Physics Sample Paper with General Instruction for Class - 12

8/18/2019 Physics Sample Paper with General Instruction for Class - 12

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SECTION - A

1. Give the ratio of velocities of light rays of wavelengths 4000Å and 8000 Å in vacuum.

2. How is the intensity of the central maximum affected, if the width of slit is doubled in a single slit diffraction experiment?

3. What is the significance of the negative energy of the electron in the orbit ?

4. State two disadvantages of semiconductor devices.5. Draw schematic representation of p–type material with acceptor impurity.

SECTION - B

6. Draw a graph showing the variation of electric field, as one moves from the centre of a charged metal ball to a point on its surface

and then to a far off outside point.

OR

Two metallic spheres of same radii, one solid and the other hollow are charged to the same potential. Which of the two will hold

more charge ? Give reasons.

7. Write the expression for the magnitude of force per unit length between two infinitely long parallel, straight current carrying

conductors. Hence define the SI unit of current.

8. How are electromagnetic waves produced?

9. A photon and e – have same deBroglie wavelength 10( 10 m) . Which has greater kinetic energy? Explain in brief.

10. Descirbe in brief about the sky wave propagation.

SECTION - C

11. A charge +Q is fixed at the origin of the co-ordinate system while a small electric dipole of dipole moment p

pointing away from

the charge along the x-axis is set free from a point far away from the origin.

(a) Calculate the K.E. of the dipole when it reaches to a point (d , 0).

(b) Calculate the force on the charge +Q at this moment.

GENERAL INSTRUCTIONS

1. All questions are compulsory.

2. There are 26 questions in total. Q. no. 1 to 5 carry 1 mark each, Q. 6 to 10 carry 2 marks each, Q. 11 to 22 carry 3 marks each,

Q. 23 is a value based question carry 4 marks, Qs. 24 to 26 carry 5 marks each.

3. 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 given choices in such questions.

4. Use of calculator is not allowed.

5. You may use the following physical constants wherever necessary :c = 3 × 108 ms –1, h = 6.6 × 10 –34 Js, e = 1.6 × 10 –19 C, µ0 = 4 × 10

–7 Tm A –1.

Boltzmann constant k = 1.38 × 1023 JK –1, Avogadro's number NA = 6.023 × 1023 /mole,

Mass of neutron mn = 1.6 × 10 –27 kg

SAMPLE PAPER (PHYSICS)

Time : 3 Hrs. Max. Marks : 70


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12. A solid sphere of radius R has a charge Q distributed in its volume with a charge density = k r a , where k and a are constants

and r is the distance from its centre.

If the electric field at1

is2 8

R

r times that at r = R, find the value of a.

13. Consider n cells connected in series in a row and m such rows connected in parallel. Obtain an expression for the maximum

current from such a combination.

14. Use Lenz’s law to determine the direction of induced current in the situations described by Fig.

(a) A wire of irregular shape turning into a circular shape

(b) A circular loop being deformed into a narrow straight wire.

a

b

c

d

(b)

b ' d '

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

(a)

a b

c

d a '

c '

15. A coil of inductance L, a capacitor of capacitance C and a resistor of resistance R are connected in series with an alternating

source of emf 0E E sin t . Write expressions for

(i) total impedance of circuit

(ii) frequency of source emf for which circuit will show resonance.16. Explain with the help of lens maker’s formula. Why does a convex lens behave as -

(i) a converging lens when immersed in water ( = 1.33) and

(ii) a diverging lens when immersed in CS2 solution ( = 1.6) ?

17. Two polaroids are placed at 90° to each other and the transmitted intensity is zero. What happens when one more polaroid is

placed between these two bisecting the angle between them ? What will be the direction of polarization of outcoming beam ?

18. If a n0 is unstable with half life of 917 sec. Why don’t all n0 s of a nucleus decay eventually into p+ s? How does a nucleus of

Z protons and (A – Z) neutrons remain stable?

OR

What is impact parameter and angle of scattering ? How are they related to each other ?

19. For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2V.

Determine the modulation index . What would be the value of if the minimum amplitude is zero volt ?

20. Rays of red and blue light are incident on a given prism. Explain, which will have the larger value of m the angle of minimum

deviation.

21. Derive an expression for the a.c. across a resistance R connected to an alternating source of e.m.f. E = E0 sin t. Explain the

variations of e.m.f. and current graphically and with a phasor diagram.

