physics paper 2012 1 - poornima university · the base line of the prism inside it. find the angle...

CBSE/12th Class/Physics/2012/SET-1

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S.No Questions Answers Q.1 Why must electrostatic field be normal to the surface at every point of a

charged conductor?

Ans.1 The charge is present only on the outer surface in case of conductors. The field is zero inside the conductor. And at the surface it has to be normal. If it is not normal to the surface, then it would have some nonzero component along the surface. The free charges on the surface would then experience a force and start moving. In a static situation, electrostatic field should not have tangential component, which in turn implies that the surface of a charged conductor must be normal to the surface at every point. If a conductor has no surface charge, then the field is zero.

Q.2 Under what conditions does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?

Ans.2 Under the condition, 1/f = 0 or f → ∞ A biconvex lens will act like a plane sheet of glass if it is immersed in a liquid having the same index of refraction as itself.

Q.3 State de-Broglie hypothesis. Ans.3 All matter can exhibit wave-like behaviour. For example a beam of electrons can

be diffracted just like a beam of light or a water wave. Matter waves are an example of wave–particle duality. The concept that matter behaves like a wave is also referred to as the de Broglie hypothesis due to having been proposed by Louis de Broglie in 1924. Matter waves are often referred to as de Broglie waves.

The de Broglie wavelength is the wavelength, λ, associated with a particle and is related

to its momentum, p, through the Planck constant, h:

Q.4 Name of physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum.

Ans.4 Microwaves and UV rays both are a part of the electromagnetic spectrum. Thus, the physical quantity that remains same for both types of radiation will be their speeds, equal to c.

Q.5 When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction?

Ans.5 No, When electrons drift in a metal from lower to higher potential, it does not mean that all the free electrons of the metal are moving in the same direction because locally electrons may collide with ions and may change its direction of motion.

Q.6 Predict the direction of induced current in a metal ring when the ring is moved towards a straight conductor with constant speed v. The conductor is carrying current I in the direction shown in the figure.

Ans.6 According to Lenz law, An electric current induced by a changing magnetic field will flow such that it will create its own magnetic field that opposes the magnetic field that created it.Using Lenz’s law we can predict the direction of the induce current in the ring will be clockwise.

Q.7 The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator?

Ans.7 On the equator, the values of both angle of dip (δ) and vertical component of earth’s magnetic field is zero. So, in this case, Bv = 0


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Q.8 Q8. Show on a graph, the variation of resistivity with temperature for a typical semiconductor.

Ans.8

Therefore, for a semiconductor, resistivity decreases rapidly with increasing temperature.

Q.9 Therefore, for a semiconductor, resistivity decreases rapidly with increasing temperature.

Ans.9 We know that for a point charge Q,

Thus, their corresponding plots would be

Q.10 Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.

Ans.10 Self Inductance of a Long Solenoid: If the inner core of a long solenoid coil with N number of turns per metre length is hollow, “air cored”, then the magnetic induction within its core will be given as:

---------------------(1)

µ0 = Absolute magnetic permeability of free space

N = Total number of turns in the solenoid

The magnetic flux that is produced in its inner core is equal to:

--------------------(2)

A = Area of each turn of the solenoid The self-inductance of the coil is given by,


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----------------------(3)

By cancelling out and grouping together like terms, then the final equation for the coefficient of self-inductance for an air cored coil (solenoid) is given as:

Where, μο is the Permeability of Free Space (4.π.10-7) N is the Number of turns A is the Inner Core Area (π.r 2) in m2 l is the length of the Coil in metres

If core is of any other magnetic material, µ0 is replaced by

if n is the number of turns per unit length then

This is self inductance of a long solenoid.

Q.11 A ray of light, incident on an equilateral prism moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.

Ans.11 It is given that the prism is equilateral in shape. So, all the angles are equal to 60°.Thus, the angle of prism, A = 60° The angle of refraction in case of a prism

Applying Snell’s law,

Here, µa → refractive index of air, n1 = 1

µg → refractive index of glass, i → angle of incidence so,

So, the angle of incidence is i= 600

Q.12 The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason, then, to operate the photodiode in reverse bias?

