## Saturday, 26 July 2014

### CBSE CLASS 12 PHYSICS TEST PAPERS FOR PRACTICE

CBSE CLASS XII PHYSICS
Electrostatics
Q1. A positively charged glass rod attracts a suspended ball of straw. Can we conclude that the ball is negatively charged?
Ans1. No, because the positively charged rod can attract both charged as well as uncharged objects.
Q2. What is the exact value of 1/4 p eo.
Ans2. The exact value of 1/4p eo is 8.99 x 109 NC-2m2.
Q3. What happens when an excess charge is placed on the surface of a conducting body?
Ans3. When excess charge is placed on the surface of a conducting sphere then it spreads due to force of repulsion till it becomes uniformly distributed.
Q4. Does the electric charge vary with the motion of the charged bodies as is the case with mass?
Ans4. No, charge on a body does not change with motion of the body irrespective of its velocity in comparison with the velocity of light.
Q5. Give one example of conservation of charge.
Ans5. Annihilation of matter is an example. A particle and its anti-particle combine to give two g-ray photons. e- + e+ ¾¾® g + g
Q6. Out of three charged bodies A,B and C, if A and B repel each other and B attracts C, then what is nature of force between A and C?
Ans6. From the given question, it is clear that A and C have opposite charges so A and C attract each other.
Q7. Does charge affect mass of a body?
Ans7. Charge on a body means either electrons are added or removed from the body. Since mass of electrons is very small so change in mass is not easily detectable.
Q8. Is this electric field uniform? Ans8. No, it is not uniform because direction of field is changing at every point on the lines of force.
Q9. What is the direction of electric field due to an electric dipole on the equatorial line of the dipole?
Ans9. It is along a line parallel to axis of dipole directed opposite to the dipole moment vector. Q10. What is the nature of symmetry of the dipole field?
Ans10. The dipole field has a cylindrical symmetry. The axis of the cylinder is along the axis of the dipole.

Q1. Under what conditions, a charged circular loop behaves as a point charge in respect of electric field due to it?
Ans1. When the point of observation is at a large distance from the the axis of the ring then the ring behaves like a point charge. For a ring of radius 'a' the electric field at distance 'x' is
E = (1/4
o)qx/(x2 + a2)3/2,  when x2>>a2 then E = (1/4 o) x q/x2, that is E µ 1/x2
Q2. Three chargers each of 1/3 mC are placed at the corners of an equilateral triangle of side of 3 cm. How much work is required to be done on the system to take the charges to infinite distance from each other.
Ans2. The difference in the potential energy = the amount of work to be done. The initial potential energy of the system (P.E) = (1/4o) x 3 x q x q)/3 x 10-2 = (9 x 109 x 3 x 1/3 x 1/3 x 10-12)/3 x 10-2 = 1/10 J.
Final P.E = 0, (because the separation between the charges is infinite)  so change in P.E = W.D = -1/10 J, so -1/10 J is the W.D by the person.

Q3. If electric potential is -10 V at infinity and +5 V at some point P due to a point charge otherwise then what is the potential at point P.
Ans3. Net potential at point P = potential at P - potential at infinity = 5 - (-10) = 15 V.
Q4. At what point the electric potential is zero due to the system of charges as shown?

Ans4. Let P be the point where the potential is zero (1/4
o) (10.10-6)/(x . 10-2) = (1/4o).(30.10-6)/(6 - x).10-2.
so  1/x = 3/(6 - x), (6 - x) = 3x, so potential is zero at x = 1.5 cm from +10
mc.
Q5. If dipole moment of an electric dipole is 4 x 10-6 C-m then what is the electric potential at a point 20 cm away at an angle of 60o with the dipole?
Ans5. Vp = (1/4
o) x p cos q/r2, = (9 x 109 x 4 x 10-6 x 1/2)/(400 x 10-4), Vp = 4.5 x 105 V.
Q6. What is the magnitude of the force between +10
mc and +20 mc where 6 cm is the length of copper slab.
Ans6. F = (1/4peo)(q1q2 )/(d - t)2 = 9 x 109 x [(10 x 20 x 10-12)/(9 - 3)2 x 10 -4 = 9 x 109 x [(200 x 10-12)/36 x 10-4
Therefore F = 500 N.