22. Draw an energy level diagram for hydrogen atom and with the help of an arrow, show the transition which give the H-line in the

emission spectrum of hydrogen.


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SECTION - D

23. Mr. Arvind who is a farmer by profession wanted to pay electricity bill. He realized that the consumption shown by the meter

was unbelievably high. He thought that the meter must have been faulty. He wanted to check the meter. But unfortunately he

did not have any idea as to how to do this. There came his friend Ravi to help him. He told Arvind to run only the electric heater

rated 1500W in his house for some time keeping other appliances switched off. He also calculated the power consumed in

kilowatt hour and compared the value with the meter. Arvind was happy and thanked Ravi for his timely help and the knowledge.

(i) What are the values displayed by Ravi?

(ii) Express kWh in joules. Find the resistance of the heater coil.

SECTION - E

24. In an intrinsic semiconductor, explain how current flow takes place?

OR

Describe : (i) NAND gate, and (ii) NOR gate.

25. (a) What is interference of light? In Young's double slit experiment, show that2max2

min

I (a b)

I (a b)

(b) The fringe width at a distance of 50 cm from the slits in Young's double-slit experiment for the light of wavelength 6000 Å

is 0.042 cm. What will be the fringe-width at the same distance for the light of wavelength 5000 Å?

OR

(a) Light from an ordinary source (say a sodium lamp) is passed through a polaroid sheet P1. The transmitted light is then

made to pass through a second polaroid sheet P2 which can be rotated so that the angle () between the two polaroid

sheets varies from 0° to 90°. Show graphically the variation of the intensity of light transmitted by P1 and P2, as a function

of angle . Take the incident beam intensity as I0. Why does the light from a clear blue portion of the sky, show a rise and

fall of intensity when viewed through a polaroid which is rotated ?

(b) A biconvex lens with both faces of the same radius of curvature is to be manufactured from a glass of refractive index 1.55.

What should be the radius of curvature of the focal length of the lens to be 20 cm ?

26. Derive an expression for the torque on a rectangular coil of area A, carrying a current I placed in a magnetic field B. The angle

between the direction of B and the vector perpendicular to the plane of the coil is .

OR

Describe the motion of a charged particle in a uniform magnetic field. Obtain an expression for the radius of the path of the

charged particle moving perpendicular to uniform magnetic field. Show that the time taken to complete one revolution by the

particle is independent of its speed.

v si n B

cos O

v

v


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1. The velocity of light rays of different wavelengths in vacuum is same and hence the ratio of their velocities is 1. (1 mark)

2. The intensity of the central maximum increased by four times. (1 mark)

3. The negative energy of the electron in the orbit signifies the fact that the nucleus and the electron form a bound system.(1 mark)

4. 1. The semiconductor devices are temperature sensitive. Even a small overheating may cause damage to them. (½ mark)

2. The semiconductor devices cannot handle as much power as ordinary vacuum tubes can do. (½ mark)

5. Common p–type semiconductor with negatively charged acceptor atom and associated hole is shown as. (1 mark)

Hole

6. Charged Metal ball

E

x

O

(2 marks)

OR

Both the spheres will hold the same amount of charge. (1 mark)It is because, the two spheres possess equal capacitance. The capacitance of a sphere depends only on its radius. It does notmatter, whether the sphere is hollow or solid. (1 mark)

7. For two infinitely long parallel straight conductors kept at a distance r apart in vacuum and carrying currents I1and I2,

Force per unit length of the conductor = 0 1 2µ I IF

2 r

(1 mark)

S.I. unit of current is 1 A.

If I1 = I2 = 1A, r = 1m then7F 2 10 N / m

70µ 104

1 A is the amount of current flowing through two parallel wires separated by a distance of 1m, carrying a force of interactionof 2 × 10 –7 N/m between them. (1 mark)8. Electromagnetic waves are produced by accelerated charges. An oscillating charge produces an oscillating electric field

in space, which produces an oscillating magnetic field. Both the fields regenerate each other as the wave propagates throughthe space. (2 marks)

9. For e – of wavelength , K.E. = E1 =1

2mv2

or mv2 = 2E1 (1½ marks)

or m2v2 = 2mE1 mv = 12mE

As

1

h h

mv 2mE or

2

1 2

hE

2 m

SOLUTI ONS


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For photon of wavelength , Energy is 2hc