Ans.12 The reason to operate the photodiode in reverse bias that in reverse bias the width of the depletion layer increases which reduces the capacitance across the junction, thereby increasing response time. The sensitivity of a photodiode is thus very high, a


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property that is certainly desired. Q.13 Distinguish between ‘Analog and Digital signals’

OR Mention the function of any two of the following used in communication system: (i) Transducer (ii) Repeater (iii) Transmitter (iv) Bandpass Filter

Ans.13 Analog Signal: It varies continuously with variable may be time or distance etc. E.g. Sound of human Digital Signal:It is a type of signal which has only two values high or low. In digital high mean 1 and low means zero.E.g. Temperature of day - Maximum 30°C ⇒ 1 Minimum 15°C ⇒ 0

OR (i) Transducer: A device that converts variations in a physical quantity, such as pressure or brightness, into an electrical signal, or vice versa. (ii) Repeater: It is an electronic device used in transmission system to regenerate the signal. It picks up a signal amplifies it and transmits it to receiver. (iii) Transmitter: It is a set of equipment used to generate and transmit electromagnetic waves carrying messages or signals, especially those of radio or television. (iv) Bandpass Filter: A bandpass filter is an electronic device or circuit that allows signals between two specific frequencies to pass, but that discriminates against signals at other frequencies.

Q.14 Define mutual inductance between two long coaxial solenoids. Find out the expression for the mutual inductance of inner solenoid of length l having the radius r1 and the number of turns n1 per unit length due to the second outer solenoid of same length and r2 number of turns per unit length.

Ans.14 Mutual Inductance: The ability of production of induced emf in one coil, due to varying current in the neighboring coil is called mutual inductance. Magnetic flux, Φ = MI Where, M is called coefficient of mutual induction Mutual Inductance of Two Long Solenoids:

Let us Consider two long solenoids S1 and S2 of same length l, such that solenoid S2 surrounds solenoid S1 completely as shown in figure. Φ21 = M21I1 Where, M21 is the coefficient of mutual induction of the two solenoids Magnetic field produced inside solenoid S1 on passing current through it, B1 = µ0n1I1 Magnetic flux linked with each turn of solenoid S2 will be equal to B1 times the area of cross section of solenoid S1. Magnetic flux linked with each turn of the solenoid S2 = B1A Therefore, total magnetic flux linked with the solenoid S2, Φ21 = B1A n2l = µ0n1I1 A n2l Φ21 = μ0n1n2lAI1 ∴ M21 = µ0n1n2Al


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Similarly, the mutual inductance between the two solenoids, when current is passed through solenoid S2 and induced emf is produced in solenoid S1, is given by M12 = µ0n1n2Al ∴M12 = M21 = M (say) Hence, coefficient of mutual induction between the two long solenoids

Q.15 A cell of emf E and internal resistance r is connected to two external

resistance R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations: (i) without any external resistance in the circuit (ii) with resistance R1 only (iii) with R1 and R2 in series combination (iv) with R1 and R2 in parallel combination The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.

Ans.15 (i) Without any external resistance in the circuit –

The current in this case would be maximum, so I1 = 4.2A (ii) With resistance R1 only –

The current in this case will be the second smallest value, so I2 = 1.05 A (iii) With R1 and R2 in series combination –

The current in this case will be minimum as the resistance will be maximum, so I3= 0.42 A. (iv) With R1 and R2 in parallel combination−

The current in this case would be the second largest value, so I4 = 1.4 A.

Q.16 Q16.Two identical circular loops, P and Q, each of radius r and carrying equal currents are

Kept in the parallel planes having a common axis passing through O. The direction of current in P is clockwise and in Q is anticlockwise as seen from O which is equidistant from the loops P and Q. Find the magnitude of the net magnetic field at O.

Ans.16 We know that, formula for field at an axial point is given as

=> The direction of current in P is clockwise, so, by using right hand thumb’s rule the direction of the magnetic field will be towards left. The direction of current in Q is anticlockwise, so, direction of magnetic field is towards left. Thus, the net magnetic field at point O will be the sum of the magnetic fields due to loops P and Q.


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Also, as the fields produced are at an equal distance to O, BP = BQ, So, net field

Q.17 A metallic rod of ‘L’ length is rotated with angular frequency of ‘ω’ with

one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.

