Question 1 5 / -1
When a transistor is used in common emitter configuration, the variation of the _____________ voltage is called the output characteristic.
Solution
CONCEPT:
Transistor: A semiconductor device used to amplify or switch electronic signals and electrical power is called a transistor .It is made of semiconductor material with at least three terminals for connection to an external circuit hence they are called a bipolar transistor. The three basic single-stage bipolar junction transistor which is used as a voltage amplifier is called CE amplifier . The input (Vin ) of CE amplifier is taken from the base terminal and the output (Vout ). is collected from the collector terminal The emitter terminal is common for both base and collector terminals and the input is applied across the emitter-base junction with the forward-biased this amplifies the signal. Thus the base current is amplified by applying suitable input voltage across the base-emitter junction and we get amplified output voltage across collector-emitter junction
Explanation-
From the above explanation, we can see that, in common emitter configuration, the base current is amplified and we get amplified output signal.
This configuration can be used as switching circuit, in which by applying suitable base-emitter current the circuit can be switched on or off.
Hence when a transistor is used in common emitter configuration the variation is output can be made by varying base current and the variation of the collector current with the collector-emitter voltage is called the output characteristic.
Question 2 5 / -1
The area of the plates of a parallel plate capacitor is A. If the capacitor is having a charge Q, then the force between the plates is:
Solution
CONCEPT :
Capacitor:
The capacitor is a device in which electrical energy can be stored . In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium. The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric. Parallel plate capacitor:
A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance. The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric. The electric field intensity at the outer region of the parallel plate capacitor is always zero whatever be the charge on the plate. The electric field intensity in the inner region between the plates of a parallel plate capacitor remains the same at every point. The electric field intensity in the inner region between the plates of a parallel plate capacitor is given as, \(\Rightarrow E=\frac{σ}{\epsilon_o}=\frac{Q}{A\epsilon_o}\)
The potential difference between the plates is given as, \(\Rightarrow V=\frac{Qd}{A\epsilon_o}\)
The capacitance C of the parallel plate capacitor is given as, \(\Rightarrow C=\frac{Q}{V}=\frac{A\epsilon_o}{d}\)
Where A = area of the plates, d = distance between the plates, Q = charge on the plates, and σ = surface charge density
EXPLANATION:
We know that the electric field intensity due to one plate of the capacitor on the other plate is given as, \(\Rightarrow E=\frac{Q}{2A\epsilon_o}\) -----(1)
The magnitude of electric force experienced by a charged particle in an electric field is given as, ⇒ F = Eqo -----(2)
Where E = electric field intensity, and qo = charge on the particle
By equation 1 and equation 2, the force on a plate due to the charge Q on the other plate is given as (qo = Q),
⇒ F = Eqo
\(\Rightarrow F=\frac{Q}{2A\epsilon_o}\times Q\)
\(\Rightarrow F=\frac{Q^2}{2A\epsilon_o}\)
Hence, option 2 is correct. Additional Information
If the dielectric medium of dielectric constant K is filled between the plates:
When the dielectric medium is filled in the space between the plates of the parallel plate capacitor, its capacitance increases. The electric field intensity in the inner region between the plates of a parallel plate capacitor is given as,
\(\Rightarrow E'=\frac{σ}{\epsilon_oK}=\frac{Q}{A\epsilon_oK}\)
The potential difference between the plates is given as,
\(\Rightarrow V'=\frac{Qd}{A\epsilon_oK}\)
The capacitance C of the parallel plate capacitor is given as,
\(\Rightarrow C'=\frac{Q'}{V'}=\frac{A\epsilon_oK}{d}\)
Question 3 5 / -1
Choke coil:
Solution
CONCEPT:
Choke coil:
A choke coil is a device that is used in the electrical circuits to allow DC current to flow through it while blocks AC current from passing . A choke consists of a coil of insulated wire that is wound around a magnetic core. The low electrical resistance allows DC and AC to pass with little power loss but it limits the AC current from passing through it due to its reactance . The choke coil works because it can act as an inductor. When the current passes through will change as AC currents create a magnetic field in the coil that works against the AC current. This is known as inductance and blocks most of the AC current from passing through. This results in the currents that do not change such as DC currents can continue passing through while those currents are blocked by the magnetic field. EXPLANATION:
A choke coil is a low resistance inductor coil used to suppress or limit the flow of alternating current without affecting the flow of direct current , as in applications that require conversions of AC to DC . Hence, option 1 is correct.
