Board Question Paper: October 2015
Physics
Time: 3 Hours | Total Marks: 70
State an expression for the moment of inertia of a solid uniform disc, rotating about an axis passing through its centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration:
- about a tangent in the plane of the disc, and
- about a tangent perpendicular to the plane of the disc.
Solution:
Expression for M.I. about central axis:
The moment of inertia ($I_c$) of a solid uniform disc of mass $M$ and radius $R$, about an axis passing through its centre and perpendicular to its plane is given by:
$$I_c = \frac{1}{2}MR^2$$
i. About a tangent in the plane of the disc ($I_T$):
First, consider the M.I. about any diameter ($I_d$). By the theorem of perpendicular axes:
$$I_z = I_x + I_y$$
Here, $I_z = I_c$ and due to symmetry $I_x = I_y = I_d$.
$$I_c = 2I_d \implies I_d = \frac{I_c}{2} = \frac{1}{2}\left(\frac{1}{2}MR^2\right) = \frac{1}{4}MR^2$$
Now, apply the theorem of parallel axes. The tangent in the plane is parallel to a diameter and at a distance $h = R$ from it.
$$I_{T,plane} = I_d + Mh^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$$
Radius of Gyration ($K$):
$$MK^2 = \frac{5}{4}MR^2 \implies K = \sqrt{\frac{5}{4}}R = \frac{\sqrt{5}}{2}R$$
ii. About a tangent perpendicular to the plane of the disc ($I_{T, \perp}$):
This axis is parallel to the central axis ($I_c$) passing through the centre, separated by distance $h = R$.
Using the theorem of parallel axes:
$$I_{T, \perp} = I_c + Mh^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$$
Radius of Gyration ($K$):
$$MK^2 = \frac{3}{2}MR^2 \implies K = \sqrt{\frac{3}{2}}R$$
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In a set, 21 turning forks are arranged in a series of decreasing frequencies. Each tuning fork produces 4 beats per second with the preceding fork. If the first fork is an octave of the last fork, find the frequencies of the first and tenth fork.
Solution:
Given:
Number of forks ($N$) = 21.
Beat frequency ($x$) = 4 beats/sec.
The frequencies are in decreasing order ($n_1 > n_2 > \dots$).
$n_1$ = Frequency of 1st fork.
$n_{21}$ = Frequency of last fork.
Relation: $n_1 = 2 \times n_{21}$ (Octave condition).
Formula:
For a series of decreasing frequencies, the frequency of the $p^{th}$ fork is given by:
$$n_p = n_1 - (p-1)x$$
Therefore, for the last fork ($p=21$):
$$n_{21} = n_1 - (21-1) \times 4$$
$$n_{21} = n_1 - 80$$
Calculation:
Substitute $n_1 = 2n_{21}$ into the equation:
$$n_{21} = 2n_{21} - 80$$
$$n_{21} = 80 \text{ Hz}$$
Now, find the frequency of the first fork ($n_1$):
$$n_1 = 2 \times 80 = 160 \text{ Hz}$$
Find the frequency of the tenth fork ($n_{10}$):
$$n_{10} = n_1 - (10-1)x$$
$$n_{10} = 160 - (9 \times 4)$$
$$n_{10} = 160 - 36 = 124 \text{ Hz}$$
Answer:
Frequency of first fork = 160 Hz
Frequency of tenth fork = 124 Hz
Discuss the composition of two S.H.M.s along the same path having same period. Find the resultant amplitude and initial phase.
Solution:
Consider two S.H.M.s of the same period ($T$) and same frequency ($\omega = 2\pi/T$) acting along the same path (say x-axis). Let their amplitudes be $A_1$ and $A_2$ and initial phases be $\alpha_1$ and $\alpha_2$.