Ans.17

Let us Consider the infinitesimally small length dx at a distance x. So speed of this part is ωx. Induced small emf = Bωx dx (since emf = vBl) The emf between the ends of the rotating rod is

Q.18 When an ideal capacitor is charged by a dc battery, no current flows.

However, when an ac source is used, the current flows continuously. How does one explain this, based on the concept of displacement current?

Ans.18 The ideal capacitor is a purely reactive device, containing absolutely zero resistive (power dissipative) effects. I.e. Ideal capacitor means infinite resistance for dc.

When an ac source is used, the current flows continuously, but we know that the capacitor has dielectric (air) between its plates. So, there is no current and circuit would be incomplete. Thus, capacitor is charged due to contribution of varying electric field. The current between the capacitor plates is given by displacement current.

Q.19 In the figure a long uniform potentiometer wire AB is having a constant

potential gradient along its length. The null points for the two primary cells of emfs ε1 and ε2 connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε1/ ε2 and (ii)

Ans.19 Apply Kirchhoff’s law in loop ACFGA: Φ(120) = ε1 − ε2 =>ε1 = ε2 + Φ(120)(1) Where Φ = potential drop per unit length.


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position of null point for the cell ε1.How is the sensitivity of a potentiometer increased?

OR

Using Kirchhoff’s rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Ω resistance. Also find the potential difference between A and D.

Now, Apply Kirchhoff’s law in loop AEHIA: Φ(300) = ε2 + ε1 => ε2 +(ε2 + Φ (120)) = Φ (300) (By substituting value of ε1 from equation (1)) => 2ε2 = (300 − 120)Φ => ε2 = 90Φ … (2) Thus, ε1 = 90Φ + 120Φ ε1 = 210Φ … (3) so,

(ii) As we know, ε = Φl Thus, from equation (2) and (3) Null point for cell ε2 is 90 cm and for cell ε1, it is 210 cm. Sensitivity of the potentiometer can be increased by: (a) Increasing the length of the potentiometer wire (b) Decreasing the resistance in the primary circuit.

OR Apply Kirchhoff’s In loop ABCFA: I+ I + 4I1 = 9 − 6 2I + 4I1 = 3 … (1) As there is no current flowing through the 4Ω resistance, I1 = 0 From (1), 2I = 3 => I = 1.5A Thus the current through resistances R is 1.5A. As there is no current through branch CF, thus equivalent circuit will be, By applying Kirchhoff’s loop law we get, 1.5 + 1.5 + R (1.5) = 9 − 3 R = 2Ω Potential difference between A and D = (9 − 3) = 6V

Q.20 The figure shows a series LCR circuit with L = 10.0 H, C = 40 µF, R = 60 Ω connected to a variable frequency 240 V source, calculate (i) the angular frequency of the source which drives the circuit at resonance, (ii) the current at the resonating frequency, (iii) the rms potential drop across the inductor at resonance.

Ans.20 (i) the angular frequency of the source which drives the circuit at resonance,


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(ii) the current at the resonating frequency, we know that the inductive reactance cancels out the capacitive reactance. The impedance = Z = 60Ω the value of resistance The current amplitude at resonant frequency

(iii) the rms potential drop across the inductor at resonance. For R.M.S value of current,

For R.M.S potential drop across inductor

=2000V Q.21 (a) Why are coherent sources necessary to produce a sustained

interference pattern? (b) In Young’s double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3.

Ans.21 (a) Two sources of emitting light wave of same frequency or wavelength and of stable phase difference are required to see interference pattern and we can obtain such nature of light wave from coherent source. So, coherent sources necessary to produce a sustained interference pattern. (b)

At path difference λ,


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Q.22 A rectangular loop of wire of size 2 cm × 5 cm carries a steady current of

1 A. A straight long wire carrying 4 A current is kept near the loop as shown. If the loop and the wire are coplanar, find (i) the torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.

