Note: For this part students must share their own experience.

Size: px

Start display at page:

Download "Note: For this part students must share their own experience."

Virgil Fleming
5 years ago
Views:

1 26. White travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm. It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenly he noticed a child walking alone on the road. He asked the boy to come inside the car till the thunderstorm stopped. Dr. Pathak dropped the boy at his residence. The body insisted that Dr. Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for his concern for safety of the child. Answer the following questions based on the above information: (a) Why is it safer to sit inside a car during a thunderstorm? Which two values are displayed by Dr. Pathak in his action? (c) (d) Which values are reflected in parents' response to Dr. Pathak? Give an example of similar action on your part in the past from everyday life. Ans. (a) It is safer to be inside a car during thunderstorm because the car acts like a Faraday cage. The metal in the car will shield you from any external electric fields and thus prevent the lightning from traveling within the car. (c) (d) Awareness and Humanity Gratitude and obliged I once came across to a situation where a puppy was struck in the middle of a busy road during rain and was not able go cross due to heavy flow, so I quickly rushed and helped him. Note: For this part students must share their own experience. 27. (a) State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf's of two primary cells. Obtain the required expression used for comparing the emfs. Write two possible causes for one sided deflection in a potentiometer experiment. OR (a) State Kirchhoff's rules for an electric network. Using Kirchhoff's rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge. In the meterbridge experimental set up, shown in the figure, the null point D is obtained at a distance of 40 cm from end A of the meterbridge wire. If a resistance of 10 is connected in series with R 1, null point is obtained at AD = 60 cm. Calculate the values of R 1 and R 2. Ans. (a) Working principle of Potentiometer:

When a constant current is passed through a wire of uniform area of crosssection, the potential drop across any portion of the wire is directly proportional to the length of that portion.

2 When a constant current is passed through a wire of uniform area of crosssection, the potential drop across any portion of the wire is directly proportional to the length of that portion. Applications of Potentiometer for comparing emf s of two cells: The following Figure shows an application of the potentiometer to compare the emf of two cells of emf E 1 and E 2 E 1, E 2 are the emf of the two cells. 1, 2, 3 form a two way key. When 1 and 3 are connected, E 1 is connected to the galvanometer (G). Jokey is moved to N 1, which is at a distance l 1 from A, to find the balancing length. Applying loop rule to AN 1 G31A, Φ l E 1 = 0..(1) Where, Φ is the potential drop per unit length Similarly, for E 2 balanced against l 2 (AN 2 ), Φ l E 2 = 0..(2) From equations (1) and (2), E E l.. (3) l Thus we can compare the emf s of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from Eq. (3).

3 (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared. (ii) The positive ends of all cells are not connected to the same end of the wire. OR (a) Kirchhoff s First Law Junction Rule The algebraic sum of the currents meeting at a point in an electrical circuit is always zero. Let the currents be I 1, I 2 I 3, and I 4 Convention: Current towards the junction positive Current away from the junction negative I 3 + ( I 1 ) + ( I 2 ) + ( I 4 ) = 0 Kirchhoff s Second Law Loop Rule In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and current flowing through them.

For closed part BACB, E 1 E 2 = I 1 R 1 + I 2 R 2 I 3 R 3 For

Bridge: The Wheatstone Bridge is an arrangement of four

. R 1, R 2, R 3, and R 4 are the four resistances.

4 For closed part BACB, E 1 E 2 = I 1 R 1 + I 2 R 2 I 3 R 3 For closed part CADC, E 2 = I 3 R 3 + I 4 R 4 + I 5 R 5 Wheatstone Bridge: The Wheatstone Bridge is an arrangement of four resistances as shown in the following figure.. R 1, R 2, R 3, and R 4 are the four resistances. Galvanometer (G) has a current I g flowing through it at balanced condition, I g = 0 Applying junction rule at B, I 2 = I 4

5 Applying junction rule at D, I 1 = I 3 Applying loop rule to closed loop ADBA, Applying loop rule to closed loop CBDC, From equations (1) and (2), This is the required balanced condition of Wheatstone Bridge. Considering both the situations and writing them in the form of equations Let R' be the resistance per unit length of the potential meter wire,

6 28. (a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field. A proton and a deuteron having equal momenta enter in a region of a uniform magnetic field at right angle to the direction of a the field. Depict their trajectories in the field. OR (a) A small compass needle of magnetic moment m is free to turn about an axis perpendicular to the direction of uniform magnetic field B. The moment of inertia of the needle about the axis is I. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period. A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of (i) horizontal component of earth's magnetic field and (ii) angle of dip at the place. Ans. (a) Consider a rectangular loop ABCD carrying current I.

Case I - The rectangular loop is placed such that the uniform

No force is exerted by the magnetic field on the arms AD and BC.

F 1 = IbB Magnetic field exerts a force F 2 on arm CD.

7 Case I - The rectangular loop is placed such that the uniform magnetic field B is in the plane of loop. No force is exerted by the magnetic field on the arms AD and BC. Magnetic field exerts a force F 1 on arm AB. F 1 = IbB Magnetic field exerts a force F 2 on arm CD. F 2 = IbB = F 1 Net force on the loop is zero. The torque on the loop rotates the loop in anti-clockwise direction. Torque, τ =

= I(ab)B τ = BIA If there are n such turns the torque will be niab where, b Breadth of the rectangular coil a Length of the rectangular coil A = ab Area of the coil Case II - Plane of the loop is not

8 = I(ab)B τ = BIA If there are n such turns the torque will be niab where, b Breadth of the rectangular coil a Length of the rectangular coil A = ab Area of the coil Case II - Plane of the loop is not along the magnetic field, but makes angle with it. Angle between the field and the normal is θ. Forces on BC and DA are equal and opposite and they cancel each other as they are collinear. Force on AB is F 1 and force on CD is F 2. F 1 = F 2 = IbB Magnitude of torque on the loop as in the figure:

We know, Lorentz force, F = Bqv sinθ where θ = angle between velocity of particle and magnetic field = 90 o So, Lorentz force, F = Bqv Thus the particles will move in circular path.