The displacement equations are:
$$x_1 = A_1 \sin(\omega t + \alpha_1)$$ $$x_2 = A_2 \sin(\omega t + \alpha_2)$$By the principle of superposition, the resultant displacement $x$ is:
$$x = x_1 + x_2 = A_1 \sin(\omega t + \alpha_1) + A_2 \sin(\omega t + \alpha_2)$$Expanding the sine terms:
$$x = A_1(\sin \omega t \cos \alpha_1 + \cos \omega t \sin \alpha_1) + A_2(\sin \omega t \cos \alpha_2 + \cos \omega t \sin \alpha_2)$$ $$x = \sin \omega t (A_1 \cos \alpha_1 + A_2 \cos \alpha_2) + \cos \omega t (A_1 \sin \alpha_1 + A_2 \sin \alpha_2)$$Let us substitute:
$$R \cos \delta = A_1 \cos \alpha_1 + A_2 \cos \alpha_2 \quad \dots(1)$$ $$R \sin \delta = A_1 \sin \alpha_1 + A_2 \sin \alpha_2 \quad \dots(2)$$Where $R$ is the resultant amplitude and $\delta$ is the resultant phase.
Substituting (1) and (2) back into the equation for $x$:
$$x = R \sin \omega t \cos \delta + R \cos \omega t \sin \delta$$ $$x = R \sin(\omega t + \delta)$$This shows the resultant motion is also an S.H.M. with the same period.
Resultant Amplitude ($R$): Squaring and adding (1) and (2):
$$R^2 = (A_1 \cos \alpha_1 + A_2 \cos \alpha_2)^2 + (A_1 \sin \alpha_1 + A_2 \sin \alpha_2)^2$$ $$R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\alpha_1 - \alpha_2)}$$Initial Phase ($\delta$): Dividing (2) by (1):
$$\tan \delta = \frac{A_1 \sin \alpha_1 + A_2 \sin \alpha_2}{A_1 \cos \alpha_1 + A_2 \cos \alpha_2} \implies \delta = \tan^{-1}\left(\frac{A_1 \sin \alpha_1 + A_2 \sin \alpha_2}{A_1 \cos \alpha_1 + A_2 \cos \alpha_2}\right)$$A sonometer wire is in unison with a tuning fork of frequency 125 Hz when it is stretched by a weight. When the weight is completely immersed in water, 8 beats are heard per second. Find the specific gravity of the material of the weight.
Solution:
Given:
Frequency of fork $n = 125$ Hz.
Initial condition: Unison, so wire frequency $n_1 = 125$ Hz.
When weight is immersed in water, tension decreases, so frequency decreases. Beats = 8/sec.
New frequency $n_2 = 125 - 8 = 117$ Hz.
Formulae:
Frequency of a stretched string $n \propto \sqrt{T}$.
Tension in air $T_1 = Mg = V \rho g$.
Tension in water $T_2 = Mg - \text{Upthrust} = V \rho g - V \rho_w g$.
Specific gravity (S.G) = $\frac{\rho}{\rho_w}$.
Calculation:
From the relation $n \propto \sqrt{T}$:
$$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}}$$
Substitute tensions:
$$\frac{n_2}{n_1} = \sqrt{\frac{V \rho g - V \rho_w g}{V \rho g}} = \sqrt{1 - \frac{\rho_w}{\rho}}$$
Let Specific Gravity $S = \frac{\rho}{\rho_w}$, so $\frac{\rho_w}{\rho} = \frac{1}{S}$.
$$\frac{117}{125} = \sqrt{1 - \frac{1}{S}}$$
Squaring both sides:
$$\left(\frac{117}{125}\right)^2 = 1 - \frac{1}{S}$$
$$\frac{1}{S} = 1 - \left(\frac{117}{125}\right)^2$$
$$\frac{1}{S} = \frac{125^2 - 117^2}{125^2}$$
Using $a^2 - b^2 = (a-b)(a+b)$:
$$125^2 - 117^2 = (125-117)(125+117) = 8 \times 242 = 1936$$
$$\frac{1}{S} = \frac{1936}{15625}$$
$$S = \frac{15625}{1936} \approx 8.07$$
Answer:
The specific gravity of the material of the weight is 8.07.
i. Which of the following substances is ductile?
Reason: Copper is highly ductile and can be easily drawn into thin wires. (Note: While steel is ductile, 'Copper' is the textbook example for high ductility compared to high carbon steel or glass).
ii. Angle of contact for the pair of pure water with clean glass is _______.
iii. A seconds pendulum is suspended in an elevator moving with constant speed in downward direction. The periodic time (T) of that pendulum is _______.
Reason: When moving with constant speed (velocity), acceleration $a=0$. The effective gravity remains $g$. Thus, $T$ remains unchanged.
iv. The equation of a progressive wave is $y = 7 \sin (4t - 0.02x)$, where x and y are in cms and time t in seconds. The maximum velocity of a particle is _______.