Q7. What is the magnitude of electric field at ground level and give its direction?
Ans7. The magnitude of the electric field is 100 V/m and it is directed towards the surface of the earth.
Q8. What is the flux through a hollow sphere, which has two electric dipoles inside it?
Ans8. Zero, net charge on each dipole is zero and hence according to Gauss's Law E.ds = q/
eo, when q is zero so E is zero and flux is also zero.

Q1. Explain electrostatic shielding? Give one example.
Ans1. The method of protecting certain region from the effect of external electric field is called electrostatic shielding. The electrostatic shielding can be achieved by protecting and enclosing the sensitive instrument inside a hollow conductor because inside a hollow conductor electric field is zero. For example, a car or a bus during the lightning acts as an electrostatic shielding.
Q2. Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?
Ans2. When a vehicle moves, its body gets charged on account of friction due to air. The tyres also accumulate the charges on account of friction between the tyres and the road. The metallic ropes from the vehicle touching the ground enable the accumulated charges to flow to earth. This would otherwise be hazardous to the inflammable materials.
Q3. The earth is receiving 9 protons, 9 positrons per minute per cm2 of its area. Find the current received by the earth. Take radius of earth = 6400 km.
Ans3. Number of protons received per sec per cm2 = 9/60
Number of positrons received per second per cm2 = 9/60
Surface area of earth = 4
p R2 =4 x 3.14 x (6.4 x 108)2 cm2.
So, number of protons received by earth per second
= 4 x 3.14 x(6.4 x 108)2 x 9/60
Number of positrons received by earth per second
= 4 x 3.14 x (6.4 x 108)2 x 9/60
Since charge on proton or positron is same
Hence, total charge received by the earth
= 2 x 4 x 3.14 x (6.4 x 108)2 x 9/60 x (1.6 x 10-19) C = 0.2468 C
By definition, charge received per second is defined as the current. I = 0.2468 amp.

