# Sample Model Papers For Physics CBSE Board for students online

* Back To Papers List |

General instructions:
^{°}
and the angle of dip is 90^{°}.
^{-3}m.
1. The algebraic sum of the electric currents that meet at any point in a network is zero. 2. In any closed electric circuit the algebraic sum of the products of current and resistance in each part of the network is equal to the algebraic sum of the electromotive forces in the circuit.
The conditions are: 1. The waves should be coherent. 2. The beams should be of nearly same intensity. 3. The beams should not intersect at too large an angle.
If _{ 2}m_{1} is the refractive index from medium 1 to
medium 2 then,_{2}m_{1 }= sin i/sin rWhere i and r are the angles of incidence and reflection respectively. From the principle of reversibility, if the ray of light is reversed its original path will be retraced. _{1}m_{2 =} sin r/sin i or, _{ 1}m_{2} = 1 / _{2}m_{1}
f = (½) * R f = 0.25 m P = 1/f P = 1/0.25 = 4 diaptors. Q 13 Derive a relation between surface temperature of the sun with
the solar constant. Consider sun to be a black body at temperature T and radius
RAns. 13 _{0} at the center of a hollow sphere of radius r, r R_{0}.Surface area of the hollow sphere = 4pr ^{2}.All the energy emitted by the sun falls normally on the inner surface of the hollow sphere. If S is the solar constant, the solar luminosity is given by, L _{0} = 4pr^{2}SAccording to Stephen's law, the energy emitted per sec per unit area is given by, E = s T ^{4}Surface area of sun = 4pR0 ^{2}Total energy emitted per second by the sun = 4pR _{0}^{2}E
= 4pR_{0}^{2}s T^{4}As solar luminosity is also equal to the total energy radiated per second by the sun. Therefore, 4pR ^{02}sT^{4} = 4pr^{2}ST = [r ^{2}S/R_{0}^{2}s]^{1/4}Knowing the value of S, s, r and R0, T can be determined.
Ans. 14 _{1} = ± 4pR_{1}^{2}E
(r)Flux through S1 = ± 4pR _{2}^{2}E (r)Ratio = (R _{1}/R_{2})^{2}
2. When the circuit is closed current flows in the direction of electrostatic field outside and in the opposite direction of electrostatic field inside the cell. There is a net field inside the cell opposite to the electrostatic field.
R = 2 Wl = 10m d = 0.4mm A = 0.1256 mm ^{2}From the following formula, r is easily obtained. R = pl/A r = 2.512 * 10 ^{-8} Wm
q = [mg/E] t _{g}[1/t_{e} + 1/t_{g}] where q is the charge, E is the electric field, t _{g} is the time
taken by the droplet during free fall and t_{e} is the time taken
by the droplet when moving in the electric field.
Ansistor has its emitter-base junction forward biased by connecting the emitter to the negative terminal of V _{EB}
battery and the p-type collector is reverse biased by connecting it to
the positive terminal of V_{CB} battery as shown in the figure
below.The forward biasing of emitter base circuit repels the electrons from
the emitter towards the base, setting up emitter current I
1/G = NAB/k where k is the constant of suspension B is the magnetic field. A is the area of the coil {length * breadth} N is the no. of turns of the coil. From the relation it is clear that the sensitivity depends on N, B and A. V
^{2} = eV = 1.6 * 10^{-19}
* 500 = 8.0 * 10^{-17} eVv = [2eV/m] ^{1/2}= [2 * 1.6 * 10 ^{-19} * 500 /9.1 * 10^{-31}]^{1/2}v = 1.3 * 10 ^{7} m/secMomentum = mv =9.1 * 10 ^{-31} * 1.3 * 10^{7} kg m/sec.Q 23 A series LCR circuit consists of R = 10 W, Xe = 60 W and an inductance
coil. The circuit is found to resonate when put across 300 V, 100 Hz supply.