Question 4 5 / -1
Which among the following is not related to the conservation of charge?
Solution
CONCEPT :
Law of conservation of charge:
The total charge of an isolated system remains constant . The electric charges can neither be created nor destroyed , they can only be transferred from one body to another . The law of conservation of charge is obeyed both in large scale and microscopic processes . In fact, charge conservation is a global phenomenon i.e., the total charge of the entire universe remains constant . EXPLANATION :
From above it clear that charge can neither be created nor be destroyed , but it can be transferred from one object to another by using some methods like induction and conduction . Therefore option 1 is incorrect and options 2 and 4 is correct. If the charges are distributed in a system, then the net charge of the system remains constant . Therefore option 3 is correct
Question 5 5 / -1
Angular momentum of an electron revolving in 2nd orbit of radius r:
Solution
CONCEPT :
Bohr model : In 1913, Niels Bohr gave the Bohr's atom model which retained essential features of Rutherford's model and at the same time took into account its drawbacks.The electrons revolve around the nucleus in circular orbits which is called the stationary orbit. According to Bohr's atom model, the electrons of an atom revolve around the nucleus only in those orbits in which the angular momentum of the electron is an integral multiple of h /2π. Which is given by \(\Rightarrow L = \frac{nh}{2\pi}\)
Where n = orbit, h = Planck constant
CALCULATION :
Given n = 2
According to Bohr's atom model, the angular momentum is given by \(\Rightarrow L = \frac{nh}{2\pi}\)
Substituting the value of n in the above equation
\(\Rightarrow L = \frac{2h}{2\pi} = \frac{h}\pi{}\)
Hence, option 2 is the answer
Question 6 5 / -1
A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 8 : 1. The ratio of radii of the fragments is
Solution
CONCEPT:
Conservation of Momentum: In a system or body, if no external force acts on it then total momentum is conserved. The linear momentum of a body is given mass × velocity. In the nucleus, neutrons are closely packed in a small radius.The cube of the radius (R) of the nucleus is proportional to the mass number (M). R3 α M or (R1 /R2 )3 = M1 /M2 R1 /R2 = (M1 /M2 )1/3 CALCULATION:
given that v1 /v2 = 8/1
Momentum = mv
By conservation of momentum m1 v1 = m2 v2
v1 /v2 = m2 /m1 = 8/1 ----(1)
Also from R1 /R2 = (M1 /M2 )1/3 = (1/8)1/3 = 1/2
So, The answer is option 1 i.e. 1 : 2
The volume of the nucleus is proportional to Mass.V α M 4/3 π R3 α M from here formula R3 α M (or radius of nucleus α M1/3 )
Question 7 5 / -1
All capacitors used in the diagram are identical and each is of capacitance C. Then the effective capacitance between the point A and B is_______
Solution
CONCEPT :
In the Parallel circuit, the equivalent capacitance is the algebraic sum of all the capacitance.And in the Series circuit , the reciprocal of the equivalent capacitance is the algebraic sum of all the reciprocal of the capacitance.
Ceq = C1 + C2 + C3 +...... (In parallel)
1/Ceq = 1/C1 + 1/C2 + 1/C3 +...... (in series)
CALCULATION :
Here First three capacitors are in parallel combination with each other and the next three are in parallel with each other. And both are in series with each other For parallel combination:
Cnet = Ceq = C1 + C2 + C3 = C + C + C = 3 C
C' net = Ceq = C1 + C2 + C3 = C + C + C = 3 C
For Series combination :
Now 1/Cresultant = 1/3C + 1/3C = 2/3C
Cresultant = 3C/2 = 1.5 C
So the correct answer is option 3
Additional Information
Wheatstone bridge: The Wheatstone bridge works on the principle of null deflection , i.e. the ratio of the resistances in the circuit is equal and there is no current flow in the circuit.\({R_x \over R_3} = {\frac{{{R_2}}}{{{R_1}}}} \)
It works on the principle of null deflection. Under normal conditions, when the bridge is in an unbalanced condition the current flows through the galvanometer. When the bridge is in a balanced condition, there will be no current flow through the galvanometer.
