Physics: Sample Question Paper- 1, Class 12 | Physics Class 12 - NEET PDF Download

Ques 1: Define the electric dipole moment. What is the direction of this vector? (1)
Ans: It is defined as the product of magnitude of charge and the distance between them, that is, 2qa and the direction is from –q to +q.

Ques 2: What is the SI unit of electric flux and electric field? (1)
Ans: Electric flux: V.m; Electric field: V/m.

Ques 3: The number of turns in the secondary of a transformer is 100 times the number of turns in the primary. If 1kW of power is supplied to the primary, how much power is obtained at the secondary? (1)
Ans: The input power is equal to the output power if 100% efficiency is assumed. So the power at the secondary will be 1kW.

Ques 4: Arrange these in decreasing order of frequency: microwaves, infrared, x-ray, visible rays. (1)
Ans: X-ray, infrared, visible, microwaves

Ques 5: Which of these quantities remains unchanged when light is refracted: wavelength, frequency and velocity? (1)
Ans: Frequency

Ques 6: Does the frequency of incident radiation affect the stopping potential or saturation current or both? (1)
Ans: It affects the stopping potential but not the saturation current.

Ques 7: In an unbiased p-n junction why do holes diffuse from the p region to the n region? (1)
Ans: Because the hole concentration is more in the p-region as compared to the n-region.

Ques 8: What is noise in a communication system? (1)
Ans: Noise refers to any unwanted signals that tend to disturb the transmission and processing of message signals in a communication system.  

Ques 9: An electron, alpha particle and proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength? (2)
Ans: de Broglie wavelength,

For the same kinetic energy, de Broglie wavelength depends on the mass of the particle.
As alpha particle is the most massive of the three its de Broglie wavelength will be the shortest.

Ques 10: Is it correct to say that a potentiometer is equivalent to an ideal voltmeter? (2)
Ans: Yes, because the potentiometer does not draw any current from the circuit but it still measures the potential difference.
An ideal voltmeter has infinite resistance, and hence it does not draw any current from the circuit.

Ques 11: A magician during a magic show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? What is the focal length of the glass lens in the liquid? (2)
Ans: The refractive index of the liquid must be equal to 1.47 for the lens to disappear. Now the focal length of the lens is 1/f = 0 or f tends to infinity. In other words, the lens in the liquid, acts like a plane sheet of glass.

Ques 12: A short bar magnet placed with its axis at 60 degrees with a uniform external magnetic field of 1.0 T experiences a torque of magnitude 5x10 2 N m. What is the magnitude of magnetic moment of the magnet? (2)
Ans: 


Ques 13: Why does a paramagnetic sample display greater magnetism when cooled? Why is diamagnetism, in contrast, almost independent of temperature? (2)
OR
Why are sky waves not used for transmission of TV signals? State two factors that can increase the range of transmission.
Ans: The random thermal motion is reduced at low temperatures, so the tendency to disturb the alignment of dipoles is reduced at lower temperatures. However, in a diamagnetic sample the alignment is opposite to the magnetising field, so temperature does not affect diamagnetism.
OR
TV signals have a frequency between 100 and 200 MHz, so they are not reflected by the ionosphere. Thus their transmission via sky waves is not possible. The range can be increased by using, i) taller antenna, and ii) geostationary satellites.

Ques 14: What is power factor? Why does a low power factor imply large power loss in transmission? (2)
Ans: Power factor is cos∅ where P = VIcos∅.
If power factor is small, the current has to be increased to maintain the same power. This leads to greater I²R losses in transmission.

Ques 15: How does the resolving power of a compound microscope change on
i) decreasing the wavelength of light used
ii) decreasing the diameter of the objective lens? (2)
Ans: Resolving power =
i) The resolving power increases on decreasing the wavelength of light.
ii) It does not change on decreasing the diameter of the objective.

Ques 16: Find the maximum frequency and minimum wavelength of X-rays produced by 30keV electrons. (2)
Ans: Energy of electrons
E 30keV 30* 10³* 1.6* 10^-19 joule
From Einstein equation,
E= hv


Ques 17: Define the term electric potential. Give the dependence of electric potential due to a small electric dipole at a far off point lying on
i) the axial line and
ii) equatorial line (3)
Ans: Electrostatic potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point.
At a far off point lying on the axial line the electric potential is

On the equatorial plane V = 0 everywhere.