Ans.22

torque

(as M and B have same direction)

(ii)we know that Magnetic force on line AB and Magnetic force on line CD are equal and opposite. So they cancel out each other. Magnetic force on line AD

l= 5cm=0.05m

(Attractive) Magnetic force on line CB


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=ilB’

(Repulsive) Thus, net force

Given I = 4A I = 1 A R = 1 cm = 0.01 m r′ = (1 + 2) cm = 0.03 m l = 5 cm = 0.05 m Plugging in the values

Q.23 Write Einstein’s photoelectric equation. State clearly how this equation is

obtained using the photon picture of electromagnetic radiation. Write the three salient features observed in photoelectric effect which can be explained using this equation.

Ans.23 Einstein’s photoelectric equation, hv = Kmax + Φ where, h = Planck’s constant v = frequency Φ = Work function

According to Plank's quantum theory, light is emitted from a source in the forms of bundles of energy called photons. Energy of each photon is .


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Einstein made use of this theory to explain how photo electric emission takes place. Einstein postulated that a photon of energy hv is absorbed by the electron of the metal surface, then the energy equal to Φ is used to liberate electron from the surface and rest of the energy hv −Φ becomes the kinetic energy of the electron. The minimum energy required by the electron of a material to escape out of it, Φ is work functions. The additional energy acquired by the electron appears as the maximum kinetic energy ‘Kmax’ of the electron. Kmax = hv – Φ hv = Kmax + Φ This is Einstein's photoelectric equation. Where Kmax = eV0. Salient features observed in photoelectric effect: — • There exists a minimum cut − off frequency v0, for which the stopping potential is zero. • The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the frequency of incident radiation. • Photoelectric emission is instantaneous.

Q.24 (a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n times the de Broglie wavelength associated with it. (b) The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?

Ans.24 (a) According to Bohr’s second postulate, the electron revolves around the nucleus only in circular energy levels called orbits in which the angular momentum of the electron is integral multiple of h/2π (where h = Planck’s constant = 6.62 × 10−34 Js). So, if m is the mass of electron v is the velocity of electron in permitted quantized orbit with radius r ,then

------------(1) Where n is the principle quantum number and can take integral values like N=1,2,3……… This is the Bohr’s quantization condition. Now, de-Broglie wavelength is given as

Where λ → is wavelength associated with electron. v is the velocity of electron. h − is the velocity of electron. m − mass of electron

------(2)


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Putting value of v from (2) in (1)

Hence, the circumference of the electron in the nth orbital state in hydrogen atom is n times the de Broglie wavelength associated with it. (b)If n is the quantum number of the highest energy level involved in the transitions, then the total number of possible spectral lines emitted is

The electron in hydrogen atom is initially in the third excited state. And, Third excited state means fourth energy level i.e. n = 4. The electron makes transition from n = 4 to n = 1 so highest n is n = 4. Therefore, possible spectral lines

N= 6 Thus, 6 is the maximum number of spectral lines which can be emitted when it finally moves to the ground state.

Q.25 Use Huygens’s principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light. When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band?

Ans.25 Let us consider a parallel beam of monochromatic light is incident normally in a slit of width b as shown if figure. According to Huygens’s principle every point of slit acts as a secondary source of wavelets spreading in all directions. Screen is placed at a far distance. A particular point P on the screen receives waves from all the secondary sources. All these waves start from different point of the slit and interface at point P to give resultant intensity.

At P0, all waves are in same phase thus interface constructively with each other and


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maximum intensity is observed. As we move from P0, the wave arrives with different phase and intensity is changed. Intensity at point P is given by

For central maxima α = 0 thus, I = I0 When the width of slit is made double the original width then intensity will get four times of its original value. Width of central maximum is given by,

Where, D = Distance between screen and slit, λ = Wavelength of the light, b = Size of slit. So, as size of slit increases, the width of central maxima decreases. Hence, double the size of the slit would result in half the width of the central maxima.

Q.26 At P0, all waves are in same phase thus interface constructively with each other and maximum intensity is observed. As we move from P0, the wave arrives with different phase and intensity is changed. Intensity at point P is given by

For central maxima α = 0 thus, I = I0 When the width of slit is made double the original width then intensity will get four times of its original value. Width of central maximum is given by,

Where, D = Distance between screen and slit, λ = Wavelength of the light, b = Size of slit. So, as size of slit increases, the width of central maxima decreases. Hence, double the size of the slit would result in half the width of the central maxima.