Question 8 5 / -1
Which one of the following circuit elements is an active component?
Solution
A passive element is an electrical component that does not generate power but instead dissipates, stores, and/or releases it ex : resistor, capacitor, and inductor Active components include amplifying components such as transistors , Zener diode and tunnel diodes.Either all semi conducts are considered to be an active component
Question 9 5 / -1
According to Bohr's model, the radius of the second orbit of the helium atom is
Solution
CONCEPT:
Bohr's Orbits (for Hydrogen and H2-like Atoms)-
Radius of the orbit : For an electron around a stationary nucleus the electrostatic force of attraction provides the necessary centripetal force .Mathematically the radius of an orbit can be given as
\({r_n} = \frac{{0.53{n^2}}}{Z}{\rm{{\dot A}}}\)
Where Z = atomic number and n = orbit number
EXPLANATION:
Given – n = 2 and Z = 2
Mathematically the radius of an orbit can be given as
\({r_n} = \frac{{0.53{n^2}}}{Z}{\rm{{\dot A}}}\)
\({r_2} = \frac{{0.53{{\left( 2 \right)}^2}}}{2}{\rm{{\dot A}}} = \frac{{0.53 \times 4}}{2}{\rm{{\dot A}}}\)
r
2 = 1.06 Å
Question 10 5 / -1
Which of the following statement is true about de-Broglie wavelength of moving ball in cricket:
Solution
CONCEPT:
According to de Broglie matter has a dual nature of wave-particle. The wave associated with each moving particle is called matter waves . Characteristics of Matter waves: The lighter the particle, the greater is the de Broglie wavelength. The higher the velocity of the particle, the smaller is its de Broglie wavelength. The de Broglie wavelength of a particle is independent of the charge or nature of the particle. The matter waves are not electromagnetic in nature. Only charged particles produce electromagnetic waves. CALCULATION:
de Broglie wavelength associated with the particle \(⇒ \lambda = \frac{h}{P}\) -----(1)
Where, h = Planck's constant and P = Linear momentum of a particle
If, m = Mass of the particle, v = Velocity of the particle, and E = Energy of the particle
\(⇒ E=\frac{1}{2}\times m\times v^{2}\) -----(2)
⇒ P = mv -----(3)
By equation 2 and equation 3:
⇒ \(P=\sqrt{2\times m \times E}\)
Hence,
⇒ \(\lambda = \frac{h}{\sqrt{2mE}}\)
Hence, option 2 is correct.
Question 11 5 / -1
A thin spherical conducting shell has charge Q on it. The potential inside the conductor has a constant value throughout. The electric field inside the conductor is?
Solution
CONCEPT :
Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field. \(⇒ V=\frac{W}{q}\)
Potential due to a single charged particle Q at a distance r from it is given by: \(⇒ V=\frac{Q}{4\piϵ_{0}r}\)
Where, ϵ0 is the permittivity of free space and has a value of 8.85 × 10-12 F/m in SI units
An Equipotential surface is a surface with a constant value of the potential at all points on the surface. The following figures show equipotential surfaces due to a point charge and an electric dipole.
The relation between electric potential and the electric field is, \(\Rightarrow E=-\frac{dV}{dx}\)
Where E is the electric field, V is the electric potential, and x is the direction of the electric field in space
The relation between electric potential and the electric field is, \(\Rightarrow E=-\frac{dV}{dx}\)
Since the potential inside the conductor is constant with any variation in dx, then dV = 0, \(\Rightarrow \frac{dV}{dx} = 0\)
\(\Rightarrow E=0\)
Therefore option 3 is correct.
Additional Information
Potential at a point P due to a system of charged particles Q1 , Q2 , Q3 , ... Qn having distances r1 , r2 , r3 , ... rn respectively from point P is given by: \(⇒ V=\frac{Q_1}{4\piϵ_{0}r} + \frac{Q_2}{4\piϵ_{0}r} + \frac{Q_3}{4\piϵ_{0}r} + ...+\frac{Q_n}{4\piϵ_{0}r}\)
\(⇒ V=\sum_{i=1}^{n}\frac{Q_i}{4\piϵ_{0}r_i}\)
Question 12 5 / -1
To a fish under water viewing obliquely a fisherman standing on the bank of a lake, the man looks
Solution
CONCEPT :
Refraction of Light: The bending of the ray of light passing from one medium to the other medium is called refraction .
