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KENDRIYA VIDYALAYA NO. : 1, BHOPAL Unit Test –II SEPT- 2016 Class XII (Physics)-Solution Time : 1½ hr. M.M. 40 General Instructions 1. 2. 3. All questions are compulsory. There are 16 questions in all. This question paper has five sections. Section 'A' contains five questions of one mark each. Section 'B' contains three question of two marks each. Section 'C' contain five questions of three marks each. Section 'D' contain one value based question of four marks and Section 'E' contains two questions of five marks each. 4. There is no overall choice. However, an internal choice has been provided, one question of two marks, one question of three marks and both the questions of five marks. You have to attempt only one of the choices in such question. 1. Name an electromagnetic radiation used for detecting fake currency note write its frequency range. Ans:- U- V Radiations ; Frequency range - 3×1016 – 7.5 × 1014 Hz. How are the figure of merit and current sensitivity of a galvanometer related to each other? Ans:- Reciprocal to each other. What is the angle of dip at a place where vertical and horizontal components of earth magnetic fields are equal. Ans:- 45° Predict the polarity of the capacitor in the situation described in the figure below. SECTION 'A' 2. 3. 4. 5. Ans:- Plate “a” will be negative with respect to “b”. Why core of a transformer is laminated? Ans:- To reduce electric energy loss due to eddy currents. SECTION 'B' 6. A galvanometer of resistance 120 Ω gives full scale deflection for a current of of 5 mA. How can it be converted to an ammeter of range 0 to 5 A. Also determine the net resistance of ammeter. Ans:- By using shunt resistor of 0.12 Ω it can be converted into required range of ammeter. 1 𝑅𝑛𝑒𝑡 1 1 G S 1 1 = 120 0.12 = = 1 1000 » 120 Rnet = 0.1198= 0.12 Ω 7. 8. OR Write the four measures that can be taken to increase the sensitivity of a galvanometer. Ans: - 1. Strength of magnetic field 2. Shape of pole of magnet used 3. Number of turns in a coil 4. Area of coil in magnetic field. Represent the EM wave propagating along the x-axis.In which electric and magnetic fields are along y-axis and Z-axis respectively. Ans:- Circuit shown here uses an airfield parallel plate capacitor. A mica sheet is now introduced between the plates of a capacitor. Explain with reason the effect of brightness of bulb. ANS:- SECTION 'C' 9. Derive the expression for the force between two infinitely parallel straight wires carrying current in the same Ans:- direction. Hence, define “ampere” on the basis of above derivation. Figure above shows two long parallel conductors a and b separated by a distance d and carrying (parallel) currents Ia and Ib , respectively. The conductor ‘a’ produces, the same magnetic field B at all points along the conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards (when the conductors are placed horizontally). Its magnitude is given by Eq. The conductor ‘b’ carrying a current I will experience a sideways force due to the field Bab. The direction of this force is towards the conductor ‘a’. We label this force as F, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by Force per unit length The above expression is used to define the ampere (A). The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 × 10–7 newtons per metre of length. 10. State Ampere, circuital law. Use this law to obtain an expression for the magnetic field due to a toroid. Ans:- It states that the line integral of magnetic field B along a closed path is equal to µ0 Times the current (I) passing through the closed path. Three circular Amperian loops 1, 2 and 3 are shown by dashed lines. By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop. The circular areas bounded by loops 2 and 3 both cut the toroid: so that each turn of current carrying wire is cut once by the loop 2 and twice by the loop 3. the loop 2 and twice by the loop 3. Let the magnetic field along loop 1 be B in magnitude. Then in Ampere’s circuital law , However, the loop encloses no current, so Ie=0 thus B1 = 0 Thus, the magnetic field at any point P in the open space inside the toroid is zero. Let the magnetic field inside the solenoid be B. We shall now consider the magnetic field at S. Once again we employ Ampere’s law we get : Let r be the average radius of the toroid and n be the number of turns per unit length OR State Ampere, circuital law. Use this law to obtain an expression for the magnetic field due to a current carrying solenoid. Ans:- It states that the line integral of magnetic field B along a closed path is equal to µ0 Times the current (I) passing through the closed path. We shall assume that the field outside is zero. The field inside becomes everywhere parallel to the axis. Consider a rectangular Amperian loop abcd. Along cd the field is zero as argued above. Along transverse sections bc and ad, the field component is zero. Thus, these two sections make no contribution. Let the field along ab be B. Thus, the relevant length of the Amperian loop is, L = h. Let n be the number of turns per unit length, then the total number of turns is nh. The enclosed current is, Ie = I (n h), where I is the current in the solenoid. From Ampere’s circuital law 11. An alternating voltage of frequency f is applied across a series LCR circuit. Let fr be the resonance frequency