Reason: Comparing with $y = A \sin(\omega t - kx)$, $A=7$, $\omega=4$. $v_{max} = A\omega = 7 \times 4 = 28$ cm/s.
v. The dimensions of emissive power are
Reason: Emissive power $E = \frac{\text{Energy}}{\text{Area} \times \text{Time}} = \frac{[ML^2T^{-2}]}{[L^2][T]} = [M^1L^0T^{-3}]$.
vi. The pressure (P) of an ideal gas having volume (V) is $\frac{2E}{3V}$, then the energy E is _______.
vii. The fundamental frequency of transverse vibration of a stretched string of radius r is proportional to _______.
Reason: $n = \frac{1}{2Lr}\sqrt{\frac{T}{\pi \rho}}$. Frequency is inversely proportional to radius.
i. Draw a neat labelled diagram of conical pendulum. State the expression for its periodic time in terms of length.
Expression for periodic time:
$$T = 2\pi \sqrt{\frac{L \cos \theta}{g}}$$ Where:$L$ = Length of the pendulum string
$\theta$ = Angle made by string with the vertical
$g$ = Acceleration due to gravity
ii. A raindrop of diameter 4 mm is about to fall on the ground. Calculate the pressure inside the raindrop. [Surface tension of water T = 0.072 N/m, atmospheric pressure = $1.013 \times 10^5$ N/m²]
Solution:
Diameter $d = 4$ mm $\implies$ Radius $r = 2$ mm $= 2 \times 10^{-3}$ m.
Surface Tension $T = 0.072$ N/m.
Atmospheric Pressure $P_0 = 1.013 \times 10^5$ N/m².
Excess pressure inside a liquid drop $P_{ex} = \frac{2T}{r}$.
$$P_{ex} = \frac{2 \times 0.072}{2 \times 10^{-3}} = 0.072 \times 10^3 = 72 \text{ N/m}^2$$
Total Pressure inside = $P_0 + P_{ex}$
$$P_{in} = 101300 + 72 = 101372 \text{ N/m}^2$$
$$P_{in} = 1.01372 \times 10^5 \text{ N/m}^2$$
iii. Discuss the weightlessness experienced by an astronaut in an orbiting satellite.
Answer:
1. An astronaut in a satellite orbiting the Earth is in a state of free fall towards the Earth.
2. The gravitational force acting on the astronaut provides the necessary centripetal force for the circular motion. Mathematically, $mg = \frac{mv^2}{R}$.
3. The normal reaction ($N$) exerted by the floor of the satellite on the astronaut is given by the equation: $mg - N = \frac{mv^2}{R}$.
4. Since $mg = \frac{mv^2}{R}$, it follows that $N = 0$.
5. Since the apparent weight of a body is measured by the normal reaction force, the astronaut feels zero weight, i.e., weightlessness.
iv. The periodic time of a linear harmonic oscillator is $2\pi$ second, with maximum displacement of 1 cm. If the particle starts from extreme position, find the displacement of the particle after $\frac{\pi}{3}$ seconds.
Solution:
$T = 2\pi$ s $\implies \omega = \frac{2\pi}{T} = 1$ rad/s.
Amplitude $A = 1$ cm.
Particle starts from extreme position, so displacement equation is:
$$x = A \cos(\omega t)$$
At $t = \frac{\pi}{3}$:
$$x = 1 \cdot \cos\left(1 \cdot \frac{\pi}{3}\right) = \cos(60^\circ) = 0.5 \text{ cm}$$
v. State and prove : Law of conservation of angular momentum.
Statement: Angular momentum of a rotating body remains constant if the resultant external torque acting on the body is zero.
Proof:
Angular momentum is given by $\vec{L} = \vec{r} \times \vec{p}$.
Differentiating with respect to time:
$$\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p})$$
$$\frac{d\vec{L}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} + \frac{d\vec{r}}{dt} \times \vec{p}$$
Since $\frac{d\vec{r}}{dt} = \vec{v}$ and $\vec{p} = m\vec{v}$, and $\vec{v} \times \vec{v} = 0$, the second term vanishes.