The refraction of light takes place on going from one medium to another because the speed of light is different in the two media . The greater the difference in the speeds of light in the two media, the greater will be the amount of refraction . A medium in which the speed of light is more is known as an optically rarer medium and a medium in which the speed of light is less is known as an optically denser medium . EXPLANATION :
The fish is in water which is a denser medium compared to air present on the banks of river where the fisherman is standing. In simple terms the light is travelling from a rarer medium to denser medium then according to laws of refraction the light must bend towards the normal or the object on the banks appears to be bigger than the actual height. Hence, option 1 is the answer.
Question 13 5 / -1
Our sky looks like red at the time of sunrise and sunset because in this position the scattering will be
Solution
The correct answer is Maximum .
During sunrise and sunset , the light rays have to travel a longer distance so scattering is maximum .During the day , the scattering is minimum .Key Points
During sunrise, the light rays coming from the Sun have to travel a greater distance in the earth’s atmosphere before reaching our eyes. In this journey, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. Since blue color has a shorter wavelength and red color has a longer wavelength, the red color is able to reach our eyes after the atmospheric scattering of light. Therefore, the Sun appears reddish early in the morning. Additional Information
Scattering of light is the phenomenon in which light rays get deviated from their straight path on striking an obstacle like dust or gas molecules, water vapors, etc.Scattering of light gives rise to many spectacular phenomena such as the Tyndall effect and the “red hues of sunrise and sunset ”.
Question 14 5 / -1
An electron in a TV tube has a speed of 6 × 107 m/s. What is the de Broglie wavelength associated with the electron (m = 9.11 × 10-31 kg)?
Solution
Concept:
De Broglie proposed that the wavelength λ associated with a particle of momentum p is given by:
\(\lambda = \frac{h}{p} = \frac{h}{{mv}}\)
where m is the mass of the particle and v its speed. Planck’s Constant, h = 6.623 ×× 10-34 Js,
Above equation is known as the de Broglie relation and the wavelength λ of the matter-wave is called de Broglie wavelength.
Thus, the significance of de Broglie equation lies in the fact that it relates the particle character to the wave character of matter.
Important Points
According to Planck’s quantum theory, the energy of a photon is described as E = hν
According to Einstein’s mass energy relation, E = mc2
Frequency ν can be expressed in terms of wavelength λ as, ν = c/λ
hν = mc2 ⇒ hν/c = mc ⇒ λ = h/mc, (This equation is applicable for a photon)
Calculation:
Given:
m = 9.11 × 10-31 kg, v = 6 × 107 m/s
\(\lambda = \frac{h}{{mv}} = \frac{{6.623 \times {{10}^{ - 34}}}}{{9.11 \times {{10}^{ - 31}} \times 6 \times {{10}^7}}} = 1.21 \times {10^{ - 11}}m = 0.121 \) Å
Question 15 5 / -1
Experimental observation in a single slit experiment show that the intensity has a central maximum at θ = 0 and other secondary maxima at θ = ______________. (where 'θ' is the angle between the central line and the line joining the midpoint of the slit of width 'a' to any point 'P' on the screen, 'λ' is the wavelength of the light and where n = ±1, ±2, ±3, ....)
Solution
CONCEPT:
Single Slit Experiment: The diffraction or bending phenomenon of light in which light from a coherent source interfere with itself and produce a distinctive pattern on the screen called the diffraction pattern.In single slit diffraction, the light that passes through a single slit is on the order of the wavelength of the light. The diffraction pattern on the screen will be at a distance D away from the slit.
For secondary maxima :
θ = (n+1/2)λ/a
and for central maxima, θ = 0
For minima :
θ = nλ/a
Where a is slit width, n = ±1, ±2, ±3, .... and λ is wavelength.
