Physics

CBSE Class 12

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State the law of radioactive decay.

Plot a graph showing the number (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half-life T_{½.}

Law of radioactive decay states that, the number of nuclei undergoing decay per unit time, at any instant, is proportional to the total number of nuclei present in the sample at that instant.

Number of undecayed nuclei at t = 3T_{1/2} is

And, at t = 5T_{1/2 }the number of undecayed nuclei is

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32.

(a) Obtain lens makers formula using the expression

Here the ray of light propagating from a rarer medium of refractive index (n_{1}) to a denser medium of refractive index (n_{2}) is incident on the convex side of spherical refracting surface of radius of curvature R.

(b) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the image formed.

Len’s Makers formula:

Consider a thick lens L. The refractive index of the material of lens is n_{2. }The lens is placed in a medium of refractive index n_{1}. R_{1} and R_{2} is the radius of curvature of the lens and poles are represented by P_{1} and P_{2}. Thickness of the lens is given by t. O is a point object which is placed on the principal axis of the lens.

Object distance from pole P_{1} is u.

Image formed is at a distance of v’ from P_{1}.

Now, using the refraction formula at spherical surface, we have

... (1)

Image I’ acts as a virtual object for second surface and after refraction at second surface, final image is formed at I.

Distance of I from pole P_{2} of second surface is v.

Distance of virtual object I’ from pole P_{2} is (v’ – t).

For refraction at second surface, the ray is going from second medium (refractive index n_{2}) to first medium (refractive index n_{1}), therefore from refraction formula at spherical surface.

i.e.,

Thickness of the lens is negligibly small as compared to v’.

Therefore,

... (2)

Now, adding equations (1) and (2), we have

Now, using the lens formula, we get

is the required Lens maker’s formula.

b) Ray diagram for the image formation of an object between focus (F) and pole (P) of a concave mirror is given by,

Magnification, m =

From fig.

Also,

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33.

Derive an expression for the impedance of a series LCR circuit connected to an AC supply of variable frequency.

Plot a graph showing variation of current with the frequency of the applied voltage.

Explain briefly how the phenomenon of resonance in the circuit can be used in the

Tuning mechanism of a radio or a TV set.a)

Expression for impedance in LCR series circuit:

Suppose a resistance R, inductance L and capacitance C are connected to series and an alternating voltage V = is applied across it.

Since L, C and R are connected in series, current flowing through them is the same. The voltage across R is V_{R}, inductance across L is V_{L} and across capacitance is V_{C. }The voltage V_{R} and current i are in the same phase, the voltage V_{L} will lead the current by angle 90° while the voltage V_{C} will lag behind the current by 90°.

Clearly V_{C }and V_{L} are in opposite directions, therefore their resultant potential difference =V_{C} -V_{L }(if V_{C} >V_{C}).

Thus, V_{R} and (V_{C} -V_{L}) are mutually perpendicular and the phase difference between them is 90°. As seen in the fig, we can say that, as applied voltage across the circuit is V, the resultant of V_{R} and (V_{C} -V_{L}) will also be V.

So,

Therefore, Impedance of the circuit is given by,

This is the impedence of the LCR series circuit.

b)

The below graph shows the variation of current with the frequency of the applied voltage.

c)

A radio or a TV set has a L-C circuit with capacitor of variable capacitance C. The circuit remains connected with an aerial coil through the phenomenon of mutual inductance. Suppose a radio or TV station is transmitted a programme at frequency f, then waves produce alternating voltage of frequency f, in area, due to which an emf of same frequency is induced in LC circuit, When capacitor C is in circuit is varied then for a particular value of capacitance, , the resonance occurs and maximum current flows in the circuit; so the radio or TV gets tuned.

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34.

(i) With the help of a labelled diagram, describe briefly the underlying principle and working of a step up transformer.

(ii) Write any two sources of energy loss in a transformer.

(iii) A step up transformer converts a low input voltage into a high output voltage.

Does it violate law of conservation of energy? Explain.

i) Underlying principle of a step-up transformer:

A transformer which increases the ac voltage is known as a step up transformer.

Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.

i) Underlying principle of a step-up transformer: A transformer which increases the ac voltage is known as a step up transformer.

Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.

Working: When an A.C source is connected to the ends of the primary coil, the current changes continuously in the primary coil. Hence, the magnetic flux which is linked with the secondary coil changes continuously. So, the emf which is developed across the secondary coil is same as that in the primary coil. The emf is induced in the coil as per Faraday’s law.

Assumption: We assume that there is no leakage of flux so that, the flux linked with each turn of primary coil is same as flux linked with secondary coil.

ii) Two sources of energy loss in the transformer:

Joule Heating – Energy is lost in resistance of primary and secondary windings in the form of heat.

H = I^{2} Rt

Flux leakage - Energy is lost due to coupling of primary and secondary coils not being perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil.

iii) Conservation of law of energy is not violated in step-up transformer. When output voltage increases, the output current automatically decreases. Thus, there is no loss of energy.

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35.

(a) Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism.

Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation.

(b) Explain briefly how the phenomenon of total internal reflection is used in fiber optics.

a) The phenomenon of refraction for a ray of monochromatic light passing through a glass prism is shown as below:

Let, PQR be the principal section of the prism.

The refracting angle of prism is A. Monochromatic light EF is incident on face PQ at angle of incidence i_{1}. This ray enters from a rarer to denser medium and hence is refracted towards the normal FN. Refracted ray is FG. Angle of refraction for this face is r_{1}. Refracted ray FG becomes incident on face PR and is refracted away from the normal GN_{2 }and emerges in the direction GH. The angle of incidence on this face is r_{2} (into prism) and angle of refraction (into air) is i_{2}. At point O, incident and the emergent ray meet. The angle between these two rays is called the angle of deviation ‘’.

.

... (1)

The normal FN_{1 }and GN_{2} on faces PQ and PR respectively, when produced meet at N.

Let,

... (2)

We can see that, for one angle of deviation we have two angles of incidence. But, angle of deviation is minimum for only one particular angle if incidence. Angle of minimum deviation is denoted by .

So let, i_{1} = i_{2} = i

And r_{1} = r_{2} = r

Therefore, we have (4) and (6), we have

b) An optical fiber is a device based on total internal reflection by which a light signal may be transmitted from one place to another with a negligible loss of energy. When a light ray is incident on one end at a small angle of incidence, it suffers refraction from air to quartz and strikes the quartz-coating interface at an angle more than the critical angle. Hence, total internal reflection happens and the ray of light strikes the opposite face again at an angle greater than critical angle. The phenomenon of total internal reflection takes place. Thus the ray within the fiber suffers multiple total internal reflections and finally strikes the other end at an angle less than critical angle for quartz-air interface and emerges in air.

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