sample paper of Physics XII 2014
This is the thoroughly revised and updated sample question paper in Physics XII-2014 specially prepared for the students appearing in the class XII final exams 2014. The questions are specially prepared covering the whole syllabus.
(Answer all the questions . Your answers should be brief and to the point).
Q1. A magnet is moving towards a coil with a uniform speed υ as shown in the
figure. State the direction of the induced current in the resistor R.
Ans 1. From X to Y.
Q2. Give one example each of a ‘system’ that uses the
(i) Sky wave (ii) Space wave
mode of propagation 1.
Ans 2. (i) Short wave broadcast services
(ii) Television broadcast (or microwave
links or Satellite communication)
Q3. A concave mirror, of aperture (an opening, hole) 4cm, has a point object placed on its principal axis (main axis) at a distance of 10cm from the mirror. The image, formed by the mirror (actual image) is not a sharp image. Give reasons.
Ans 3. The incident rays are not paraxial.
Q4. Two dipoles, made from charges ±q and ±Q, respectively, have equal dipole moments. Give the (i) ratio between the ‘separations’ of the these two pairs of charges (ii) angle between the dipole axis of these two dipoles.
Ans 4. As qa = Qa’, we have a’/a= q/Q and θ = 0°.
Q5. Define the term ‘Transducer’ for a communication system.
Ans 5. A ‘transducer’ is any device that converts one form of energy into another.
Q6. For the circuit shown here, would the balancing length increase, 2 decrease or remain the same, if (i) R1 is decreased (ii) R2 is increased without any other change, (in each case) in the restof the circuit. Justify your answers in each case.
Ans 6. (i) decreases (The potential gradient would increase).
(ii) increases (The terminal p.d across the cell would increase).
Q7. An athlete peddles a stationary tricycle whose pedals are attached to a coil having 100 turns each of area 0.1m². The coil, lying in the X-Y plane, is rotated, in this plane, at the rate of 50 rpm, about the Y-axis region where a uniform magnetic field, B ̅̅̅̅̅̂̂= (0.01)k tesla is present. Find the following:
(i) maximum emf (ii) average e.m.f
which is generated in the coil after one complete revolution.
Ans 7. (i) The maximum emf ‘ε’ generated in the coil is,
ε = N B A ω.
= N B A 2Πf
= [100 x 0.01 x 0.1 x 2Π(5/6) ] V
= (Π/6)V≈ 0.52 V
(ii) The average emf generated in the coil after one complete revolution will be = 0.
Q8. A monochromatic source, emitting light of wave length (λ) 600 nm has a power
output of 66W. Calculate the number of photons emitted by this source in
Ans 8.Energy of one photon = E = hc/λ
E = (6.6 ×10-34 × 3× 108 ) / (6 × 10-7)
E = (3.3 × 10-19).
E1 = energy emitted by the source in one
second = 66J
number of photons emitted by the source
66 / (3.3×10-19) = 2 × 1020
Total number of photons emitted by
source in 2 minutes
= N = n x 2 x 60
= 2 x 1020 x 120 = 2.4 x 1022 photons.
Q9. The galvanometer in the given figure does not show any deflection. Calculate the ratio of the resistors R1 and R2 used in these two circuits.
Ans 9. For circuit 1 we have (from the Wheatstone
bridge balance condition), R1/9 = 4/6.
R1 = 6Ω
In circuit 2, the interchange of the positions
of the battery and the galvanometer (sensitive ammeter) does
not change the balance
condition (it will not effect the diagram in any manner)
R2/8 = 6/12.
Q10. The electron in an hydrogen atom initially in a state of quantum number
makes a transition (changes its state) to a state whose excitation energy (excited state* )with respect to the
ground state is 10.2 eV. If the wavelength associated with the photon (light particles) emitted
in this transition are 487.5 mm, find the:
(i) energy in eV (ii) value of the quantum number n1 of the electron in its
Ans 10. In an hydrogen atom the energy (En) of
electron in a state having principal
quantum number ‘n’ is given by:
En= -13.6/n2 eV.
E1 = -13.6eV and E2 = -3.4 eV
It follows that the state n=2 has an
excitation energy of 10.2 eV. Hence the
electron is making a transition from n=n1 (initial stage) to
n=2 (final stage) where (n1>2).
Now En1 – E2= hc/λ.
But, hc/λ = (6.63 × 10-24 × 3 × 108) / (487.3 × 10-9 × 1.6 × 10-9).
En1 = (-3.4 + 2.55) eV.
≃ – 0.85 eV
But we also have : En1 = -13.6/n21 eV
we get n1 = 4.
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