EXPLANATION :
Thus the intensity has a central maximum at θ = 0 and other secondary maxima at θ = (n+1/2)λ/a . So option 1 is correct,
Question 16 5 / -1
Nature of electromagnetic waves is transverse. Evidence of this is given by
Solution
CONCEPT :
Polarization: This is a process by which unpolarised light is transformed into polarized light.Polarisation tells about the wave nature of light, as the light wave is polarised in a particular plane. The longitudinal waves cannot be polarised. EXPLANATION :
Polarization does not work on the longitudinal wave. Only transverse waves can be polarized. So, by the phenomenon of polarization, electromagnetic waves show transverse nature . So the correct answer is option 2. Additional Information
Diffraction It refers to various phenomena that occur when a wave encounters an obstacle or a slit . It is defined as the bending of waves around the corners of an obstacle or through an aperture into the region of the geometrical shadow of the obstacle/aperture. E.g - the rainbow pattern found on a CD .
Electromagnetic waves: The wave which is generated due to vibration between electric field and magnetic field and it does not need any medium to travel is called an electromagnetic wave. It can travel through a vacuum.Light is a form of energy which is an example of electromagnetic waves. Electromagnetic waves are transverse in nature because they propagate by varying their electric and magnetic fields so that both the fields propagate perpendicular to each other.
Question 17 5 / -1
The device used for converting AC to DC is called __________?
Solution
CONCEPT :
Rectifier : A device that is used to convert ac voltage into dc voltage is called a rectifier .
Semiconductors: The materials that have a conductivity between conductors and insulators are called semiconductors.Conductor : The material that has very high conductivity is called conductor.EXPLANATION :
The device used for converting AC to DC is called Rectifier . Hence option 4 is correct.
Question 18 5 / -1
The phenomenon utilized in an optical fibre is
Solution
CONCEPT :
Optical fibers are transparent fibers and act as a light pipe to transmit light between its two ends . They are made up of silicon dioxide .
The total internal reflection occurs when the angle of incidence is greater than the critical angle . In optical fiber glasses of a high and lower index are assembled in the precise order. The light comes in from one end of the fiber and after thousands of successive internal reflections, the light reaches the opposite end of the fiber with almost zero loss. EXPLANATION :
The refractive index of the material of the core (μ1 ) is higher than that of the cladding (μ2 ) . Therefore option 1 is correct. When the light is incident on one end of the fiber at a small angle , the light passes inside, undergoes repeated total internal reflections along with the fiber, and finally comes out. Hence, the function of the cladding is to provide a lower refractive index at the core interface in order to cause reflection within the core so that light waves are transmitted through the fiber . Thus, cladding is used in the optical fiber in order to cause Total internal reflection .
Question 19 5 / -1
Which is the best device to measure e.m.f. of a cell:
Solution
CONCEPT :
Some measuring devices are:
S.No Voltmeter Ammeter Galvanometer Potentiometer 1. It is a device used to measure potential difference and is always put in parallel with the ‘circuit element’ across which potential difference is to be measured.
It is the device to measure current and is always put in connected in series across the device the current is to be measured.
An instrument used to detect, measure, and determine the direction of electric currents is called a galvanometer .
A potentiometer is a device mainly used to measure the emf of a given cell and to compare emf's of cells .
2. Ideally, the voltmeter has infinite resistance .
Ideally, the ammeter has zero resistance . A resistance of high value is connected in series to convert the galvanometer into a voltmeter . It is also used to measure the internal resistance of a given cell .
EXPLANATION :
From the above, it is clear that a potentiometer is the best device to measure e.m.f. of a cell . Therefore option 4 is correct.