E

22 2

1

E hc / 2c m

E hh / 2 m

Substitute the values of c, , m, h, we get8 10 312

341

E 2 3 10 10 9 10E 6.6 10

2 3 9 10 90

16.6 1.1

E2 > E1 K.E. of photon is greater than e – . (½ mark)

10. Long distance communication of frequencies (few MHz to 40 MHz) is done by means of reflection of these waves fromionosphere of earth. This is sky wave propagation and is used for short wave broad -cast services. The density of ionised particle in ionosphere varies with height. There is a layer where ionised particle density is maximum (~300 km) which reflectsthese high frequency waves. (2 marks)

11. (a) Potential energy of the dipole-charge system

U i = 0 (since the charge is far away) (½ mark)

U f = – Q × 2

0

1

4

p

d (½ mark)

K.E. = | U f – U i | = 20

1

4

pQ

d (½ mark)

(b) Electric field at origin due to dipole3

0

1 2

4

p E i

d

(½ mark)

Thus, force on charge Q is given by3

0

2

4

pQF QE i

d

(1 mark)

12. Let us consider a spherical shell of radius x and thickness dx.

The volume of this shell is 4 x2(dx). The charge enclosed in this spherical shell is

2(4 ) adq x dx kx

24 adq kx dx .

x

dxFor r = R :

The total charge enclosed in the sphere of radius R is

32

0

4 43

R aa R

Q k x dx k a

.

The electric field at r = R is3

11 2

0 0

1 4 1 4

4 4 3(3 )

aakR k

E Raa R

(1 mark)

For r = R /2 :

The total charge enclosed in the sphere of radius R/2 is

/ 2 32

0

4 ( / 2)' 4

3

R aa k R

Q k x dxa

The electric field at r = R/2 is13

2 20 0

1 4 ( / 2) 1 4

4 3 4 3 2( / 2)

aak R k R

E a a R

(1 mark)

Given, 2 11

8 E E

1

1

30 0

1 4 1 1 4

4 (3 ) 2 4 32

aak R k

Ra a

1 + a = 3 a = 2 (1 mark)


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13. Total e.m.f = nE, Total resistance = R +nr

m(1 mark)

R

I

m rows

n cells

I(1½ marks)

[ Total internal resistance p

1 1 1

r nr nr + .......... m times =

m

nr r p =

nr

m

Current = I =nE

nr R

m

I =mnE

mR nr I will be maximum if mR + nr is minimum.

That is possible if mR = nr i.e. R =nr

mi.e. External resistance of the circuit = Total internal resistance of all the cells. (½ mark)

14. (a) As wire increases its area, the magnetic flux linked with the loop increases, the induced emf causes a current to oppose it.The force should be inward (to pull the wire back) hence direction of current is adcba (1½ marks)

(b) Here, the area decreases, hence current induced opposes this decrease in field and would try to increase the flux by pulling

wire outward hence current is in direction a 'd 'c 'b 'a ' . (1½ marks)

15. (i) Impedance of circuit, Z = 2 2L CR (X X ) =2

2 1R LC

(1 mark)

(ii) For resonance XL = XC 1

LC

(1 mark)

2 1

LC 1

LC

1 12 f f

LC 2 LC

(1 mark)

16. As 2

1 1 2

1 1 11

f R R

(1 mark)

for lens material 2 = 1.5

(i) If the convex lens is immersed in water ( = 1.33) its focal length will be positive hence it behaves as converging lens.(1 mark)

(ii) If the convex lens is immersed CS2 solution ( = 1.6) its focal length will be negative hence it behaves as diverging lens.(1 mark)

17. If E is amplitude of electric field component emanating from 1st polaroid, then from 2 nd polaroid at 45°, the amplitude of electric

field component is E = E cos 45° = E / 2 . (1 mark)Again amplitude of electric field component coming from 3rd polaroid at 45° to 2nd polaroid would be

E = E cos 45° =1

22 2

E E half of E (1 mark)


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As intensity E 2

Intensity transmitted from three polaroids will be 1/4th of the intensity transmitted from the first polaroid. The direction of polarization of the outcoming beam will be same as that coming from the first polaroid. (1 mark)

18. A free n0 has rest mass greater than p+. Thus – decay is allowed. But + decay of free p+ is not allowed i.e.,1 1 0

1 0 1H n e is not allowed. (1 mark)In a nucleus, p+ and n0 are not free hence + decay of p+ takes place side by side with –decay of n0 in the nucleus. The energyrequired for + decay comes from difference in binding energy of p+ and n0 individually in the nucleus.

