Subject

Physics

Class

CBSE Class 12

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

31.

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 T½.

Depict in the plot the number of undecayed nuclei at (i) t = 3 T½ and (ii) t = 5 T½.


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 = 3T1/2 is fraction numerator straight N subscript straight o over denominator space 8 end fraction
And, at t = 5T1/2  the number of undecayed nuclei is straight N subscript straight o over 32

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 Multiple Choice QuestionsLong Answer Type

32.

(a) Obtain lens makers formula using the expression

 n subscript 2 over v minus n subscript 1 over u equals fraction numerator left parenthesis n subscript 2 minus n subscript 1 right parenthesis over denominator R end fraction

Here the ray of light propagating from a rarer medium of refractive index (n1) to a denser medium of refractive index (n2) 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 n2. The lens is placed in a medium of refractive index n1. R1 and R2 is the radius of curvature of the lens and poles are represented by P1 and P2. 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 P1 is u.

Image formed is at a distance of v’ from P1.

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

fraction numerator straight n subscript 2 over denominator straight v straight apostrophe end fraction minus straight n subscript 1 over straight u equals fraction numerator straight n subscript 2 straight space minus straight space straight n subscript 1 over denominator straight R end fraction                                                 ... (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 P2 of second surface is v.

Distance of virtual object I’ from pole P2 is (v’ – t).

For refraction at second surface, the ray is going from second medium (refractive index n2) to first medium (refractive index n1), therefore from refraction formula at spherical surface.

i.e.,  n subscript 1 over v minus fraction numerator n subscript 2 over denominator left parenthesis v apostrophe minus t right parenthesis end fraction equals fraction numerator n subscript 1 space minus space n subscript 2 over denominator R subscript 2 end fraction

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

Therefore, 

n subscript 1 over v minus fraction numerator n subscript 2 over denominator left parenthesis v apostrophe right parenthesis end fraction equals negative fraction numerator n subscript 2 space minus space n subscript 1 over denominator R subscript 2 end fraction                              ... (2)

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

Error converting from MathML to accessible text.

Now, using the lens formula, we get 

Error converting from MathML to accessible text.

Error converting from MathML to accessible text. 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 =fraction numerator Size straight space of straight space image left parenthesis straight A straight apostrophe straight B straight apostrophe right parenthesis over denominator Size straight space of straight space object straight space left parenthesis AB right parenthesis end fraction 

From fig. angle space A P B space equals space angle B P Q space equals space i

Also, angle space A P B space equals space angle B P Q space equals space i
I n space increment A P B comma space tan space i space equals space fraction numerator A B over denominator B P end fraction italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic. italic. italic. italic space italic left parenthesis italic 1 italic right parenthesis
I n space increment A apostrophe P B apostrophe comma space tan space i space equals fraction numerator A apostrophe B apostrophe over denominator B apostrophe P end fraction space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space
N o w italic comma italic space f r o m italic space e q u a t i o n italic space italic left parenthesis italic 1 italic right parenthesis italic space a n d italic space italic left parenthesis italic 2 italic right parenthesis italic comma italic space w e italic space h a v e

fraction numerator A B over denominator B P end fraction equals fraction numerator A apostrophe B apostrophe over denominator B apostrophe P end fraction

rightwards double arrow space Magnification comma space fraction numerator A apostrophe B apostrophe over denominator A B end fraction equals fraction numerator B apostrophe P over denominator B P end fraction

straight i. straight e. comma space straight m space equals negative straight v over straight u
So comma space straight m space equals space minus straight v over straight u
Hence comma space proved. space

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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 = straight V subscript straight o space sinωt space 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 VR, inductance across L is VL and across capacitance is VC. 

The voltage VR and current i are in the same phase, the voltage VL will lead the current by angle 90° while the voltage VC will lag behind the current by 90°. 

Clearly VC and VL are in opposite directions, therefore their resultant potential difference =VC -V(if VC >VC).