Question 20 5 / -1
Radioactive polonium 214 Po, decays by alpha emission to
Solution
CONCEPT :
Radioactivity:
Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation . A material containing unstable nuclei is considered radioactive. A radioactive nucleus consists of an unstable assembly of protons and neutrons which becomes more stable by emitting an alpha, a beta particle, or a gamma photon. Atoms are radioactive if their nuclei are unstable and spontaneously (and random) emit various particles α, β, and/or γ radiations. Three crucial forms of Radioactivity: Gamma Decay -(Photons having high energy are throw down).Beta Decay -(Discharge consists of Electrons).Alpha Decay -(Discharge consists of a Helium nucleus).An α particle is essentially a He nucleus consisting of two protons and two neutrons . Two protons correspond to 2 units of charge on an α particle. Two protons and two neutrons correspond to 4 units of mass of an α particle. Hence, the symbol for an α particle is 4 He2 . From the definition of α particle given above, emission of an α particle will lead to a decrease in 2 units of charge and a decrease in 4 units of mass of the atom. CALCULATION:
Given Atomic number Z1 = 84 and Mass number A1 = 214
An α particle is essentially a He nucleus consisting of two protons and two neutrons . From the definition of α particle given above, emission of an α particle will lead to a decrease in 2 units of atomic number and a decrease in 4 units of mass of the atom. So when the polonium 214 Po, decays by alpha emission, then the resultant element will have,
⇒ Z2 = Z1 - 2
⇒ Z2 = 84 - 2
⇒ Z2 = 82
And,
⇒ A2 = A1 - 4
⇒ A2 = 214 - 4
⇒ A2 = 210
The Pb (Lead) has the atomic number 82. Therefore the new element will be,
⇒ 210 Pb(Z=82)
Hence, option 2 is correct.
Question 21 5 / -1
Volt per ampere is also called
Solution
CONCEPT
Ohm’s law : The Current through the metallic element is directly proportional to the potential difference when the temperature remains constant. i.e., ⇒I ∝ V or V ∝ I or V = IR
Where V = potential difference, I = current, and R = resistance
Ohm’s law is not a universal law, the substances, which obey ohm’s law are known as ohmic substance. e.g. aluminium, copper, silver conductors, etc. The substance which does not obey ohm’s law is non-ohmic substance e.g. gases, crystal rectifiers, thermionic valves, transistors, semiconductors , diode , bulb filaments, etc.
If the graph between current (I) and Voltage (V) is a straight line, they obey ohm’s law. The non-ohmic resistors show a curve in the graph. EXPLANATION :
The ohms law is written as: ⇒ V = IR
The above equation is written as,
\(\Rightarrow R = \frac{V}{I}\Rightarrow (\Omega)\,Omega=\frac{Volt\,(V)}{Ampere \,(I)}\)
Question 22 5 / -1
A particle having a charge of 15 µC is brought from infinity to a point P by doing a work of 3.0 mJ. The potential at point P is ________.
Solution
CONCEPT :
Electric potential (V):
The potential difference between two points in an electric field may be defined as the amount of work done in moving a unit positive charge from one point to the other against the electrostatic force i.e., \({\rm{Electric\;potential\;}}\left( {\rm{V}} \right) = \frac{{{\rm{Work\;done\;}}\left( {\rm{W}} \right)}}{{{\rm{Charge\;}}\left( {\rm{q}} \right)}}\)
The electric potential V at a point P due to point charge +q at a distance r is given by \(⇒ V=\frac{1}{{4{\rm{\Pi }}\varepsilon_o }}\frac{q}{r}\)
CALCULATION :
Given - Charge (q) = 15 μC = 15 × 10-6 C and work done (W) = 3 mJ = 3 × 10-3 J
The potential at point P is \(⇒ V =\frac{W}{q}=\frac{3\times 10^{-3}}{15\times 10^{-6}}=0.2\times 10^{3}\, V=200\, V\)
Question 23 5 / -1
Two charges +2q and -6q are placed at a distance r from each other. Another charge Q is placed at the midpoint between the two charges such that the resultant force on the charge -6q is zero, then the charge Q will be equal to:
Solution
CONCEPT:
The force between multiple charges:
Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed the principle of superposition. The principle of superposition says that in a system of charges q1 , q2 , q3 ,..., qn , the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q2 , q3 ,..., qn . The total force F1 on the charge q1 , due to all other charges, is then given by the vector sum of the forces F12 , F13 , ..., F1n . \(⇒ \vec{F_1}=\vec{F_{12}}+\vec{F_{13}}+...+\vec{F_{1n}}\)
CALCULATION:
Given: q1 = +2q, q2 = -6q, q3 = Q, and r = distance between q1 and q2
The force on the charge q2 due to charge q1 is given as, \(⇒ F_{21} = k\frac{{{q_1}\; \times \;{q_2}}}{{{r^2}}}\)
\(⇒ F_{21} = k\frac{{{2q}\; \times \;{-6q}}}{{{r^2}}}\)
\(⇒ F_{21} = \frac{{{-12kq^2}}}{{{r^2}}}\) ----(1)
The charges q1 and q2 are unlike so they will attract each other. Therefore the direction of the force F21 will be towards the right. Similarly, the force on q2 due to charge q3 is given as,
\(⇒ F_{23} = k\frac{{{q_3}\; \times \;{q_2}}}{{{(\frac{r}{2})^2}}}\)
\(⇒ F_{23} = k\frac{{{Q}\; \times \;{-6q}}}{{{(\frac{r}{2})^2}}}\)
\(⇒ F_{23} = \frac{{{-24kqQ}}}{{{r^2}}}\) ----(2)
The force on the charge q2 will be zero if the forces F21 and F23 are equal in magnitude and opposite in direction. ⇒ F21 = -F23
\(⇒ \frac{{{-12kq^2}}}{{{r^2}}}=-\frac{{{-24kqQ}}}{{{r^2}}}\)
\(⇒ Q=-\frac{q}{2}\)
Hence, option 2 is correct.
