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Question 1
A IC charge is placed at a point O. What is the work done in moving a 2C charge once along a circle of radius 10cm and centre O?
1) ZERO: as the 2C charge is being moved on an equipotential surface.

Question 2
What is the absolute permitivity of a medium whose dielectric constant is one?
dielectric constant = permitivity in a meduim/permitivity in vacuum=> 1=e/e₀_{=> e = e0}

Question 3
Draw an equipotential surface for a) uniform electric field b) non uniform electric field.
Ans. a) b)

Question 4
If the temperature of the cold junction in a thermocouple is decreased, what will be its effect on the neutral temperature and the temperature of inversion?
Neutral temperature remains unaffected Temperature of inversion is increased

Question 5
An electron travelling West to East enters a chamber having a uniform electrostatic field in a North to South direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
[Electrical Force (F_E) is SN dn :. the magnetic force ,F_B, should be in the NS dn] F_B is in the NS dn current is in the EW dn ==> the magneti fiel should be directed perpendicular to the horizontal plane in the downward direction.

Question 6
A charged particle is moving in a uniform magnetic field with varying speeds. Out of the following physical quantities which physical quantity remains unaltered. a) kinetic energy b) linear momentum c) radius d) angular frequency
Angular frequency

Question 7
Does a Conductor become charged when a current is flowing through it?
No. According to the Law of conservation of charges, total charge entering a certain cross sectional area is equal to the net charge leaving the area. Sq=0

Question 8
A wire of receptivity 49 x 10^-6Wcm. is stretched to three times its original length. What will be the new receptivity of the wire?
The length 'l' & area of cross section 'A' of the wire change when the wire is stretched but as the resistivity is independent of 'l' & 'A' the resistivity is unaffected ==> P = 49 x 10^-6Wcm.

Question 9
State Biot Savart's Law What is the magnetic field induction at the point 0 due to a current flowing in an infinite shaped wire as shown in the figure?
Biot Savart's Law: Accourding to Biot -Savart's Law, the magnetic field dB at point P due to a current 'i' flowing through a smll segment of length dl is given by and it magnitude is directed perpendicular to the plane containing the radius vector & the current element where t is the displacement vector of point P from the current element idl & q the angle between Magnet field is contributed by the a) quadreant AB of a circle b) two straight conductors extending to infinity i) from A & ii) from B

Question 10
The current 'i', (A) flowing through a wire depends on time 't', (sec) as i = 10 + 4t. What is the charge crossing through a section of the wire in 10 sec?

Question 11
Draw a neat well labelled diagram of a Lechlancle cell.
Ans.

Question 12
The plots for the current i (A) flowing through a metallic resistor versus the potential drop V (volts) across the resistor at two different temperatures T₁ & T₂ is shown in the figure. Which of the two temperatures T₁ & T₂ is higher and explain why?
Ans. T₂>T₁ slope of A > slope of B => R_A < R_B as resistance is inversely proportional to the slope. \ T_A < T_B as resistance is directly proportional to temperature => T₁ < T₂ R a T for metals.

Question 13
Calculate the electric force experienced by an electron drifting through a copper wire of cross sectional area 5 mm² carrying a current of 1A. The receptivity of copper is 1.5 x 10^-6Wcm.
Ans. Force = qE q = e ; E = V F = e V l l V = iR = ipL = eipL A AL F = eip e = 1.6 x 10^-19C A i = lA = 1.6 x 10^-19 x 1 x 1.5 x 10^-8 p = 1.5 x 10^-6 Wm = 1.5 x 10^-8 Wm 5 x 10^-6 A = 5mm² = 5 x 10^-6 m² F = 0.48 x 10 ^-21 N

Question 14
State two reasons why electrolytes have lower electrical conductivity than metallic conductors in general.
i) The charge carriers (ions) in electrolytes are heavier as compared to the charge carriers (electrons) in metals. Electron mobility (drift velocity) is greater than ions ii) the number of ions per unit volume in electrolytes is less than the number of electrons per unit volume in metals.

