Q 12 a) Define surface tension.
Solution: Surface tension is defined as the property of fluid by virtue of which its surface acts as a stretched membrane and it tends to acquire minimum surface area.
Q 12 b) Establish a relation between surface tension and surface energy of a liquid.
Solution: Surface energy of a fluid is defined as the ratio of work done in increasing the surface area of a fluid to the increase in surface area.$$\rm \sigma = \frac{F \times x}{\triangle A}$$$$\rm \sigma = \frac{T l \times x}{l \times x}$$$$\rm \sigma = T$$Hence, surface energy of a liquid is numerically equal to its surface tension.
[Answer is incomplete as it requires figure and a little explanation of the figure.]
Q 12 c) Two spherical rain drops of equal size are falling vertically through with a certain terminal velocity. If these two drops were to coalesce to form a single drop and fall with new terminal velocity, explain how terminal velocity of new drop compares to original terminal velocity.
Solution: Let r be the radius of rain drops of equal sizes then their total initial volume is $\rm 2 \times \frac{4}{3} \pi r^3$. Let R be the radius of the coalesced rain drop. The final volume is $\rm \frac{4}{3} \pi R^3$. Since, there is no loss in volume, the two volumes will be equal$$\rm \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3$$$$\rm R = 2^{\frac{1}{3}} r$$If v be the initial terminal velocity and V be the final terminal velocity then$$\rm v \propto r^2$$$$\rm V \propto R^2$$From above two equations, we have,$$\rm \Rightarrow \frac{V}{v} = \frac{R^2}{r^2}$$$$\rm \Rightarrow \frac{V}{v} = \frac{2^{\frac{2}{3}} \cdot r^2}{r^2}$$$$\rm \Rightarrow \frac{V}{v} = \frac{2^{\frac{2}{3}}}{1}$$$$\rm \therefore V = 2^{\frac{2}{3}}v$$Hence, the new terminal velocity will be greater than the original terminal velocity.
Q 13 a) Define torque.
Solution: Torque is defined as the vector product of force and the perpendicular distance between the line of application of force and the rotation axis.
Q 13 b) Establish a relation between torque and angular acceleration of a rigid body.
Solution: Let a rigid body contain n number of particles namely 1,2,3, …, n. Let the mass, velocity, and distance from the axis of each particle be denoted by $m_i$, $v_i$, and $r_i$, respectively where i = 1,2,3, …, n. Each particle is accelerated by the same angular acceleration $\alpha$. We know, $a = \alpha r$.
Hence, force on each particle $\rm F_i = m_i a_i = m_i r_i \alpha$. Since, $\rm \tau_i = F_ir_i$
$$\rm \tau = F_1r_1 + F_2r_2 + … + F_nr_n$$$$\rm \tau = m_1r_1 \alpha r_1 + m_2r_2 \alpha r_2 + … + m_nr_n \alpha r_n$$$$\rm \tau = \left ( m_1r_1^2 + m_2r_2^2 + … + m_nr_n^2 \right ) \alpha$$$$\rm \tau = \sum_{r = 1}^n m_ir_i^2 \cdot \alpha$$$$\rm \tau = I \alpha$$
Q 13 c) In a flywheel, most of the mass is concentrated at the rim, why?
Solution: In a flywheel, most of the mass is concentrated at the rim to increase its moment of inertia. By increasing moment of inertia, the flywheel continues its state of rotational motion for a longer time.
Q 14 a) Define ‘Doppler’s effect’ in sound.
Solution: The observed frequency of a sound is chaned when there is relative motion between a source and an observer. This phenemomenon is defined as “Doppler’s effect” in sound.
Q 14 b) When a sound of source approaches us, then the pitch of sound which appears to be higher than the actual one and when the source of sound moves away from us, the pitch goes low. Explain.
Solution: When the source of sound approaches us, the waves have to travel comparatively less distance in the same time so the frequency of the wave increases and thus, the pitch of sound appears to be higher. In contrary, when the source recedes from the observer, the waves have to travel longer distance, the frequency decreases and the pitch goes low.
Q 14 c) The string of two guitars at temperature 24 oC are tuned using a tuning fork of frequency 256 Hz. Take speed of sound in air is 343 m/s. If one guitar player is moving toward the audience at a speed of 2.8 m/s and the other player is stationary. Calculate the difference in frequency detected by the audience.