Question 24 5 / -1
Light from a point source in air falls on a spherical glass surface (n = 1.5 c and radius of curvature = 10 cm). The distance of the light source from the glass surface is 50 cm. At what position the image is formed?
Solution
Concept Using the lens maker formula :
\(\frac{n_2}{v} -\frac{n_1}{u} = \frac{n_2 -n_1}{R}\)
Where n2 = refractive index of glass, n1 = refractive index of air, v = image distance, u = object distance, R = radius of curvature of the surface
Calculation
Given: n2 = 1.5, n1 = 1, R = 10 cm and u = - 50 cm
\(⇒ \frac{n_2}{v} -\frac{n_1}{u} = \frac{n_2 -n_1}{R}\)
\(⇒ \frac{1.5}{v} -\frac{1}{-50} = \frac{1.5 -1}{10}\)
\(⇒ \frac{1.5}{v} = \frac{0.5}{10}-\frac{1}{50}\)
\(⇒ \frac{1.5}{v} = \frac{3}{100}\)
\(⇒ \frac{1}{v} = \frac{1}{50}\)
⇒ v = 50cm
Hence the image is formed at a distance of 50cm from the glass surface, in the direction of light
because v is positive
Hint Sign convention for lens formula:
The length measured from the optic center to the right-hand side is “positive (+ve)” The length measured from the optic center to the left-hand side is “negative (-ve)”. All the distance is measured from the optic center only. The focal length for the concave lens is negative and for the convex lens is positive.
Question 25 5 / -1
Potential at a point due to a point charge is V. The charge is doubled and also the distance of the point from the charge is doubled. The new potential is
Solution
Concept:
Electric potential :
The amount of work done to move a unit charge particle from infinity to a point in an electric field is called as potential at that point. The SI unit of electric potential is Volt (V). The electric potential due to a charge particle at a distance r is given by:
\(Electric\;potential\;\left( V \right) = \frac{{Kq}}{r}\)
Where K is a constant = 9× 109 Nm2 /C2 , q is charge and r is distance from charge particle
Explanation :
Initial distance = r
Initial charge = q
\(Initial\;electric{\rm{\;}}potential{\rm{\;}}\left( V \right) = \frac{{Kq}}{r}\)
Given that ;
Final distance (r’) = 2 × initial distance = 2r
Final charge (q’) = 2 × initial charge = 2q
\(New\;electric\;potential\;\left( {V'} \right) = \frac{{Kq'}}{{r'}} = \frac{{K2q}}{{2r}} = \frac{{Kq}}{r} = V\)
Question 26 5 / -1
Which of the following statement is correct regarding the eddy current?
