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Grade XII ALL INDIA SET 2
Physics (Theory)
Time allowed: 3 hours]
[Maximum marks:70
General Instructions:
(i)
All questions are compulsory.
(ii)
There are 30 questions in total. Question Nos. 1 to 8 are very short answer type questions and
carry one mark each.
(iii)
Question Nos. 9 to 18 carry two marks each, question 19 to 27 carry three marks each and
question 28 to 30 carry five marks each.
(iii)
There is no overall choice. However, an internal choice has been provided in one question of
two marks; one question of three marks and all three questions of five marks each. You have to
attempt only one of the choice in such questions.
(iv)
Use of calculators is not permitted. However, you may use log tables if necessary.
(v)
You may use the following values of physical constants wherever necessary:
c = 3 × 108 ms–1
h = 6.626 × 10 –34 Js
e = 1.602 × 10–19 C
0 = 4 × 10–7 Tm A–1
1
 9 109 Nm 2C2
40
Mass of electron me = 9.1 × 10–31 kg
Q 1. For the same value of angle of incidence, the angles of refraction in three media A, B and C are
15O, 25O and 35O respectively. In which medium would the velocity of light be minimum?
1
Solution:
As light travels from a rarer to denser medium it bends towards the normal as its speed decreases. So, if
the bending is more, the speed of the light would be less in that medium, compared to other media. As
the angle of refraction is measured with respect to the normal, the ray making the least angle of
refraction would bend more and the speed of light would be minimum in that case. So, the correct
option is medium A where refracting angle is 15°.
Q 2. A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through
two opposite faces of the cube?
1
Solution:
We can calculate the flux passing through the opposite faces by using the Gauss’s Law.
It is given as
Grade XII ALL INDIA SET 2
  q

E
  ds  0
Now, the flux passing through all the six surfaces would be
q
  6 
0
And the flux passing through any one surface would be
q

6 0
And finally, the flux passing through opposite two faces would be
 q 
 '  2  2 

 6 0 
q
Or,  ' 
3 0
Q 3. A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD.
Predict the directions of induced current in each coil.
1
Solution:
According to Lenz’s law the polarity of the induced emf is such that it opposes the change in magnetic
flux responsible for its production.
Since North Pole of bar magnet is receding away from the coil so the right end of the coil will develop
South Pole i.e. induced current as seen from the left end will be anticlockwise.
Again, since South Pole is pushing towards the second coil so the left end of the coil will develop
Grade XII ALL INDIA SET 2
South Pole in order to repel it and decrease the flux i.e. induced current as seen from the left end will
be clockwise.
Q 4. Two wires of equal length, one of copper and the other of manganin have the same resistance.
Which wire is thicker?
1
Solution:
Since the resistivity of alloy is greater than the resistivity of its constituents. We have resistivity of
manganin greater than resistivity of copper metal.
l
So, Resistance R  
A
Where l is length and A is area of cross-section of material.
Also  is the resistivity of the material.
For copper, R c  c
lc
Ac
and for magnanin R m  m
We have
m  c
m
1
c
As lc  lm
or
R c c A m

Rm  m A c
Again Rc = Rm
Dividing
lm
Am
Grade XII ALL INDIA SET 2
 c A m   m A c
m A m

1
c A c
Hence Am > Ac
Or, Manganin wire is thicker.
Q 5. A proton and an electron have same velocity. Which one has greater de-Broglie wavelength and
why?
1
Solution:
The de-Broglie wavelength  is given for mass m and velocity v of a particle as
h

mv
Now for an electron and proton velocity v is given same.
Since mp > me
We have inverse relations of  and mass
So  p  e
Thus electron would have greater wavelength than wavelength of proton accelerated with same
velocity.
Q 6. Mention the two characteristic properties of the material suitable for making core of a transformer.
1
Solution:
Since due to iron core we have two types of losses,
1. Iron loss i.e. energy loss in the form of heat due to the induced eddy current that can be
minimized by using thin laminated core.
2. Hysteresis loss due to alternately passing a current in the coil to magnetize and demagnetize the
iron core. This can be reduced by using a special alloy of iron core with silicon.
So material with less iron loss and less hysteresis loss is used for making core of transformer.
Q 7. What are the directions of electric and magnetic field vectors relative to each other and relative to
the direction of propagation of electromagnetic waves?
1
Solution:
Since electromagnetic waves are transverse in nature. We have electric and magnetic fields associated
with an electromagnetic wave perpendicular to each other and perpendicular to the direction of
propagation of electromagnetic waves.
Grade XII ALL INDIA SET 2
Let the direction of electric field and magnetic field is along Y- and Z- axis then the direction of
propagation of EM waves will be in positive X-axis.
Q 8. In a single-slit diffraction experiment, the width of the slit is made double the original width. How
does this affect the size and intensity of the central diffraction band?
1
Solution:
In single-slit diffraction experiment fringe width is given as
2 D

d
Where  is fringe width
 is wavelength of light used
D is the distance between screen and slit
d is the width of the slit.
Thus, if d is doubled the width of the central maximum is halved. Thus size of the central diffraction
band is reduced.
Central intensity of the diffraction pattern varies square of the slit width so as the slit get double
intensity will get four times.