Ques 18: The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. A force causes acceleration in accordance with Newton’s second law. Why then do the electrons acquire a steady average drift speed but no acceleration?
What is the path of electrons along a straight line between successive collisions in the presence of an electric field? (3)
Ans: Each free electron accelerates due to the force that acts on it. This increases its drift speed until it collides with a positive ion of the metal. It loses some energy after each collision, but it again accelerates and then collides and once again loses some energy. This sequence of events continues.
Hence, on an average the electrons acquire only a drift speed but they are not able to accelerate.
In the presence of an electric field the path of the electrons is generally curved, rather than straight.

Ques 19: With the help of a labelled diagram, briefly describe the construction of a coaxial cable. What is the upper limit of frequency up to which a coaxial cable can be used? (3)
Ans:
 
A coaxial cable is as shown in the figure above. It consists of an inner conductor that is made up of a copper wire and the outer conductor can be either a solid or braided mesh of fine wires. The outer conductor is externally covered with a polymer jacket for protection.
Upper limit of the frequency that can be used is 20MHz.

Ques 20: Two devices A and B are connected independently to a variable frequency alternating voltage source. The current in A leads the applied voltage whereas the current in B lags the applied voltage.
i) Identify the circuit element A and B.
ii) How will the current in these circuit elements change if the applied frequency is decreased? (3)
Ans: A is a capacitor and B is an inductor.
On decreasing the frequency of the applied voltage the inductive reactance decreases, so the current through the inductor will increase.
For this change in the frequency, the capacitive reactance increases, so the current through the capacitor decreases.

Ques 21: An ac voltage E = E₀ sinωt is applied across a pure inductor of inductance L. Show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of π/2. (3)
Ans:
 

Hence current lags behind the voltage by a phase angle of π/2.

Ques 22: Briefly explain the principle of working of an ac generator. What is the maximum emf produced by it? (3)
Ans: A wire loop of area A is free to rotate about an axis which is perpendicular to a uniform magnetic field B.

If the normal to the loop n makes an angle θ with  , then flux through the loop is 

If this loop rotates with a constant angular velocity  the flux through it changes at the rate

where C₀ is a constant
∴ emf is induced between ends A and B given by

V_m = BAω = Peak value of emf generated.

Ques 23: Explain, with the help of a labelled diagram, the principle and construction of a cyclotron. (3)
Ans: Principle: A beam of charged particles describe a circular path when subjected to a uniform magnetic field directed perpendicular to their plane of motion. The beam can be accelerated time and again by a high frequency electric field of correctly adjusted frequency applied between two dees.

Construction: It consists of a two semi circular disc like hollow metal containers called dees D₁ and D₂ connected to high frequency alternating voltage.
A magnetic field is applied perpendicular to the plane of the dees.

Ques 24: A semiconductor has equal electron and hole concentrations of 2 x 10 ⁸ /m³. On doping with a certain impurity, the hole concentration increases to 4 x 10¹⁰ /m³.
(i) What type of semiconductor is obtained on doping?
(ii) Calculate the new electron hole concentration of the semiconductor.
(iii) How does the energy gap vary with doping?  (3)
OR
(a) It is often heard “The domestic supply is at a voltage of 220V”. What voltage value is this referring to? Express both peak & rms values.
 (b) What is the instantaneous and average power per cycle supplied to the Capacitor? (3)
Ans: Given: n_i = 2 x 10⁸/m³ 
On doping n_h = 4 x 10¹⁰/m³
(i) On doping hole concentration will increase, hence, it is p-type semiconductor.

(iii) Energy gap decreases with doping because acceptor level gets created between the valance band and conduction band.
OR
(a) It is referring to the RMS value. V_rms=220V
So, the peak value for voltage is  
(b) The instantaneous power is always P(t ) = V (t )I (t ) , and in this case, it is


Ques 25: (a) Draw a labeled ray diagram to show the formation of an image by a compound microscope. Write expression for its magnifying power.
(b) How does the resolving power of a compound microscope change, when
(i) refractive index of the medium between the object and the objective lens is increased and
(ii) wavelength of the radiation used is increased? (3)
Ans: (a)

Here,
O = objective lens
E = Eye piece
AB = object, A” B” = final image
Magnifying power ( m ) of a compound microscope is given by

where L = distance between C₁ and C₂
f_e ,f_o are focal length of eye-piece and objective respectively
d = distance of final image formed from eye-piece.
(b) Resolving power of a compound microscope is given by

where μ is the refractive index of the medium between object and objective lens, θ is half the angle of cone of light from the point object and λ is the wavelength of radiations used.
(i) on increasing  μ , resolving power will increase.
(ii) on increasing λ , the resolving power will decrease.