Question 15
Derive an expression for the electric potential at a point along the axial line of an electric dipole.
Consider an electric dipole, of dipole moment 'm', consisting of a pair of equal and opposite charges of strength 'q' separated by a distance of '2a' placed in vacuum. Let P be a point on the axial line at a A.<--------------2a-------------->.B------------.P -q +q <---------------x-------------> distance 'r' from the centre of the dipole The potential V₁ at P due to the positive charge (+q) V₁ = kq = +kq BP (r-a) The potential V₂ at P due to the negative charge (-q) V₂ = kq = -kq AP (r+a) The total potential V P is V = V₁ + V₂ = kq - kq r-a r+a = kq[(t+a) - (t-a)] r²-a² V = kq2a r²-a² m = 2aq V = km r²-a² V = km r² if r>> a

Question 16
State Gauss's Law. A charge of 42mc is located at the centre of a cube & a charge of 36mc is located outside the cube. What is the electric flux linked with any one face of the cube?
Gauss's Law states that the closed surface integrat of the electrostatic field due to any source is directly proportional to the total charge contained inside the surface S in the volume V

Question 17
A proton and an alpha particle accelerated through the same potential difference enter normally into a magnetic field of strength B. Determine the ratio of x to rp where rp & r are their respective radii of their circular paths.
When a particle moves in a circular path in magnetic field when accelerated through a potential differrence V

Question 18
Two bulbs of rating 100w-220V and 40w-220V are connected in series across a 220 V supply. Justify with necessary calculations which bulb will glow brighter.

Question 19
With the help of a neat well labelled circuit diagram describe the method to determine the internal resistance of a primary cell.
Principle: The potential drop 'V' across any length 'e' of the potentioneter wire is directly proportional to the length'e' Theory & Formula Used r = R (e₁ - e₂)/e_{2 where r is the internal resistance of the primary cell. e ₁ is the balancing length of the primary cell without shunt e ₂ is the balancing length of the primary cell without shunt R is the shunt resistance. PROCEDURE & CALCULATIONS 1. Set the circuit as shown in the figure & check for a two sided deflection. 2. Insert key K1 & remove key K2 & locate the balancing length for the lechlanche cell without shunt. Let it be l1 3. Both the keys K1 & K2 are inserted such that the cell is shunted by a resistance R. Locate the balancing length for the shunted cell Let it be e2. 4. Compute the internal resistance r using the formula r = R (e1 - e2)/e1 5. The observation can be repeated for different values of R & the mean value of r calculated.}

Question 20
The voltmeter connected across the 600W resistor has a resistance of 600W. Determine the potential drop read by the voltmeter.
Ans.

Question 21
In a triangle ABC, the sides AB, BC and CA are 5cm, 4cm and 3cm respectively. If the charges 12pc and 16pc are placed at A and B respectively, determine the electric field intensity at C. What will be the force experienced by HOC charge placed at C.

Question 22
An isolated hollow conducting spherical shell of radius 'R' carries a charge 'Q'. Plot the variation of the electrostatic field intensity, the corresponding potential and the electric flux with distance r, as measured from the centre of the sphere.
Ans.

Question 23
a) On a neat well labelled diagram depict the elements of the earth's magnetic field. b) It is observed that the neutral points lie along the equatorial line of a magnet placed on the Table. What is the orientation of the magnet with respect to the earth's magnetic field?
Ans. PQRS - geographic meridian PNMS - magnetic meridian BH - earth's horizontal component of the magnetic field. BV - earth's vertical component of the magnetic field.} B - earth's magnetic field. d - angle of declinatin f - angle of depth. North pole of the magnet faces the magnetic south (geopraphical North )

Question 24
A thin magnetic needle vibrates, in a vibration magnetometer, in the horizontal plane with a period of 16 seconds. The needle is cut into two halves by a plane normal to the magnetic axis of the needle. Determine the period of vibration of each half needle.

Question 25
In the circuit, the charge on the capacitor c2 is 2mC and the charge on the capacitor c3 is 3mC. If the capacitance of c2 is 2mF. Determine a) the potential drop across C1 b) the equivalent capacitance of the circuit c) the energy dissipated in the curcuit.

Question 26
Derive an expression for the torque on a rectangular coil of area 'A', carrying a current 'i' and placed in a uniform magnetic field B, such that the normal to the plane of the coil makes an angle q with the direction of B. If the coil is left on its own, how does it tend to orient itself relative to the magnetic field.
Consider a rectangular coil of length 'e' & breadth 'b' carrying a current is placed in a uniform magnetic field of strength B placed such that the normal to the coil makes an angle q wrt the field B. Force experienced by the arm AB Force experienced by the arm CD According to Flemmings Left hand rule the forces are in opposite directions as the forces are equal & opposite they set up a The coil will align itself perpendicular to the filed B such that the angle between the normal & the field is O0 The coil is equilibrium

Question 27
Write the expression for the force between two parallel straight conducting wires carrying a current 'i' & separated by a distance by a distance 'a' in vacuum, b) Two parallel wires carrying current in the same direction attract each other, while two beams of electrons travelling in the same direction repel each other. Explain.
The Force between two wires of length l carrying currents l1 & l2 separated by a distance of 'a' in vacuum is given by F = Bil sina, a=90 = m_oL₁L₂L/2pa, B = m0i/2pa b) Two parallel straight wire attract each other if they are carrying currents in the same direction, as they experience a magnetic force which according to Flemming's right hand rule is attractive. Electrostatic forces are zero where as between the two eletron beams travelling in the same direction repel each other as they experience both an electrostatic force F_E of repulsion & a magnetic force F_B of attraction but as F_B/F_E = 10^-27 => F_B is less than F_E net force is repulsive.

