5}) \times T_2 $$
$$ 1.8 \times 200 = 1.2 \times T_2 $$
$$ 360 = 1.2 T_2 $$
$$ T_2 = \frac{360}{1.2} = 300 \text{ K} $$
Answer: Temperature is 300 K.
(vii) The co-efficient of mutual induction between primary and secondary coil is 2H. Calculate induced e.m.f. if current of 4A is cut off in \(2.5 \times 10^{-4}\) seconds.
Given: \(M = 2\) H, \(dI = 4 - 0 = 4\) A, \(dt = 2.5 \times 10^{-4}\) s.
Formula: \(e = M \frac{dI}{dt}\)
$$ e = 2 \times \frac{4}{2.5 \times 10^{-4}} $$
$$ e = \frac{8}{2.5} \times 10^4 = 3.2 \times 10000 = 32000 \text{ V} $$
Answer: Induced e.m.f is 32000 V (32 kV).
(viii) The decay constant of radioactive substance is \(4.33 \times 10^{-4}\) per year. Calculate its half life period.
Given: \(\lambda = 4.33 \times 10^{-4}\) /year.
Formula: \(T_{1/2} = \frac{0.693}{\lambda}\).
$$ T_{1/2} = \frac{0.693}{4.33 \times 10^{-4}} = \frac{6930}{4.33} \approx 1600 \text{ years} $$
Answer: Half life is 1600 years.
(i) If the polarising angle for a given medium is \(60^\circ\), then the refractive index of the medium is ............
Reason: Brewster's Law: \(\mu = \tan(i_p) = \tan(60^\circ) = \sqrt{3}\).
(ii) The resolving power of a telescope depends upon the ............
Reason: \(R.P. = \frac{D}{1.22\lambda}\), where D is the aperture/diameter.
(iii) Electric intensity due to a charged sphere at a point outside the sphere decreases with ............
Note: \(E = \frac{1}{4\pi \epsilon_0 K} \frac{q}{r^2}\). E decreases as K increases. Also E decreases as \(r^2\) increases (or \(1/r^2\) decreases?). Options (c) and (d) refer to distance changes which would Increase intensity if distance decreases. So (b) is the most logical fit for "decreases with".
(iv) In potentiometer experiment, if \(l_1\) is the balancing length for e.m.f. of cell of internal resistance r and \(l_2\) is the balancing length for its terminal potential difference when shunted with resistance R then :
Reason: \(r = R(\frac{l_1}{l_2} - 1) \Rightarrow \frac{r}{R} + 1 = \frac{l_1}{l_2} \Rightarrow \frac{R+r}{R} = \frac{l_1}{l_2}\).
(v) The energy of photon of wavelength \(\lambda\) is ............
(vi) Which logic gate corresponds to the truth table given below?
| A | B | Y |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
Reason: Output is high only when both inputs are low. This is the inverse of OR.
(vii) The process of superimposing a low frequency signal on a high frequency wave is ............
State the principle on which transformer works. Explain its working with construction. Derive an expression for ratio of e.m.f.s and currents in terms of number of turns in primary and secondary coil.
Principle: Mutual Induction.
Construction/Working: Consists of a soft iron core with two coils (Primary \(P\) and Secondary \(S\)). AC voltage applied to \(P\) creates varying flux, which links to \(S\) via the core, inducing EMF in \(S\).
Derivation:
Induced EMF in Primary: \(E_p = -N_p \frac{d\phi}{dt}\)
Induced EMF in Secondary: \(E_s = -N_s \frac{d\phi}{dt}\)
Taking ratio: \(\frac{E_s}{E_p} = \frac{N_s}{N_p}\).
Assuming ideal transformer (Power input = Power output): \(E_p I_p = E_s I_s\).
So, \(\frac{E_s}{E_p} = \frac{I_p}{I_s}\).
Combined: \(\frac{E_s}{E_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = K\) (Transformation Ratio).
A conductor of any shape, having area \(40 \text{ cm}^2\) placed in air is uniformly charged with a charge \(0.2 \mu C\). Determine the electric intensity at a point just outside its surface. Also, find the mechanical force per unit area of the charged conductor. \([\epsilon_0 = 8.85 \times 10^{-12} \text{ S.I. units}]\)
Given:
Area \(A = 40 \text{ cm}^2 = 40 \times 10^{-4} \text{ m}^2\)
Charge \(q = 0.2 \mu C = 0.2 \times 10^{-6} \text{ C}\)
Step 1: Surface Charge Density (\(\sigma\))
$$ \sigma = \frac{q}{A} = \frac{0.2 \times 10^{-6}}{40 \times 10^{-4}} = \frac{2 \times 10^{-7}}{4 \times 10^{-3}} = 0.5 \times 10^{-4} \text{ C/m}^2 $$
Step 2: Electric Intensity (\(E\))
$$ E = \frac{\sigma}{\epsilon_0} = \frac{0.5 \times 10^{-4}}{8.85 \times 10^{-12}} $$
$$ E \approx 5.65 \times 10^6 \text{ V/m (or N/C)} $$
Step 3: Mechanical Force per unit area (\(f\))
$$ f = \frac{1}{2}\epsilon_0 E^2 \quad \text{or} \quad f = \frac{\sigma^2}{2\epsilon_0} $$
$$ f = \frac{1}{2} \sigma E = \frac{1}{2} (0.5 \times 10^{-4})(5.65 \times 10^6) $$
$$ f = 0.5 \times 0.5 \times 5.65 \times 100 \approx 141.25 \text{ N/m}^2 $$
Answer: Electric Intensity is \(5.65 \times 10^6\) N/C and Force per unit area is \(141.25\) N/m².
