Solution

Wave theory cannot explain the following laws of photoelectric effect:
(i)The instantaneous emission of photoelectrons.
(ii)Existence of threshold frequency for metal surface.
(iii)K.E. of emitted electrons is independent of the intensity of light and depends on frequency.
The concept of photon explained that energy is not only emitted and absorbed in discrete energy quanta, but also propagates through space in definite quanta with the speed of light. It can explain all the above photoelectric effect, which wave theory cannot explain.

Solution

Solution

Solution

In loop ABCFA
+80 – 20 I2 + 40 I1 = 0
4 = I2 - 2 I1
In loop FCDEA
-40 I1 -10(I1 + I2) + 40 = 0
-50 I1 - 10 I2 + 40 = 0
5 I1 + I2 = 4
Solving these two equations
I1 = 0A
& I2 = 4A
OR
The end error, in a meter bridge, is the error arising due to :
(i)Ends of the wire not coinciding with the 0 cm / 100 cm marks on the meter scale.
(ii)Presence of contact resistance at the joints of the meter bridge wire with the
metallic strips.
It can be reduced/overcome by finding balance length with two interchanged
positions of R and S and taking the average value of ‘S’ from two readings.
For a meter bridge ,
R/S=L/100-L
From the given conditions,
5/S=L/100-L (i)
5/(S/2)=1.5L/100-1.5L (ii)
Dividing (i) and (ii)
L=100/3 cm
Solving for S we get S=10Ω

Solution

(a)(i)Microwaves (10¹⁰-10¹² Hz)
(ii)Ultraviolet rays (10¹⁵ to 10¹⁷ Hz)

Solution

The wavefront is a locus of points that oscillate in phase.
Consider a plane wave AB incident at an angle ‘I’ on a reflecting surface MN

let t = time taken by the wave front to advance from B to C.
∴ BC = vt
Let CE represent the tangent plane drawn from the point C to the sphere of
radius ‘vt’ having A as its center.
then AE= BC= vt
it follows that
triangle EAC is congruent to triangle BAC
Hence,The angle of incidence = angle of reflection
OR
The refractive index of medium 2, w.r.t medium 1 equals the ratio of the sine of
angle of incidence (in medium 1) to the sine of angle of refraction (in medium 2)
Refractive index of radium 2 w.r.t medium 1
=sin i/sin r=velocity of light in medium 1/velocity of light in medium 2

Solution

Mutual inductance is property of two coils due to which each coil opposes any change of current flowing in the other. The mutually induced e.m.f, in one coil produces the opposition in other coil. Its SI unit is henry (H).
(b)

Solution

(a) Torque on current-carrying loop in a magnetic field :

Consider a coil PQRS placed in a magnetic field. Let 0 be the angle between the plane of the coil and direction of B. When current flow through coil each side experiences a force. The forces on vertical side will constitute a couple.
Moment of torque = One of forces × Perpendicular distance between lines of action of force

(b)A magnetic field, in which the plane of the coil in all positions remains parallel to the direction of magnetic field is called radial magnetic field. In a radial magnetic field, magnetic torque remains maximum for all the positions of the coil.

Solution

Solution

(a) Gauss law for magnetism : If a closed surface is imagined in a magnetic field, the number of lines of force emerging from the surface must be equal to the number entering it. That is, the net magnetic flux out of any closed surface is zero.Gauss law signifies that magnetic mono poles does not exist.
(b)(i) In a bar magnet, each lines of force, starts from a north pole and reaches the south pole externally and then goes from south pole to a north pole internally. Thus, magnetic line of force forms a closed loop.
(ii)No two lines of force will never intersect each other.
(iv )In a uniform field, the lines are parallel and equidistant from each other.
(v) The lines of force are crowded near the poles.
OR