(1 mark)In a stable nucleus of Z p+ s and (A – Z) n0 s both decays occur in dynamic equlibrium. (1 mark)

OR

Impact parameter is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the central lineof the nucleus, when the particle is far away from the nucleus of the atom.Fig. shows the path or trajectory of an alpha particle in the coulomb field of a heavy nucleus. The impact parameter b and scattering angle are also shown in the diagram.

(1 mark)

For large impact parameters, force experienced by the alpha particle is weak, because F varies as1/(distance)2. Therefore, when impact parameter is large, an alpha particle will deviate through a much smaller angle. However,when impact parameter is small, force experienced is large and hence the alpha particle will scatter through a large angle. For thecase of head on collision, impact parameter b tends to zero. The alpha particle will rebound like a ball thrown against a wall,scattering through 180°.Fig. shows theoretically calculated paths of alpha particles moving with a speed of 1.63 × 107 m/s in the coulomb field of a gold nucleus. The gold nucleus is supposed to be at the origin O of the co-ordinate system. The values of impact parameters chosenfor alpha particles 1, 2, 3, 4 respectively are 2.5 fm, 10 fm, 20 fm and 100 fm.

(1 mark)

Rutherford calculated analytically, the relation between the impact parameter b and scattering angle , which is given by2

0

1 cot / 2

4

Zeb

E

(1 mark)

where21

2 E mv is kinetic energy of alpha particle, when it is far away from the atom.

19. Let Ac and Am be the amplitude of carrier wave and message signal wave. Given, Amax = Ac + Am = 10 V ...(i) Amin = Ac – Am = 2 V ...(ii)

On solving, we get, Ac = 6 V; Am = 4 V (1 mark)


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Amax = 6 + 4 = 10 V Amin = 6 – 4 = 2 V

Modulation index, =

max min

max min

10 2 8 2

10 2 12 3

A A

A A

(1 mark)

When, Amin = 0, then =max

max

1 A

A (1 mark)

20. For a prism of small angle, the angle of deviation is given by = A ( – 1) (1 mark)

r = A (r – 1)and b = A (b – 1) (1 mark)Since b > r it follows that b >r . (1 mark)Thus, blue light will have larger value of the angle of minimum deviation.

21.

VR

R

Let VR be the instantaneous voltage drop across R.E is the applied alternating e.m.f. to the circuit. E0 is the maximum voltage.

0E E sin t IR

0 0E

I sin t I sin tR

(1½ marks)

where 00E

IR

= Maximum value of current.


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n =n = 6n = 5n = 4

n = 3

n = 2

n = 1

H line(Balmer Series)

0

–0.5 –0.85

–1.51

–3.4

–13.6

n

E

(1 mark)

H line corresponds to transitions from n = 3 to n = 2 level i.e., it is the first member of Balmer series. (1 mark)23. (i) Use of scientific knowledge, open mindedness, helpful (1 mark)

(ii) 1 Wh= 3.6 × 106 (1 mark)

Resistance of heater coil, R = V2 /220 220

P 32.251500

(2 marks)

24. In an intrinsic semiconductor, each of the 4 valence electrons is between two atoms (Si or Ge) in a shared covalent bond. It is bound at low temperatures, but at high temperatures it can pick thermal energy and move out of the valence bond and into theinterstitial space. This electron is free to conduct. The vacancy left behind by it is called a hole and it also conducts charge.

(1 mark)Another electron from a different bond can come and occupy this vacancy, hence creating a vacancy elsewhere and causingmotion of bound electrons. These holes move towards the negative potential giving rise to hole current Ih. (1 mark)The total current I, is sum of hole current Ih and electron current Ie. (1 mark)

Si

Si

Si

Si

Si

Si

Si

Si Si

free electron (I )e

hole site

bound e movesinto a hole

–

(I )h

e –

(2 marks)

OR

(i) The NAND gate. If we connect the output ‘Y’ of AND gate to the input of a NOT gate, shown in Fig.(1) (a), the gate soobtained is called NAND gate. The logic symbol of NAND gate is shown in Fig. 1 (b)

In Boolean expression, the NAND gate is expressed as . ,Y A B and is being read as ‘ A AND B negated’. The truth tableof NAND gate can be obtained by logically using the truth table of AND and NOT gates as shown in Fig 1 (c) and its finaltruth table is shown in Fig. 1 (d). (1 mark)

(½ mark)

(½ mark)

Fig (1)


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Circuit for NAND gate is as shown in Fig. 2.