With the help of a neat labelled diagram, describe the Geiger-Marsden experiment. What is mass defect?
Geiger-Marsden Exp (Rutherford's Alpha Scattering): Alpha particles from a source (Bismuth) are collimated and strike a thin gold foil. A detector (Zinc Sulfide screen + Microscope) rotates to detect scattered alpha particles. Observations led to the nuclear model of the atom.
Mass Defect: The difference between the sum of masses of nucleons (protons and neutrons) constituting a nucleus and the actual mass of the nucleus is called mass defect.
$$ \Delta m = [Z m_p + (A-Z) m_n] - M_{nucleus} $$
The photoelectric work function for a metal surface is 2.3 eV. If the light of wavelength 6800Å is incident on the surface of metal, find threshold frequency and incident frequency. Will there be an emission of photoelectrons or not?
Given: \(\phi_0 = 2.3 \text{ eV}\), \(\lambda = 6800 \text{ \AA} = 6.8 \times 10^{-7} \text{ m}\).
Constants: \(h = 6.63 \times 10^{-34} \text{ Js}\), \(c = 3 \times 10^8 \text{ m/s}\), \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).
1. Threshold Frequency (\(\nu_0\)):
$$ \phi_0 = h \nu_0 $$
$$ \nu_0 = \frac{2.3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = \frac{3.68}{6.63} \times 10^{15} \approx 5.55 \times 10^{14} \text{ Hz} $$
2. Incident Frequency (\(\nu\)):
$$ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{6.8 \times 10^{-7}} = \frac{30}{6.8} \times 10^{14} \approx 4.41 \times 10^{14} \text{ Hz} $$
Conclusion: Since incident frequency (\(4.41 \times 10^{14}\)) is less than threshold frequency (\(5.55 \times 10^{14}\)), NO emission will take place.
(i) Determine the change in wavelength of light during its passage from air to glass. If the refractive index of glass with respect to air is 1.5 and the frequency of light is \(3.5 \times 10^{14}\) Hz, find the wave number of light in glass. [Velocity of light \(c = 3 \times 10^8\) m/s]
Given: \(\mu = 1.5\), \(\nu = 3.5 \times 10^{14} \text{ Hz}\), \(c = 3 \times 10^8 \text{ m/s}\).
1. Wavelengths:
\(\lambda_{air} = c/\nu = (3 \times 10^8) / (3.5 \times 10^{14}) \approx 8.57 \times 10^{-7} \text{ m} = 8570 \text{ \AA}\).
\(\lambda_{glass} = \lambda_{air} / \mu = 8570 / 1.5 \approx 5713 \text{ \AA}\).
Change in Wavelength: \(\Delta \lambda = 8570 - 5713 = 2857 \text{ \AA}\).
2. Wave number in glass (\(\bar{\nu}\)):
$$ \bar{\nu} = \frac{1}{\lambda_{glass}} = \frac{1}{5.713 \times 10^{-7} \text{ m}} \approx 1.75 \times 10^6 \text{ m}^{-1} $$
(ii) In biprism experiment, 10th dark band is observed at 2.09 mm from the central bright point on the screen with red light of wavelength 6400 Å. By how much will fringe width change if blue light of wavelength 4800Å is used with the same setting?
Given: Red \(\lambda_1 = 6400 \text{ \AA}\), 10th Dark \(x_{10d} = 2.09 \text{ mm}\).
Condition for dark band: \(x_n = (2n-1)\frac{\lambda D}{2d}\).
$$ 2.09 = (2(10)-1) \frac{\lambda_1 D}{2d} = 19 \frac{\lambda_1 D}{2d} $$
So, ratio \(\frac{D}{d} = \frac{2.09 \times 2}{19 \times 6.4 \times 10^{-7}} \approx 343750\).
Fringe width \(\beta = \lambda \frac{D}{d}\).
\(\beta_{red} = 6.4 \times 10^{-7} \times 343750 = 0.22 \text{ mm}\).
\(\beta_{blue} = 4.8 \times 10^{-7} \times 343750 = 0.165 \text{ mm}\).
Change: \(\Delta \beta = 0.22 - 0.165 = 0.055 \text{ mm}\).
(iii) Describe Kelvin’s method to determine the resistance of galvanometer by using metre bridge.
Circuit: Standard Metre Bridge circuit, but the Galvanometer is placed in the gap where unknown resistance usually goes (Left Gap), and a Resistance Box in the Right Gap. The jockey is connected directly to the junction between gaps (B) and moved on wire (D).
Method (Half Deflection/Equal Deflection): Adjust Resistance Box to \(R\). Move jockey to find point D where deflection in Galvanometer remains same whether the jockey touches the wire or not (implies potential at B = potential at D). However, specifically for Kelvin's method (measure G): We adjust R such that the deflection in G is some value. We touch jockey at null point such that deflection remains unchanged (no current flows through the path BD - wait, in Kelvin's method for G, the galvanometer acts as its own detector).
Formula: \(G = R \frac{l_g}{100-l_g}\).
(iv) Explain the elementary idea of an oscillator with the help of block diagram.
An electronic oscillator converts DC energy from a power supply into AC energy of required frequency. It consists of:
- Amplifier: Increases signal strength.
- Feedback Network: Returns a portion of output to input in phase (Positive Feedback).
- Tank Circuit: Determines frequency.