Also, $\frac{d\vec{p}}{dt} = \vec{F}$ (Force).
$$\frac{d\vec{L}}{dt} = \vec{r} \times \vec{F} = \vec{\tau} \text{ (Torque)}$$
If external torque $\vec{\tau} = 0$, then $\frac{d\vec{L}}{dt} = 0$.
Therefore, $\vec{L} = \text{constant}$.
vi. A pinhole is made in a hollow sphere of radius 5 cm whose inner wall is at temperature 727°C. Find the power radiated per unit area. [Stefan’s constant $\sigma = 5.7 \times 10^{-8}$ J/m²s K⁴, emissivity (e) = 0.2]
Solution:
Temperature $T = 727 + 273 = 1000$ K.
Concept: A small pinhole in a hollow cavity acts as a perfect Black Body ($e_{eff} \approx 1$) regardless of the material's emissivity. The radiation escaping the hole depends only on temperature.
Power radiated per unit area ($E_b$) = $\sigma T^4$
$$E_b = 5.7 \times 10^{-8} \times (1000)^4$$
$$E_b = 5.7 \times 10^{-8} \times 10^{12} = 5.7 \times 10^4 \text{ W/m}^2$$
Note: If the question implies calculating radiation based on the material's emissivity ($e=0.2$) ignoring the blackbody nature of the pinhole (which is physically incorrect for a "pinhole" but possible in exam context):
$$E = e \sigma T^4 = 0.2 \times 5.7 \times 10^4 = 1.14 \times 10^4 \text{ W/m}^2$$
Correct Physics Answer (Black Body Pinhole): $5.7 \times 10^4$ W/m².
vii. Draw a neat labelled diagram showing forces acting on the meniscus of water in a capillary tube.
viii. Compute the temperature at which the r.m.s. speed of nitrogen molecules is 832 m/s. [Universal gas constant, R = 8320 J/k mole K, molecular weight of nitrogen = 28.]
Solution:
$v_{rms} = 832$ m/s.
$R = 8320$ J/kmol K.
Molecular weight $M_0 = 28$ kg/kmol (Since R is per kmol).
Formula: $v_{rms} = \sqrt{\frac{3RT}{M_0}}$
$$832 = \sqrt{\frac{3 \times 8320 \times T}{28}}$$
Squaring both sides:
$$832^2 = \frac{3 \times 8320 \times T}{28}$$
$$T = \frac{832 \times 832 \times 28}{3 \times 8320}$$
$$T = \frac{832 \times 28}{3 \times 10} \quad (\text{Since } 8320 = 10 \times 832)$$
$$T = \frac{23296}{30} = 776.53 \text{ K}$$
i. Discuss the behaviour of wire under increasing load.
Answer:
When a wire is subjected to gradually increasing load, the stress-strain relationship goes through the following stages:
- Proportional Limit (A): Hooke's law is obeyed. Stress is directly proportional to strain.
- Elastic Limit (B): If load is removed, wire regains original dimensions. Beyond this, deformation is plastic.
- Yield Point (C): Strain increases without significant increase in stress. Wire begins to flow.
- Necking: Local constriction forms.
- Breaking Point: The wire eventually breaks at the fracture point.
ii. Determine the binding energy of satellite of mass 1000 kg revolving in a circular orbit around the Earth when it is close to the surface of Earth. Hence find kinetic energy and potential energy of the satellite.
Solution:
Mass $m = 1000$ kg, Radius $R = 6400$ km $= 6.4 \times 10^6$ m.
For satellite close to Earth surface, $r \approx R$.
Binding Energy (B.E.):
$$B.E. = \frac{GMm}{2R}$$
We know $GM = gR^2$. So, $B.E. = \frac{gR^2 m}{2R} = \frac{mgR}{2}$.
Using $g \approx 9.8$ m/s²:
$$B.E. = \frac{1000 \times 9.8 \times 6.4 \times 10^6}{2}$$
$$B.E. = 31.36 \times 10^9 = 3.136 \times 10^{10} \text{ J}$$
Kinetic Energy (K.E.):
K.E. is numerically equal to B.E.
$$K.E. = 3.136 \times 10^{10} \text{ J}$$
Potential Energy (P.E.):
$$P.E. = -2 \times K.E. = -6.272 \times 10^{10} \text{ J}$$
iii. Show that all harmonics are present on a stretched string between two rigid supports.