Question 28
a) Plot the variation of thermo e.m.f E(mV) with the temperature of the hot junction, temperature of the cold junction of the thermocouple being maintained at 0⁰C. b) The following plot represents the variation of Thermoelectric power S(mV⁰C^-1) with the temperature of the hot junction qoC. for a Silver-Iron thermocouple, the temperature of the cold junction is maintained at O⁰C. Determine a) the thermoelectric coefficients b) the neutral temperature and temperature of inversion. c) If the cold junction was maintained at 40⁰C what be the neutral temperature and temperature of inversion.
a) E - thermo emf across the junctions of a thermocouple. q - temperature of the hot junction q_n - neutral temperature q_i - temperature of inversion b) E = aq +1/2bq₂ => S= dE/Dq =a + bq but the equation of a straight line is y = mx + c => b =slope of graph a = y intercept a) b=8.8-0/5000-300=4.4 x 10^-2 mv⁰C^-2 a = 13.2 mv⁰C¹ b) At neutral temperature(q_n)dE/dq =0 => q_{n=300⁰C temperature of inversion q}^{i= 2q_n - q_{c =}}_600⁰C as ^q_c = 0 c) neutral temperature (^q_n)_300⁰C unaffected Temperature of inversion ^q_i = 2^q_n - ^q_c = 600 - 40 =560⁰C

Question 29
a) State the principle of a tangent galvanometer. Name the two fields which influence the deflection of the magnetic needle in a tangent galvanometer b) How will the reduction factor of a tangent galvanometer be affected when the (i) number of turns of its coil is doubled? (ii) the radius of the coil is increased? (iii) the angle of dip at the location of the experiment increases?
Tangent galvanometer works on the principle of Tangent Law. If a magnetic needle is under the influence of two crossed fields BH & BV inclined at an angle q with BH then tan q= BV/BH The magnetic needle of a tangent galvanometer is under the influence of a) the earth's horizontal component of the magnetic field. b) the field, at the centre, of a current carrying circular coil b) mo Ni/2a = H tan q => i = 2aH/mo Ni tan q = k tan q the reduction factor K = 2aH/mo Ni i) as no of coils is doubled K is halved ii) as the radius of the coil is increased K increases. iii) H = B cos a as a angle of dip increases cos a decreases as H decreases K decreases.

Question 30
A parallel plate air cored capacitor is connected to a battery. The battery is disconnected and a dielectric slab is inserted between the plates. Describe qualitatively what happens to the charge, capacitance & energy stored in the capacitor? What would be the charge, capacitance & energy stored in the capacitor if the battery had not been disconnected & the dielectric inserted? OR a) Derive an expression for the total capacitance when three capacitors are connected in a) series b) parallel. b) A2mF capacitor charged to a potential difference of 100V and a 8mF capacitor charged to a potential difference of 50V are connected together in parallel. Find the loss in energy due to the connection.
As the battery is disconnected charge remains constant capacitance in a medium = eA/d = er e₀A/d = e_rC₀KC₀= as K > 1 capacitance increases. Energy stored U = Q²/2C as Q is constant C increases energy stored U decreases. capacitance of an air cored capacitor C₀ = e₀A/d capacitance when the dielectric is inserted C_m= e A/d = Ke₀A/d = KC₀ as K > 1 C_m increases Voltage across the capacitor is constant Q = CV as C inereases V is constant. charge Q increases. Energy stored U = 1/2 CV² C - increases V - constant energy stored increases. OR Consider the tree capacitors are connected in series as shown in the figure let their capacitances be C₁, C₂& C₃& the potential drops V₁, V₂, & V₃ resp. As they are in series charge (q) is the smae for all the capacitors the total potential drop = sum of individual potential drops => V = V₁+ V₂+ V₃(as V = Q/C) Q/Ceq = Q/C₁+ Q/C₂+ Q/C₃ is equal to the reciprocal of the sum of the reciprocals of the individual capacitances If the three capacitors C₁, C₂& C₃are connected in parallel the potential drop across each capacitor is the same say V. Let the charges stored by the capacitors be Q₁, Q₂, & Q₃resp Total charge Q = Q₁+ Q₂+ Q₃ but Q = CV => Ceq = C₁, + C₂ + C₃ C₃V. => Ceq = C₁ + C₂ + C₃ = sum of individual capacitances b. Loss in Energy U =

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