Calculate induction of the coil and current in the circuit at resonance. Given that as LCR circuit resonates when put across a 300
V, 100 Hz supply.Ans. 23 R = 10 W XC = 60 W At resonance, w ^{2} = 1/LCXC = 60 W = 1/wC n = 100 Hz, X _{C} = 60 WTherefore, w = 2pn = 3.14 * 2 * 100 = 628.2 C = 1/w X _{C}= 1/ 628.2 * 60 + 2.653 * 10- ^{5}Now, L = 1/w ^{2}C= 1/[ * (3.14) ^{2} * (100)2 * 2.653 * 10^{-5}]= 9.55 * 10 ^{-2}I = V/R = 300/10 = 30 amps
l = 589 nm m = 1.33 m = V _{air}/V_{water} = 1.33V _{water} = V_{air}/1.33 = 3 * 10^{8}/1.33 = 2.25
* 10^{8} m/secl _{air} = 589 nm = 589 * 10^{-9} mC = n l n = [3 * 10 ^{8}]/[589 * 10^{-9}]lwater = V _{water}/n = [3 * 10 ^{8}/1.33][589 * 10^{-9}/ 3 * 10^{8}]= 4.42 * 10 ^{-7} mTherefore, Speed of reflected light = 3 * 10 ^{8} m/secSpeed of refracted light = 2.25 * 10 ^{8} m/sec
Ans after 276 days, only one fourth of the initial podium remains undecayed.
Ansistor as a common emitter amplifier. The emitter is common to both input and output circuits. The emitter is forward biased by battery V _{BB} and the collector
is reversed biased by the battery V_{CC}. This decreases the resistance
R_{in} of the input circuit and increases the resistance Rout
of the output circuit. The low a.c. input signal V_{I} is superimposed
on the forward bias V_{BE}. A load resistance R_{L} is
connected between the collector and the d.c. Supply and the amplified
output is obtained between the collector and the ground.
When current I
M = image distance/object distance = v/u = v[1/v - 1/f] In this case v is negative and is equal in magnitude to D. M = 1 + D/f
1. Electrons move in circular orbits around the nucleus called the stationary states. It restricts the stationary states to those circular orbits in which the angular momentum is an integral multiple of [h/2p]. 2. An additional postulate is that : in violation to electrodynamics, accelerated motion of electronic charge in these orbits is not accompanied by emission of electromagnetic radiation. 3. When a hydrogen atom makes a tr Ansition from a state with quantum number n _{I} to the state with quantum number n_{f} (n_{f}
< n_{I}), then it does by emitting a photon of energy hn which
is the difference in energies of the two states.The energy of electron in the stationary state is the sum of its kinetic and potential energies. K.E. = ½ mv ^{2}P.E. = - e ^{2}/4pe_{0}rThe total energy E = ½ m v ^{2} - e^{2}/4pe_{0}rAccording to Bhor, a stationary state has its circumference equal to an integral number of De Broglie wavelengths. So, 2pr = nl = nh/mv mvr = nh/2p v = nh/2pmr Also, mv ^{2}/r = e^{2}/4pe_{0}rTherefore, v = 1/n[e ^{2}/4pe_{0}r][2p/h]And r = n ^{2}/m [h/2p]2[4pe_{0}r/ e^{2}]a is the fine structure constant, c is the velocity of light, h is Planck's constant, r is the radius of the orbit, e is the elctronic charge.
_{2} 2pf each are in series.
This gives the equivalent capacitance,1/C = 3/C _{2} = 3/2 pf. C = 2/3 pfThis is in parallel with the next C2 capacitance, which gives an equivalent capacitance of 8/3 pf. Now the remaining three capacitors are again in series, which gives an equivalent capacitance of 1/(1/3 + 1/3 + 3/8) or 24/25 pf. Q 30 A small city having a demand of 880 kW of electric power at 220
V is situated 15 km away from electric plant generating power at 440 V.
The resistance of the cable is 15 W. The city gets power through a 4400-220
V step down tr The current in the line can be given by, 880 * 1000/4400 =
200 AAnsformer at the substation. Calculate, 1) Power loss in the form of heat. 2) How much power the plant must supply (assuming the power loss in the form of losses is zero). 3) Voltages drop in the line. Ans. 30 1. The power loss is given by H = I ^{2}R = (200)^{2} *
15 = 600 Kw2. The plant should supply a total power of 880 + 600 = 1480 Kw 3. The voltage drop is given by, 200 * 15 = 3000 V. |