Answer:
For a string of length $L$ fixed at both ends:
1. Fundamental Mode: Loop length = $L = \lambda/2$. Frequency $n = v/2L$. (1st Harmonic)
2. First Overtone: Loop length = $L = \lambda$. Frequency $n_1 = v/L = 2(v/2L) = 2n$. (2nd Harmonic)
3. Second Overtone: Loop length = $L = 3\lambda/2$. Frequency $n_2 = 3v/2L = 3n$. (3rd Harmonic)
In general, for $p^{th}$ overtone, frequency is $(p+1)n$.
Since frequencies are $n, 2n, 3n, \dots$, integral multiples of the fundamental frequency are present.
Thus, all harmonics (even and odd) are present.
iv. A stone of mass 100 g attached to a string of length 50 cm is whirled in a vertical circle by giving velocity at lowest point as 7 m/s. Find the velocity at the highest point. [Acceleration due to gravity = 9.8 m/s²]
Solution:
$m = 0.1$ kg, $r = 0.5$ m, $v_L = 7$ m/s.
By Law of Conservation of Energy (Bottom to Top):
$$\text{Total Energy at Bottom} = \text{Total Energy at Top}$$
$$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2r)$$
$$v_L^2 = v_H^2 + 4gr$$
$$7^2 = v_H^2 + 4(9.8)(0.5)$$
$$49 = v_H^2 + 19.6$$
$$v_H^2 = 49 - 19.6 = 29.4$$
$$v_H = \sqrt{29.4} \approx 5.42 \text{ m/s}$$
Obtain an expression for average power dissipated in a purely resistive A.C. circuit. Define power factor of the circuit and state its value for purely resistive A.C. circuit.
Solution:
Average Power:
In a purely resistive circuit, Voltage $e = e_0 \sin \omega t$ and Current $i = i_0 \sin \omega t$ are in phase.
Instantaneous power $P = e \cdot i = e_0 i_0 \sin^2 \omega t$.
Average power over one cycle:
$$P_{avg} = \frac{\int_0^T P dt}{T} = \frac{e_0 i_0}{T} \int_0^T \sin^2 \omega t dt$$
Since average value of $\sin^2 \omega t$ over a cycle is $1/2$:
$$P_{avg} = \frac{e_0 i_0}{2} = \left(\frac{e_0}{\sqrt{2}}\right)\left(\frac{i_0}{\sqrt{2}}\right) = E_{rms} I_{rms}$$
Power Factor:
Power factor is the cosine of the phase angle ($\phi$) between voltage and current. ($PF = \cos \phi$).
For a purely resistive circuit, $\phi = 0$.
Therefore, Power Factor = $\cos 0 = 1$.
A rectangular coil of a moving coil galvanometer contains 50 turns each having area 12 cm². It is suspended in radial magnetic field 0.025 Wb/m² by a fibre of twist constant $15 \times 10^{-10}$ N-m/degree. Calculate the sensitivity of the moving coil galvanometer.
Solution:
Given:
$N = 50$
$A = 12 \text{ cm}^2 = 12 \times 10^{-4} \text{ m}^2$
$B = 0.025 \text{ Wb/m}^2$
$C = 15 \times 10^{-10}$ N-m/degree
Formula:
Current Sensitivity ($S_i$) is the deflection per unit current ($\theta/I$).
From equilibrium condition: $NIAB = C\theta$.
$$S_i = \frac{\theta}{I} = \frac{NAB}{C}$$
Calculation:
$$S_i = \frac{50 \times (12 \times 10^{-4}) \times 0.025}{15 \times 10^{-10}}$$
$$S_i = \frac{600 \times 10^{-4} \times 0.025}{15 \times 10^{-10}}$$
$$S_i = \frac{15 \times 10^{-4}}{15 \times 10^{-10}}$$
$$S_i = 10^6 \text{ degrees/Ampere}$$
State Bohr’s third postulate for hydrogen (H2) atom. Derive Bohr’s formula for the wave number. Obtain expressions for longest and shortest wavelength of spectral lines in ultraviolet region for hydrogen atom.