Q4. Calculate the ratio of the electrostatic force to the gravitational force between two protons. Also calculate the value of the specific charge (q/m) of a particle so that these two forces become equal in magnitude.
Ans4. For electrostatic repulsive force
Fe = (1/4
p eo)(q.q/r2) = 9 x 109 x (1.6 x 10-19)2/r2. ...........(i)
The gravitational force of attraction Fg = G.mm/r2 = 6.67 x 10-11 x (1.67 x 10-27)2/r2 ...............(ii),
On calculation we get
Fe/Fg = 1038:1. To make the magnitude of forces equal 1/4
p eo.(q.q)/r2 = G. (m.m/r2) or q2/m2 = (4p eo x G)
Q5. Two similar balls each having mass 'm' and charge 'q' are hung from silk threads of length 'l', prove that equilibrium separation x = [q2l/4 p eo mg]1/3, when each thread makes small angle q with the vertical.
Ans5. According to Coulomb's law, the force of repulsion between the two balls is given by F = (1/4
p eo) (q2/x2).
As shown in the figure the restoring force on each ball = mg Sin
q = mg q (if q is small) = mg[(x/2)/l](as angle = arc/radius) = mg x/2l
For equilibrium, the restoring force must be equal to the force of repulsion between the two balls.
So, (mg x)/2l = q2/4
p eo x2, x3 = q2l/4p eo mg
x = [q2l/4
p eo mg]1/3.
Q6. Suppose there are two pieces of copper each weighing 10 gm. If two electrons are transferred for every 1000 atoms from one piece to another, find electrostatic force between them if distance between them is 10 mm.
Ans6. Number of atoms in 10 gm of copper are = (6 x 1023) x 10/63.5 = 9.45 x 1022 ..................(i)
Number of electrons transferred/1000 atoms = 2
So total number of electrons transferred (n) = (2/1000) x 9.45 x 1022.
Or n = 1.89 x 1020, hence charge transferred = n x e = 1.89 x 1.6 x 10-19 x 1020 = 3.024 x 10 = 30.24 coulombs.
The force of attraction between the two pieces [one having charge + 30.24 C and another having charge -30.24 C]. F = 1/4
p eo .(q.q)/r2 =9 x 109 x (30.24)2/(10-2)2 = 8.23 x 1016 N.
Q7. Two identically charged spheres are separated by strings of equal lengths. The strings make an angle of 30o with each other. When suspended in a liquid of density 0.8 gm/cm3,the angle remains the same. What is the dielectric constant of the liquid? (The density of the material of the sphere is 1.6 gm/cm3).
Ans7. Let the masses of the spheres be 'm' kg and let the charge on each of them be 'q' coulomb. Let the forces of separation exerted by them on each other in air be F.
F is horizontal since the threads are of equal length. The weights mg of the spheres are vertical.
Let the threads make an angle
q with the vertical. Since the spheres are at rest, we get by applying Lami's theorem.
F/Sin(180 -
q) = mg/Sin(90 + q)
F/sin
q = mg/cos q ....................(i)
When the spheres are suspended in the liquid, the force of repulsion between them becomes F and their weights become m'g. The angle q remains the same as before.
Again, applying Lami's theorem, we get
F'/Sin
q = m'g/Cosq .................(ii)
Dividing Equation (i) by Equation (ii), we get
F/F' = m/m' ...............(iii)
Let the distance between the spheres be 'r'. This remains unaltered in the two cases.
Therefore, F= q2/4
peor2
F' =  q2/4
peoerr2
where
er is the dielectric constant of the liquid.
Therefore F/F' =
er
Let the volume of each sphere be V. Let the density of their material be
rs and the density of the liquid be rL, we have m = V rs
m' = V
rs - mass of displaced liquid
= V
rs - VrL
=V(
rs - rL)
m/m' =
rs/(rs - rL)
Substituting the expressions for F/F' and m/m' in equation (iii) we get
er = rs/(rs -rL)
= 1.6/(1.6 - 0.8)
= 1.6/0.8 = 2

Q8. Find the potential at the centre 'O' of the equilateral triangle. Ans8. Electric potential at O is the algebraic sum of the potential due to each of the charges placed at the vertices of the triangle.
VA = (1/4
p eo)(q/OA) = [9 x 109 x 3 x 10-6]/[(6/Ö3) x 10-2] = 7.8 x 105 V. The potential due to each of 6 mc charge at B and C = 15.6 x 105 V. So, the total potential at the centre is
= (7.8 + 15.6 + 15.6)105 V = 3.9 x 106 V