Ans.26 (i) The constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170 is saturation property of nuclear force. (ii) Let us consider an atom with mass number A. (Here, we are neglecting mass of the orbital electrons) Mass of the nucleus of the atom of the mass number A.= A a.m.u = A × 1.660565 × 10−27 kg Let radius of nucleus is R.

Now, we know R0 = 1.1 × 10−15 m

Density of the nucleus


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Hence, the density of nucleus over a wide range of nuclei is constant independent of mass number A.

Q.27 Name the three different modes of propagation of electromagnetic waves. Explain, using a proper diagram the mode of propagation used in the frequency range from a few MHz to 40 MHz.

Ans.27 There are three modes of propagation of electromagnetic waves, (a) Ground waves, (b) Sky waves, (c) Space waves. The mode of propagation used in the frequency range from a few MHz to 40 MHz is sky wave.

The ionosphere is situated about 65 km – 400 km above the surface of the Earth. In the ionosphere of the Earth’s atmosphere, there are a large number of charged particles called ions. The ionosphere acts as a reflecting layer for certain range of frequencies (3MHz30MHz). The ionization of molecules occurs due to the absorption of the ultraviolet rays and high energy radiation from the Sun. The transmitting antenna sends the EM signals of this frequency range towards the ionosphere. When the EM waves strikes the ionosphere, it is reflected back to the Earth. A receiving antenna at a remote location on the Earth receives these reflected signals.

Q.28 Explain the principle of a device that can build up high voltages of the order of a few million volts. Draw a schematic diagram and explain the working of this device. Is there any restriction on the upper limit of the high voltage set up in this machine? Explain. OR

Ans.28 The device used for building up high potential differences of the order of a few million volts is Van de Graff generator.A Van de Graaff generator operates by transferring electric charge from a moving belt to a terminal. The high voltages generated by the Van de Graaff generator can be used for accelerating subatomic particles to high speeds. Construction:


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(a) Define electric flux. Write its S.I. units. (b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it. (c) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?

It consists of a large spherical conducting shell (S) supported over the insulating pillars. A long narrow belt of insulating material is wound around. two pulleys P1 and P2. B1 and B2 are two sharply pointed metal combs. B1 is called the spray comb and B2 is called the collecting comb. Working: Working of the generator is based on two principles:

Discharging action of sharp points, ie., electric discharge takes place in air or gases readily, at pointed conductors.

If the charged conductor is brought in to internal contact with a hollow conductor, all of its charge transfers to the surface of the hollow conductor no matter how high the potential of the latter may be. The spray comb is given a positive potential by high tension source. The positive charge gets sprayed on the belt. As the belt moves and reaches the sphere, a negative charge is induced on the sharp ends of collecting comb B2 and an equal positive charge is induced on the farther end of B2. This positive charge shifts immediately to the outer surface of S. Due to discharging action of sharp points of B2, the positive charge on the belt is neutralized. The uncharged belt returns down and collects the positive charge from B1, which in turn is collected by B2. This is repeated. Thus, the positive charge on S goes on accumulating. In this way, voltage differences of as much as 6 or 8 million volts (with respect to the ground) can be built up. The main limiting factor on the value of high potential is the radii of sphere.If the electric field just outside the sphere is sufficient for dielectric breakdown of air, no more charge can be transferred to it. For a conducting sphere, Electric field just outside sphere.

and electric potential


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So, E = VR Now, for E = 3 × 106 V/m (dielectric breakdown) Radius of should be 1 m. Thus, the maximum potential of a sphere of radius 1 m would be 3 × 106V. OR (a)Electric Flux:

It is the number of electric field lines passing through a surface normally. Where E = Electric Field A = Area S.I. Unit of of flux = Nm2C−1 (b) Consider a uniformly charged infinite plane sheet of charge density σ.Now, we construct a Gaussian surface as shown in figure in the form of cylinder. Applying Gauss’s Law,

Hence, the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it (c)

Direction of field will be away from the sheet if sheet is positively charged.

Direction of field will be towards from the sheet if sheet is negatively charged.

physics paper 2012 1 - poornima university · the base line of the prism inside it. find the angle...

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