Thus, VR and (VC -VL) 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 VR and (VC -VL) will also be V. 

So, 

V squared space equals space V subscript R squared space plus space left parenthesis V subscript c minus V subscript L right parenthesis squared italic space

rightwards double arrow space V space equals space square root of V subscript R to the power of italic 2 space italic plus space italic left parenthesis V subscript c italic minus V subscript L italic right parenthesis to the power of italic 2 end root

B u t italic comma italic space V subscript R space equals space R i comma space V subscript c space equals space capital chi subscript c i space a n d space V subscript L space equals space capital chi subscript L i

w h e r e italic comma italic space capital chi subscript c space equals space fraction numerator italic 1 over denominator omega C end fraction space a n d space capital chi subscript L space equals space omega L
S o italic comma italic space V space equals space square root of italic left parenthesis R i italic right parenthesis to the power of italic 2 italic plus italic left parenthesis capital chi subscript c i space italic minus space capital chi subscript L i space italic right parenthesis to the power of italic 2 end root italic comma italic space 

Therefore, Impedance of the circuit is given by, 

Error converting from MathML to accessible text.
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, C italic comma italic space f space equals space fraction numerator 1 over denominator 2 pi square root of L C end root end fraction, 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 = I2 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 i1. 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 r1. Refracted ray FG becomes incident on face PR and is refracted away from the normal GN2 and emerges in the direction GH. The angle of incidence on this face is r2 (into prism) and angle of refraction (into air) is i2. At point O, incident and the emergent ray meet. The angle between these two rays is called the angle of deviation ‘straight delta’. 

angle OFG space equals straight i subscript 1 space minus space straight r subscript 1 space space and space angle OGF space equals space straight i subscript 2 space minus space straight r subscript 2
increment F O G comma space delta space i s space t h e space e x t e r i o r space a n g l e.

delta space equals space angle O F G plus angle O G F space

italic space italic space equals space left parenthesis i subscript 1 space minus space r subscript 1 right parenthesis plus left parenthesis i subscript 2 space minus space r subscript 2 right parenthesis space equals space left parenthesis i subscript 1 plus i subscript 2 right parenthesis minus left parenthesis r subscript 1 plus r subscript 2 right parenthesis                    ... (1) 

The normal FN1 and GN2 on faces PQ and PR respectively, when produced meet at N.

Let, angle space F N G space equals space theta

r subscript 1 space plus space r subscript 2 space plus theta space equals space 180 to the power of o                              ... (2) 
I n space q u a d r i l a t e r a l space P F N G comma space angle P F N space equals space 90 to the power of o

therefore space A plus space 90 degree plus theta plus 90 degree space equals space 360 degree

space italic space italic space italic space A plus theta space equals space 180 degree space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis

Comparing space equations space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis comma space we space get space

straight r subscript 1 space end subscript plus space straight r subscript 2 space end subscript equals space straight A space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis space space space space space space space space space space space space space space
Putting space this space in space equation space left parenthesis 1 right parenthesis comma space we space get

delta space equals space i subscript 1 plus i subscript 2 space minus space A italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space space space... space left parenthesis 5 right parenthesis space space space

rightwards double arrow space i subscript 1 plus i subscript 2 space equals space A plus delta italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space... space left parenthesis 6 right parenthesis

Now comma space using space Snell ’ straight s space law comma space we space get

n space equals space fraction numerator sin i subscript 1 over denominator sin r subscript 1 end fraction equals fraction numerator sin i subscript 2 over denominator sin r subscript 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 7 right parenthesis space

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 delta subscript m.

So let,                            i1 = i2 = i

And                               r1 = r2 = r

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

space space space space space straight r plus straight r space equals space straight A space space

rightwards double arrow space r space equals space bevelled A over 2 a n d space i plus i equals space A plus delta subscript m

rightwards double arrow space i equals fraction numerator A italic plus delta subscript m over denominator italic 2 end fraction 

Error converting from MathML to accessible text.

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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