Solution:
Bohr's Third Postulate: An electron radiates energy only when it jumps from an outer stationary orbit of higher energy ($E_n$) to an inner stationary orbit of lower energy ($E_p$). The energy of the photon emitted is equal to the difference in energies. $$h\nu = E_n - E_p$$
Derivation of Wave Number:
Energy of electron in $n^{th}$ orbit: $E_n = -\frac{me^4}{8\epsilon_0^2 n^2 h^2}$.
Difference $E_n - E_p = \frac{me^4}{8\epsilon_0^2 h^2} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$.
Since $\nu = c/\lambda = c \bar{\nu}$:
$$hc\bar{\nu} = \frac{me^4}{8\epsilon_0^2 h^2} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$
$$\bar{\nu} = R \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$
Where Rydberg constant $R = \frac{me^4}{8\epsilon_0^2 ch^3}$.
Ultraviolet Region (Lyman Series):
Corresponds to jumps to $p=1$.
Longest Wavelength: Min energy jump ($n=2 \to n=1$).
$$\frac{1}{\lambda_{max}} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(\frac{3}{4}\right) \implies \lambda_{max} = \frac{4}{3R}$$
Shortest Wavelength: Max energy jump ($n=\infty \to n=1$).
$$\frac{1}{\lambda_{min}} = R\left(\frac{1}{1^2} - \frac{1}{\infty}\right) = R \implies \lambda_{min} = \frac{1}{R}$$
The photoelectric current in a photoelectric cell can be reduced to zero by a stopping potential of 1.8 volt. Monochromatic light of wavelength 2200Å is incident on the cathode. Find the maximum kinetic energy of the photoelectrons in joules. [Charge on electron = $1.6 \times 10^{-19}$ C]
Solution:
Given:
Stopping potential $V_s = 1.8$ V.
Charge $e = 1.6 \times 10^{-19}$ C.
Note: The wavelength is extra information not needed for finding Max KE if stopping potential is given.
Formula:
$$K.E._{max} = e \times V_s$$
Calculation:
$$K.E._{max} = 1.6 \times 10^{-19} \times 1.8$$
$$K.E._{max} = 2.88 \times 10^{-19} \text{ Joules}$$
i. Which one of the following particles cannot be accelerated by a cyclotron?
Reason: Electrons have very small mass, so they quickly reach relativistic speeds, falling out of sync with the oscillating electric field.
ii. In biprism experiment two interfering waves are produced due to division of _______.
iii. The output of NOR gate is high, when _______.
Reason: NOR is NOT OR. Output is high only if (0 OR 0)' = 1.
iv. Light of a certain wavelength has a wave number $\bar{\nu}$ in vacuum. Its wave number in a medium of refractive index n is _______.
Reason: $\lambda' = \lambda/n$. Wave number $\bar{\nu}' = 1/\lambda' = n/\lambda = n\bar{\nu}$.
v. If the radius of a sphere is doubled without changing the charge on it, then electric flux originating from the sphere is ______.
Reason: By Gauss's Law, flux depends only on enclosed charge ($q/\epsilon_0$), not dimensions.
vi. The momentum of a photon of de Broglie wavelength 5000Å is _______.
Calculation: $p = h/\lambda = (6.63 \times 10^{-34}) / (5 \times 10^{-7}) = 1.326 \times 10^{-27}$.
vii. Ionosphere mainly consists of _______.
i. State any ‘two’ possible sources of errors in meter-bridge experiment. How can they be minimised?
Errors: 1. Contact resistance at the ends of the wire. 2. Non-uniformity of the wire.
Minimisation: 1. By measuring balancing length for both gaps (interchanging gaps). 2. By keeping the balance point near the center of the wire.
ii. A potentiometer wire has resistance of per unit length of 0.1 Ω/m. A cell of e.m.f. 1.5V balances against 300 cm length of the wire. Find the current in the potentiometer wire.
Solution:
Resistance per unit length $\sigma = 0.1 \Omega/m$.
Balance length $L = 300$ cm = 3 m.
EMF balanced $E = 1.5$ V.
Potential drop across length L is $V = I \times R_L$.
$R_L = \sigma \times L = 0.1 \times 3 = 0.3 \Omega$.
At balance, $V = E$.
$I \times 0.3 = 1.5 \implies I = 5$ A.
iii. Give any ‘two’ points of differences between diamagnetic and ferromagnetic substances.