Solution: The observed frequency will be more when the guitar player moves toward the audience at speed of $\rm v_s = 2.8 m/s$. Let f’ be the frequency heard by the audience then, $$\rm f’ = \frac{v}{v – v_s} f$$$$\rm f’ = \frac{343}{343 – 2.8} \cdot 256 = 258.1 Hz$$The audience will hear the same frequency from the guitar of stationary guitarist. Hence, the difference in frequency = 258.1 – 256 = 2.1 Hz.
Q 15) Refrigerator is a heat pump that removes the heat from lower temperature body to higher temperature surrounding.
a) Does it violate the principle of second law of thermodynamics?
Solution: No, it does not violate the principle of second law of thermodynamics because external work is done by electricl force to transfer heat from a lower temperature body to higher temperature surrounding.
b) Show the schematic diagram of working of refrigerator.
Solution:
c) At what condition, the coefficient of performance is infinity. Can it be realized in practice?
Solution: When the temperature difference between the source and the sink is zero, the coefficient of performance of refreigerator is infinity. This can be known from the formula$$\rm \beta = \frac{T_2}{T_1 – T_2}$$.If the temperature inside the refrigerator is set equal to the surrounding temperature, its coefficient of performance is infinity. However, the refrigerator becomes useless in this case. So, it cannot be realized in practice.
Q 16 a) The variation of potential difference V with length L in case of two potentiometer P and Q is shown in diagrams. Which one of these two will you prefer for comparing emf of two cells. Explain.
Solution: I will prefer potentiometer Q for comparing emf of two cells because its potential gradient is lesser than that of P. With lesser potential gradient, the potential drop is uniform. The device becomes more sensitive and results more accurate reading.
Q 16 b) A potentiometer wire is 10 m long. It has a resistance of 20 $\Omega$. It is connected in series with a battery of 3 V and resistance of 10 $\Omega$. What is the potential gradient along the wire.
Solution:
Q 17) Electron is deviated in electric and magnetic fields.
a) What path does the electron follow in electric field?
Solution: In an electric field, an electron follows parabolic path.
b) An electron passes through a space without deviation. Does it mean, there is no field?
Solution: It does not necessarily mean there is no field. It points the absence of electric field but the absence of manetic field cannot be guaranteed.
c) Is there any condition that an electron does not experience any force in magnetic field?
Solution: Yes, if an electron moves parallel or antiparallel to the direction of the magnetic field, it experiences no magnetic force. $$\rm F = Bev \sin \theta$$
Q 18) An electron and proton are those particles in which charges are equal and opposite in nature but different in mass.
a) If they move with same speed in a uniform magnetic field, compare the radii of their orbits.
Solution: The mass of a proton is much larger than that of an electron. If they move with same speed in uniform magnetic field, the radii of orbit of proton will be much larger than the electron because $$\rm r = \frac{mv}{Be}$$.
b) Beams of the electron and protons having the same initial kinetic energy enter normally into an electric field, which beam will be more curve? Explain.
Solution:
c) Define cross field.
Solution: When electric and magnetic fields is set perpendicular to each other such that the effect on one field is nullified by the other, it is defined as cross field.
Q 19 a) Derive an expression for the radius of the Bohr’s orbit for hydrogen atom.
Solution: Bohr’s orbit is the first orbit (n = 1) for hydrogen atom (Z = 1). According to Rutherford, the required centripetal force for electrons is provided by the electrostatic force of attraction between electron and proton so,$$\rm m \frac{v^2}{r} = \frac{1}{4 \pi \epsilon _o} \cdot \frac{(Ze)(e)}{r^2}$$$$\rm or, m \frac{v^2}{r} = \frac{1}{4 \pi \epsilon _o} \cdot \frac{e^2}{r^2}$$$$\rm or, v^2 = \frac{e^2}{4 \pi \epsilon _o m r}$$According to Bohr, electron orbits are quantized as follows:$$\rm mvr = n \frac{h}{2 \pi}$$$$\rm or, v = \frac{h}{ 2 \pi m r}$$$$\rm or, v^2 = \frac{h^2}{4 \pi ^2 m^2 r^2}$$From above two equations, we have,$$\rm \frac{e^2}{4 \pi \epsilon _o m r} = \frac{h^2}{4 \pi ^2 m^2 r^2}$$$$\rm \therefore r = \frac{\epsilon _o h^2}{\pi me^2}$$
Q 19 b) The energy levels of an atom are as shown in the figure. Calculate which one of these transitions will result in the emission of a photon of wave length 275 nm. (h = 6.62 $\times$ 10$^{-34}$ Js, c = 3 $\times$ 10$^8$ m/s)
Solution:
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