9 We know, Lorentz force, F = Bqv sinθ where θ = angle between velocity of particle and magnetic field = 90 o So, Lorentz force, F = Bqv Thus the particles will move in circular path. 2 mv mv Bqv r r Bq Let m p = mass of proton, m d = mass of deuteron, v p = velocity of proton and v d = velocity of deuteron The charge of proton and deuteron are equal. Given that m p v p = m d v d mv p p rp... 1 Bq r d mv d Bq d... 2 As 1 and 2 are equal, so r r r p d Thus, the trajectory of both the particles will be same.

10 Or, The torque on the needle is = m B In magnitude = mb sin Here is restoring torque and is the angle between m and B. 2 d Therefore, in equilibrium I mbsin 2 dt Negative sign with mb sin implies that restoring torque is in opposition to deflecting torque. For small values of in radians, we approximate sin = and 2 d get I mb 2 dt 2 d Or, mb 2 dt I This represents a simple harmonic motion. The square of the angular frequency is 2 = mb/j and the time period is, T 2 I mb (i) As, Horizontal component of earth s magnetic field, B H = B cosδ Putting δ = 90 0 B H = 0 (ii) For a compass needle align vertical at a certain place, angle of dip, δ = 90 0.

29. (a) Draw a ray diagram showing the image formation by a compound microscope. Hence obtained expression for total magnification when the image is formed at infinity.

11 29. (a) Draw a ray diagram showing the image formation by a compound microscope. Hence obtained expression for total magnification when the image is formed at infinity. Distinguish between myopia and hypermetropia. Show diagrammatically how these defects can be corrected. (a) OR State Huygen's principle. Using this principle draw a diagram to show how a plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snell's law of refraction. When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons: (i) Is the frequency of reflected and refracted light same as the frequency of incident light? (ii) Does the decrease in speed imply a reduction in the energy carried by light wave? Ans. (a) A compound microscope consists of two convex lenses co-axially separated by some distance. The lens nearer to the object is called the objective. The lens through which the final image is viewed is called the eyepiece. Magnifying power, when final image is at infinity: The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective.

12 Where, M e and M 0 are the magnifying powers of the eyepiece and objective respectively. If u 0 is the distance of the object from the objective and v 0 is the distance of the image from the objective, then the magnifying power of the h ' L h h ' objective is M0 Using, tan h f 0 f0 L Where, h, h' are object and image heights respectively and f 0 is the focal length of the objective. L is the tube length i.e. the distance between the second focal point of the objective and the first focal point of the eyepiece. When the final image is at infinity, M e D f e L D Magnifying power of compound microscope, M M0Me f f 0 e Nearsightedness or Myopia A person suffering from myopia can see only nearby objects clearly, but cannot see the objects beyond a certain distance clearly. Myopic eye: Correction In order to correct the eye for this defect, a concave lens of suitable focal length is placed close to the eye so that the parallel ray of light from an object at infinity after refraction through the lens appears to come from the far point of the myopic eye.

13 Farsightedness or Hypermetropia A person suffering from hypermetropia can see distant objects clearly, but cannot see nearby objects. Hypermetropic eye: Correction To correct this defect, a convex lens of suitable focal length is placed close to the eye so that the rays of light from an object placed at the point N after refraction through the lens appear to come from the near point of the hypermetropic eye. OR (a) Huygens Principle: Each point on the primary wave front acts as a source of secondary wavelets, sending out disturbance in all directions in a similar manner as the original source of light does. The new position of the wave front at any instant (called secondary wave front) is the envelope of the secondary wavelets at that instant. Refraction On The Basis Of Wave Theory

$Suppose when disturbance from point P on incident wave front reaches point on the refracted wave front, the disturbance from point Q reaches Q' on the refracting surface XY.$ $Since represents the refracted wave front, the time taken by light to travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same.$

14 Consider any point Q on the incident wave front. Suppose when disturbance from point P on incident wave front reaches point on the refracted wave front, the disturbance from point Q reaches Q' on the refracting surface XY. Since represents the refracted wave front, the time taken by light to travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same. Now, time ' QK KQ taken by light to go from Q to Q' will be t...( i) c In right-angled ΔAQK, QAK = i QK = AK sin i (ii) In right-angled ΔP'Q'K, ' ' Q P K r and ' KQ KP ' sin r. (iii) Substituting (ii) and (iii) in equation (i), The rays from different points on the incident wave front will take the same time to reach the corresponding points on the refracted wave front

15 i.e., t given by equation (iv) is independent of AK. It will happen so, sin sin sin sin if i r 0 i c i c sin r sin r This is the Snell s law for refraction of light. i. The frequency of reflected and refracted light remains same as the frequency of incident light because frequency only depends on the source of light. ii. Since the frequency remains same, hence there is no reduction in energy.

CBSE Annual Examination Question Paper 2013

CBSE Annual Examination Question Paper 2013 PHYSICS (THEY) Time allowed: 3 hours] [Maximum marks: 70 General Instructions: (i) All questions are compulsory. (ii) There are 29 questions in total. Question