Y

(½ mark)

Fig (2)

(ii) The NOR gate. If we connect the output Y of an OR gate to the input of the NOT gate as shown in Fig 3 (a). The gate soobtained is called the NOR gate .

In Boolean expression, the NOR gate is expressed as Y = A B and is being read as ‘ A OR B negated’. The logic symbolof NOR gate is shown in Fig. 3 (b). (1 mark)

(½ mark)

Fig. 3

The truth table of NOR gate can be obtained by logically using the truth table of OR and NOT gates as shown in Fig. 4 (a)

& (b)

(½ mark)

Fig. 4

Circuit for NOR gate is as shown in Fig. 4 (c)

(½ mark)

Fig. 4 (c)


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25. (a) The phenomenon of redistribution of light energy in a medium due to the superposition of light waves from two coherentsources is called interference.Suppose two waves of amplitudes a and b respectively and a phase difference , travelling in a medium, superimpose eachother.y1= a sint; y2= b sin(t + )According to superposition principle,y = y1 + y2 = a sint + b sin(t + )

= a sint + b sint .cos + b cos t. sin (1 mark)y = (a + b cos sint + b sin .cost ....(i)Put, (a + b cos) = R cos ....(ii)

b cos = R sin ....(iii) y = R cos.sint + R sin .costor y = R sin (t + ) ...(iv)This is the displacement of the resultant wave where R is the amplitude of the wave.Squaring and adding eqn. (ii) and (iii)

R = 2 2a b 2ab cos (1 mark)

As, intensity (amplitude)2 I R 2

I = a2 + b2 + 2ab cosIf cos = + 1, the intensity is maximum, Imax = (a+b)

2

If cos = – 1, the intensity is minimum.Imin= (a – b)

2

2max

2min

I (a b)

I (a b)

(1 mark)

(b) =1D

d

and 2=

2D

d

2 21 1

(1 mark)

2=2

11

=5000

0.0426000

= 0.035 cm (1 mark)

OR

(a) If the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P1, it is observed that its intensityis reduced by half. Rotating P1 has no effect on the transmitted beam and transmitted intensity remains constant.Let an identical piece of polaroid P2 be placed before P1, the light from the lamp is reduced in intensity on passing throughP2 alone. But rotating P1 has effect on the light coming from P2.In one position, the intensity transmitted by P2 followed by P1 is nearly zero. When turned by 90° from this position, P1transmits nearly the full intensity emerging from P2 as shown in Fig. (1 mark)

P2

P1

P2

P1

P2

P1

(1 mark)

(Angle 0°) (Angle 90°)


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Graph : The graph between intensity of light and the angle between polariser and analyser is shown as : (½ mark)

90° 180°0°

I0

I

When the polaroid is rotated in the path of plane polarised light, its intensity will vary from maximum (When thevibrations of the plane polarised light are parallel to the axis of the polaroid) to minimum (When the direction of vibrations becomes perpendicular to the axis of the crystal). (1 mark)

(b) Here, µ = 1.55 ; f = + 20 cm ;Suppose that R 1 = + R, then R 2 = – R

Now,1 2

1 1 1( 1)

f R R

(½ mark)

1 1 1 2

(1.55 1) 0.5520 R R R

or R = 40 × 0.55 = 22.0 cm (1 mark)

26.

P II

I

I

S

R

B

F1

F2

F3

F4

B

B

B

Q

90°

b cos

T

P N

(F )=BIl3S

B

(F )=BIl1

(1½ marks)

Here PQ = RS = = length of the coil

QR = SP = b = breadth of the coil

is the angle between plane of the coil and B

.

1 2 3 4F ,F , F , F

are forces on four arms PQ, QR, RS and SP respectively..