Q1.(a) Discuss briefly the effect of introduction of (i) conducting slab and (ii) dielectric slab in the space between the plates of a parallel plate capacitor on electric field, force between the plates and energy density in the space between the plates of the capacitor.
(b) Define electric polarisation of a non-polar dielectric and give its units.
Ans. (Try yourself).
Q2. When a electric dipole is kept in a uniform electric field, discuss its behavior and find out an expression for torque experienced by it. Using expression for torque, derive an expression for frequency of angular S.H.M of the dipole.
Ans. (Try yourself).
Q3. (a) When two spherical charged spheres are connected by a conducting wire, find out the expression for the loss in energy and explain the reason for the loss in energy.
(b) A capacitor has a dielectric of constant 6 between its plates and connected to a battery of 100 V. Now the battery is disconnected and the dielectric is removed. What is the energy loss or gain in the capacitor.
Ans. (Try yourself).
Q4. (a) If a mica sheet of thickness 4 mm is introduced in the space between the plates of parallel plate capacitor of plate separation 12 mm. then find (i) the potential difference in open space between the plates. Given area of each plate = 100cm2  and charge on each plate = 100 mc.
(b) Find the potential across the dielectric.
Ans4: (a) b = 4mm, d = 12 mm, Vo = Eo(d - b), Vo = p.d in open space. = (q/
ÎoA)(d - b) = (100 x 10-6 x (12 - 4)x 103)/(8.85 x 10-12 x 100 x 10-4)   = 9.04 x 106 V
(b) V = Eb = (Eo/k)b = (q/k
ÎoA)b = 1.13 x 106 V
Q5. (a) A point charge 'q' produces an electric flux of 2.0 x 104 Nm2/C and this flux is made to pass through a symmetrical Gaussian surface of radius 20 cm centered on the charge.
(i) Now if the radius of the Gaussian surface is tripled then how much flux would pass through the surface.
(ii) Find the value of 'q'
(b) There are three concentric spheres S1, S2 and S3. The sphere S1encloses a charge 'q' inside it. What is the ratio of the electric flux through S1, S2 and S3.

Ans5. (a) (i)According to Gauss's theorem flux f = E.ds = q/eo. We conclude that when charge is fixed then there is no change in the flux whether the radius is tripled or made larger any number of times.
(ii)
f =q/eo, f is 2.0 x 104.So, q = f.eo = 2 x 104 x 8.85 x 10-12
=17.7 x 10-8 C
=1.77 x 10-7 C
(b) The electric flux through a closed surface depends upon the charge enclosed by it, no matter how large is the Gaussian surface. It follows Gauss's Law
f = (q/eo). The charge 'q' is enclosed by S1, S2 and S3 all, hence f1:f2:f3 = q/eo:q/eo:q/eo i.e. 1 :1 : 1.
Q6. (a) If 20 % of the area of plates is filled up with dielectric of constant 5 and 30 % by another dielectric of constant 15, find the capacitance of the capacitor if initially it was 20 mf.
(b) In a Van-de-graaff generator the spherical shell of the metal has 2 x 106 V on its electrode. The dielectric strength of the surrounding gas is 6 x 107 V/m. Find the maximum radius of the shell.

Ans6. (a) C1 = (k1eo/d)(20/100)A, C2 = (k2eo/d)(30/100)A, C3 = (k3eo/d)(50/100)A
C = C1 + C2 + C3 =
eoA/100.d [ 20 k1 + 30 k2 + 50k3]
C= 1/100. Co.[ 20 k1 + 30 k2 + 50k3]
= 20 x 10-6/100 [20 x 5 + 30 x 10 + 50 x 15]
= 2 x 10-7 [100 + 300 + 750] = 2 x 10-7 
= 2300 x 10-7.
C = 2.3 x 10-4 F
(b) V = 2 x 106 V. Dielectric strength equal 6 x 107 V/m maximum field applied is 10% of the dielectric strength = (10/100)(6 x 107) = 6 x 106 V/m. Using E = V/r,
r = V/E = (2 x 106)/(6 x 106) = 1/3 m.
Q7. (a) Find the potential difference between the points A & B. (b) Find the charge on 5
mf capacitor? Ans. (Try yourself).
Q8 (a) Each capacitor is of 3 mf, find the total energy of the system.
(b) A capacitor of 6
mf is charged by 100 V battery and then disconnected. Now the positive plate on the battery is connected to negative terminal of the battery and the negative plate of the capacitor is connected to positive terminal of battery. What is the heat energy produced?
Ans. (Try yourself).
Q9. Discuss briefly the principle, construction and the working of Van-de-graaff electrostatic generator.
Ans. (Try yourself).
Q10. (a) Derive an expression for total work done in rotating an electric dipole through an angle q, when placed in a uniform electric field of strength E.
(b) Why is dielectric constant of water so high and what is the use of it?
Ans. (Try yourself).