Ques 26: In a dark night, a thief was climbing up on a wall of the building with the help of a pipeline attached to that building. A man on the roof of the adjacent building saw him and to stop him he focused the torch to the pond at the ground floor lying between both the buildings, so that the reflected light reaches the eyes of the thief. As the same happened, the thief got afraid and lost his control over the pipe and fell in the pond.
(a) What is the interesting matter in this event?
(b) Explain reflection and its laws with suitable ray diagram. (4)
Ans: (a) The interesting matter is that the king of darkness, so called thief, got afraid with a simple light reflection and lost his control over the pole and fell down in the pond.
(b) When light rays are incident on the surface of the transparent media by making an angle with the normal called angle of incidence, it reflects back by making same angle with the normal, called angle of reflection.
Laws of reflection:
(i) Angle of reflection is equal to the angle of incidence.
(ii) Incident ray, reflected ray and the normal are in the same plane.


Ques 27: (a) State the essential condition for diffraction of light to take place.
(b) Use Huygen’s principle to explain diffraction of light due to a narrow single slit and the formation of a pattern of fringes obtained on the screen.
(c) Sketch the pattern of fringes formed due to diffraction at a single slit showing variation of intensity with angle θ .
OR
(a) What are the coherent sources of light? Why are coherent sources required to obtain sustained interference pattern?
(b) State three characteristic features which distinguish the interference pattern due to two coherently illuminated sources as compared to that observed in a diffraction pattern due to a single slit. (5)
Ans: (a) Condition: Size of the obstacle or aperture should be comparable with the wavelength of the incident light.
(b) Let a plane wave front incident on slit LM be of width ‘a’. To consider diffraction effects at any point P, directed an angle θ to the incident rays, the wave front is divided into a number of parts, with each part treated as a secondary wave front.

Path difference between secondary wave front from L and M = αsinθ
For minima, α sin θ =nλ
For maxima, α sin θ =
(c)

OR
(a) Coherent Sources of light: Two sources with same frequencies having a (time independent) stable phase difference or zero phase difference.
If phase difference is variable, the interference term averages to zero. In case of stable phase difference between coherent sources, intensity at each point will vary and so sustained interference will be observed.
Difference between interference and diffraction pattern formed due to a single slit:

Ques 28: State Biot-Savart’s law. Use it to obtain the magnetic field at an axial point at a distance x from the centre of a circular coil of radius a carrying a current I. Compare the magnitude of the magnetic field at the centre of the coil and at an axial point for which x = √3a.
OR
Distinguish between the magnetic properties of para, diamagnetic and ferromagnetic substances in terms of
i) susceptibility
ii) permeability
iii) coercivity
Give one example of each of these materials. Draw magnetic field lines near a diamagnetic and paramagnetic substance. (5)
Ans: According to the Biot-Savart’s law, the magnetic field dB due to an element dl carrying a steady current I at a point P at a distance r from the current element is

By taking a circuit element we can write down the magnetic field due to this segment. Then integrate to find the magnetic field due to the circular current loop, which comes out to be

The magnetic field at the centre of the coil is found by putting x = 0

Magnetic field at an axial point for x = √3a

OR
Diamagnetic substances have small and negative susceptibility.
Their magnetic permeability is less than 1
Their coercivity is very high.  
Paramagnetic substances have positive but small values of susceptibility.
Their magnetic permeability is greater than 1.
Their coercivity is low.
Ferromagnetic substances have large and positive values of susceptibility.
Their magnetic permeability is very large.
Their coercivity is very low.  
Example of diamagnetic substance is bismuth
Example of paramagnetic substance is aluminium
Example of ferromagnetic substance is iron.



Ques 29: A transistor is used in common emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base-emitter voltage, the base current changes by 20 μA and the collector current changes by 2 mA. The load resistance is 5 kΩ . Calculate (a) the factor β , (b) the input resistance R_BE, (c) the transconductance and (d) the voltage gain.
OR
Draw a labeled diagram of a compound microscope when the image is formed at infinity. Deduce the expression for its magnifying power. How can the magnifying power be increased?
Ans:  
(b) The input resistance R_BE 
(c) Transconductance =
(d)  The change in input voltage is
The applied signal voltage = 20 mV
Thus, the voltage gain is,

OR
A labeled diagram of a compound microscope, when the final image is formed at infinity:


If m_o and m_e be the magnification produced by the objective lens and eye lens respectively, then the total magnifying power of microscope is m = m_o × m_e.
If u_o be the distance of the object AB from objective lens and v_o be the distance of image A’B’ formed by objective, then magnifying power of objective is  

The magnifying power of eyepiece when the final image is formed at infinity is

Here, object distance,  u_e = f_e 
D = Least distance of distinct vision
Now, the magnifying power of microscope is

As a first approximation
and v_o = L (distance between objective and eyepiece) then after substituting the values from above equation we get.