for the circuit. Will the current in the circuit, lag, lead or remain in the phase with the applied voltage when i) f > fr ii) f < fr ? Explain your answer in each case. 12. An EM wave travelling through a medium has electric field vector Ey = 4×105 cos (3.14×108 x- 1.57 t) N/C. Here x- is in meter and t is in second then find: 1. Wavelength 2. Frequency 3. Direction of propagation 4. Speed of wave 5. Refractive index of medium 6. Amplitude of magnetic field vector. Ans:- E0 = 4×105 N/C 1 :- λ= 4 Metre ω =3.14×108 = 2 /T ; K= 1.57= 3.14/2= 2 /λ 2:- frequency = 3.14×108 /2 = 5×107 Hz. 3. Positive X- axis 4. Speed =ω/K= 2×108 m/s 5. Refractive index of medium = 3×108/2×108 = 1.5 6. Amplitude of magnetic field vector = E 0 /C= 4×105 /3×108 = 1.33 Mt 13. A long wire first bent into a circular coil of one turn and then into a circular coil of smaller radius having ‘n’ turns. If the same current passes in both the cases, find the ratio of magnetic fields produced at the centre in the two cases. Ans:- Here a1 is radius of smaller turn SECTION 'D' 14. Raj Pal Yadav, a retired Physics Teacher was working in his field with his grandson. There was a big high tension tower carrying thick wires in their field. Grandson wanted to know as to why can’t the tower be removed from their field, so that they may get more space for crops. Raj Pal explained him the necessity of HT tower, and said it is very high voltage AC transmission line and is a lifeline of their town. • What values are displayed by Raj Pal Yadav? • Why Long distance AC transmission is done at very high voltage. • What is the principle of transformer? • What are the energy losses in transformer? Ans:- 1.Responsibity towards his profession, caring for other etc. 2. To minimize power loss in form of heat enegy. 3. Mutual induction 4. Flux Loss , Eddy current loss, copper loss etc. SECTION 'E' 15. State the working of an AC generator with the help of a labelled diagram. Deduce the expression for alternating e.m.f. generated in the coil. What is the source of energy in this device. Draw diagrams to show different positions of a coil. One method to induce an emf or current in a loop is through a change in the loop’s orientation or a change in its effective area. As the coil rotates in a magnetic field B, the effective area of the loop (the face perpendicular to the field) is A cos where is the angle between A and B. This method of producing a flux change is the principle of operation of a simple ac generator. An ac generator converts mechanical energy into electrical energy. OR Derive expression for average power in LCR series circuit. Hence explain that average power for a complete cycle for inductor and capacitor will be zero. Also define wattles current. Ans:- Given Below:- Purely capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is /2. Therefore, cos /2= 0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current. 16:- State Biot’s Savart law. Using this law find expression for magnetic field at any point of axis of an current carrying coil. According to Biot-Savart’s law, the magnitude of the magnetic field dB is proportional to the current I, the element length |dl|, and inversely proportional to the square of the distance r. The magnitude of this field is, The magnitude dB of the magnetic field due to dl is given by the Biot-Savart law:- The net contribution along x-direction can be obtained by integrating dBx = dB cosθover the loop. The summation of elements dl over the loop yields 2 R, the circumference of the loop. Thus, the magnetic field at P due to entire circular loop is : OR Describe cyclotron under following headings. 1. Principle 2. Construction 3. Working 4. Expression for K.E. 5. Limitation. Ans:- Principle:- A positively charged particle can be accelerated to a sufficiently high energy with the help of smaller values of oscillating electric field by making it cross the same electric field time and again with the use of strong magnetic field. Construction:Construction:- The cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles. As the fields are perpendicular to each other they are called crossed fields. The particles move most of the time inside two semicircular disc-like metal containers, D1 and D2, which are called dees as they look like the letter D. Working:- A high frequency alternating voltage is applied to the dees positive ions or positively charged particles (e.g., protons) are released at the centre P. They move in a semi-circular path in one of the dees and arrive in the gap between the dees in a time interval T/2; where T, the period of revolution, The frequency νa of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one half of the revolution. The requirement νa = νc is called the resonance condition. The phase of the supply is adjusted so that when the positive ions arrive at the edge of D1 D2 is at the lower potential and the ions are accelerated across the gap. Inside the dees the particles travel in a region Expression for K.E. :- Charged particles are then deflected by a magnetic field and leave the system free of the electric field. The increase in their via an exit slit with speed Limitations:1. It isis suitable onlythey accelerating heavy particles kinetic energy qV eachfor time cross from dee deuteron to anotheralpha particle etc. likeone proton, 2. The uncharged partcles can not be accelerated.