1. Magnetism: Diamagnetic are weakly repelled by magnets; Ferromagnetic are strongly attracted.
2. Susceptibility: Diamagnetic has small negative susceptibility; Ferromagnetic has large positive susceptibility.
iv. An iron rod of area of cross-section 0.1m² is subjected to a magnetising field of 1000 A/m. Calculate the magnetic permeability of the iron rod. [Magnetic susceptibility of iron = 59.9, $\mu_0 = 4\pi \times 10^{-7}$ SI units]
Solution:
Susceptibility $\chi = 59.9$.
Relative permeability $\mu_r = 1 + \chi = 1 + 59.9 = 60.9$.
Absolute permeability $\mu = \mu_0 \mu_r$.
$\mu = 4\pi \times 10^{-7} \times 60.9$
$\mu \approx 12.56 \times 60.9 \times 10^{-7} \approx 7.65 \times 10^{-5}$ T m/A.
vi. A coil of 100 turns, each of area 0.02 m² is kept in a uniform field of induction $3.5 \times 10^{-5}$ T. If the coil rotates with a speed of 6000 r.p.m. about an axis in the plane of the coil and perpendicular to the magnetic induction, calculate peak value of e.m.f. induced in the coil.
Solution:
$N=100$, $A=0.02$, $B=3.5 \times 10^{-5}$ T.
$\omega = 6000$ rpm $= \frac{6000}{60} \times 2\pi = 200\pi$ rad/s.
Peak EMF $e_0 = NBA\omega$.
$e_0 = 100 \times 3.5 \times 10^{-5} \times 0.02 \times 200\pi$.
$e_0 = 70000 \times 10^{-5} \times 0.02 \times \pi$.
$e_0 = 0.7 \times 0.02 \times 3.142 \times 200 \dots$ (Let's recalculate).
$e_0 = 100 \times 0.02 \times 3.5 \times 10^{-5} \times 628.3$
$e_0 = 2 \times 3.5 \times 10^{-5} \times 628.3 = 7 \times 10^{-5} \times 628.3$
$e_0 \approx 4.4 \times 10^{-3}$ V $= 0.044$ V.
viii. In a biprism experiment... linear magnification of the image.
Solution:
Position 1 ($d_1 = 7$mm): Lens is 30 cm from slit ($u=30$).
Total distance $D = 10 + 80 = 90$ cm.
$v = D - u = 90 - 30 = 60$ cm.
Linear magnification $m = \frac{v}{u} = \frac{60}{30} = 2$.
i. With the help of a neat circuit diagram, explain the working of a photodiode. State its any ‘two’ uses.
Working: It is operated in reverse bias. When light falls on the junction, electron-hole pairs are generated, increasing the reverse saturation current. The current is proportional to light intensity.
Uses: 1. Optical signals detection. 2. Light meters in cameras.
ii. A parallel beam of monochromatic light is incident on a glass slab at an angle of incidence 60°. Find the ratio of width of the beam in the glass to that in the air if refractive index of glass is 3/2.
Solution:
$i = 60^\circ$, $n = 1.5$.
Snell's Law: $\sin i / \sin r = n \implies \sin 60 / \sin r = 1.5$.
$\frac{\sqrt{3}}{2} / \sin r = \frac{3}{2} \implies \sin r = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.
$\cos r = \sqrt{1 - 1/3} = \sqrt{2/3}$.
Beam width ratio $W_{glass}/W_{air} = \frac{\cos r}{\cos i} = \frac{\sqrt{2/3}}{1/2} = 2\sqrt{\frac{2}{3}} = \sqrt{\frac{8}{3}} \approx 1.63$.
iv. In a single slit diffraction pattern, the distance between first minima on the right and first minima on the left... Calculate width of the slit and width of the central maximum.
Solution:
Distance between two 1st minima = Width of central max ($W_c$).
$W_c = 4$ mm $= 4 \times 10^{-3}$ m.
$W_c = \frac{2\lambda D}{a}$.
Slit width $a = \frac{2\lambda D}{W_c}$.
$a = \frac{2 \times 6 \times 10^{-7} \times 2}{4 \times 10^{-3}} = 6 \times 10^{-4}$ m $= 0.6$ mm.
Width of slit = 0.6 mm.
Width of central maximum = 4 mm.