4F I(SP B)

IB (SP) sin (180° – ) = IbB sin

2F I(QR B)

F2 = IB (QR) sin = IbB sin They are equal in magnitude but opposite in direction, and will cancel each other.

1F I(PQ B)

F1 = IB (PQ) sin 90°

= I B ( PQ B

)

This is perpendicular to the plane of the coil and directed outwards.

3F I(RS B)

F3 = I(RS) B sin 90° (1½ marks)

= I B ( RS B

)

This is perpendicular to the plane of the coil and directed inwards.They are equal, parallel but oppositely directed along their line of action, so they will form a couple which will try to rotate thecoil in anticolockwise direction about its axis.Torque on the coil = moment of the couple


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= F r

I B bcos [ perpendicular distance between the force = b cos ] (1 mark)

= IBA cos where A = b = area of the coilIf the coil has n number of turns, = nIBA cos .

If the angle between the normal on the plane of the coil and B

is then + = 90° = 90° – cos = cos (90° – ) = sin

= nIBA sin = MB sin = | M B |

(1 mark)

When M = magnetic dipole moment of the coil = nIAOR

Let a charged particle is moving in the magnetic field with a velocity v

making angle with B

.

The component of v

along B

is v cos, due to which no force will act on the particle. so it will cover a distance along the

magnetic field with a constant speed.

v sin B

cos O

v

v

The perpendicular component v sin will provide a force,F = qBvsin which provide the necessary centripetal force for the circular motion of the charged particle

2m(vsin )

Bqvsinr

(1 mark)

Bqr

vsinm

and r = radius of the path =mvsin

Bq

(1½ mark)

If sin = 1, i.e. = 90° thenmv

r Bq

Angular velocity of rotation of the particle in magnetic field =vsin Bq

r m

Time taken to complete one revolution =2

T

(1½ marks)

2 m

TBq

which is independent of its speed.

Under the combined action of both the velocities the charged particle will undergo a linear as well as a circular motion. So theresultant path will be a helix. (1 mark)


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16. Name three incurable sexually transmitted diseases and their causative organisms.

17. What are the advantages of using biofertilisers in agriculture?

18. (i) Name the substance used as a medium/matrix in gel electrophoresis.

(ii) Why does DNA move towards the anode in gel electrophoresis?

(iii) Name the compound used for staining the isolated DNA in the gel electrophoresis.

19. (i) What is cloning vector?

(ii) Explain any two methods of vectorless gene transfer.

20. (i) Give the scientific name of the soil bacterium which produces crystal (Cry) proteins.

(ii) How are these proteins useful in agriculture?

(iii) What do the differently written terms ‘Cry’ and ‘cry’ represent respectively?

21. What is meant by the term ‘hot spots’ in biodiversity? List two criteria used for determining a ‘hot spot’. Name two hot spots of

India.

22. What is eutrophication? Explain its consequence on the life of plants and animals in such water bodies.

ORWhat is biological magnification? How does DDT as a water pollutant undergo biological magnification?

SECTION - D

23. Kavita was very happy when she gave birth to her first child. But her in-laws were unhappy and blamed on Kavita to give a female

child. Kavita tried to convince them that she had no role in the child’s gender. Then Kavita’s husband took up the matter and

convinced the parents.

(i) What values did Kavita’s husband show in the above situation ?

(ii) What governs sex determination in humans?

(iii) Why can’t Kavita be blamed for not giving birth to a male child?

SECTION - E

24. Draw a labelled diagram of the sectional view of a mature pollen grain in angiosperms. Explain the functions of its different parts.

OR

(i) What is the role of cervix of the human female system in reproduction?

(ii) Distinguish between spermatogenesis and oogenesis.

25. In garden pea (Pisum sativum), a plant with red flowers was crossed with a plant with white flowers. Work out the possible

genotypes and phenotypes of F1 and F2 generations. State any one of Mendel’s law that could be derived from this cross.

OR

Mention any two autosomal genetic disorders with their symptoms.

26. (i) How normal cells get transformed into cancerous neoplastic cells? Mention the difference between viral oncogenes and

cellular oncogens.

(ii) Write a short note on detection of cancer.

OR

(i) The diagram given above is that of a typical biogas plant. Identify A, B and C. Explain the sequence of events occurring in it.

(ii) Name the institutes which helped in the development of biological fuel technology in India.

Documents

Physics Sample Paper with General Instruction for Class - 12