Solution
CONCEPT:
Eddy current:
When a changing magnetic flux is applied to a bulk piece of conducting material then circulating currents called eddy currents are induced in the material . Because the resistance of the bulk conductor is usually low, eddy currents often have large magnitudes and heat up the conductor. These are circulating currents like eddies in the water. The experimental concept was given by Focault hence also named “Focault current” . The production of eddy currents in a metallic block leads to the loss of electric energy in the form of heat. Eddy currents are minimized by using laminations of metal to make a metal core. The laminations are separated by an insulating material like lacquer. Eddy currents are undesirable in the metallic cores of transformers, electric motors, and other such devices in which a coil is to be wound over a metallic core since they heat up the core and dissipate electrical energy in the form of heat. Eddy currents are used to advantage in certain applications like:Magnetic braking in trains Electromagnetic damping Induction furnace Electric power meters EXPLANATION:
Eddy currents are undesirable in the metallic cores of transformers , electric motors , and other such devices in which a coil is to be wound over a metallic core since they heat up the core and dissipate electrical energy in the form of heat . Eddy currents are used to advantage in certain applications like magnetic braking in trains, electromagnetic damping, induction furnace, electric power meters etc. Hence, option 2 is correct.
Question 27 5 / -1
The magnetic field lines:
Solution
CONCEPT:
Magnetic field lines:
A magnetic field line is an imaginary line such that tangent to it at any point gives the direction of the magnetic field at that point in space .Magnetic field lines are drawn to represent the magnetic fields. Magnetic field lines can be drawn with the help of a magnetic compass . Magnetic field lines are also called magnetic lines of force . Properties of magnetic field lines:
The magnetic field lines of a magnet (or a solenoid) form continuous closed loops. The magnetic field lines start from the north pole and end on a south pole outside the magnet. Inside the magnet, the magnetic field lines start from the south pole and end on the north pole. The tangent to the field line at a given point represents the direction of the net magnetic field B at that point. The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field B. The magnetic field lines do not intersect with each other if they did, there will be two directions of the magnetic field at that point which is not possible.
EXPLANATION:
The magnetic field lines of a magnet (or a solenoid) form continuous closed loops . Hence, option 2 is correct.
Question 28 5 / -1
The emitter of a transistor is ______ doped
Solution
Concept:
Transistor:
A transistor is a semiconductor device used to amplify or switch electronic signals and electrical power. It is composed of semiconductor material usually with at least three terminals for connection to an external circuit. Bipolar Junction Transistor constructed with three doped Semiconductor Regions (Base, Collector, and Emitter) separated by two p-n Junctions.Emitter, Collector, and the Base is the correct order of decreasing impurities.
Doping is the intentional introduction of impurities into an intrinsic semiconductor for the purpose of modulating its electrical, optical and structural properties.
Emitter is heavily doped. Its job is to emit or inject electrons into the base. Bases are lightly doped and very thin, it passes most of the emitter-injected electrons on to the collector. The doping level of collector is intermediate between the heavy doping of emitter and the light doping of the base. The doping level and width of base are kept light so that most of the charge carriers are transferred from emitter to collector without much recombination. So that doping concentration of the Emitter(E), Base (B), Collector(C) region of a typical bipolar junction transistor varies as E > C > B.
Important Points
Transistor has three regions:
Emitter:
Collector:
Moderate doping largest area Base:
lightest doping lightest area
Explanation:
With reference to collector, Emitter is heavily doped.
So, the correct option is heavily doped
Question 29 5 / -1
Which waves are produced by using a magnetron valve?
Solution
CONCEPT :
Electromagnetic spectrum: It is a collection of a range of different waves in sequential order from radio to gamma electromagnetic waves.Frequency (ν) = speed of light (c)/wavelength (λ)
Radio waves: The lowest frequency portion comes in radio waves generally, has wavelengths range between 1 mm to 100 km or frequencies between 300 GHz to 3 kHz. There are several subcategories in between these waves like AM and FM radio. It is used for communication such as television and radio. Microwaves : The part of the electromagnetic spectrum having the frequency more than that of radio waves and less than that of infrared waves are called microwaves.It is used for cooking food and for satellite communications. Infrared Waves - It is used by electrical heaters, cookers for cooking food, and by infrared cameras that detect people in the dark.Gamma Rays - It is used in medicine (radiotherapy), industry (sterilization and disinfection), and the nuclear industry.Gamma Rays have the minimum wavelength and maximum frequency. Explanation:
A magnetron valve is used to produce microwaves . Magnetron works as a self-excited microwave oscillator and is a high-powered vacuum tube . The combined action of a magnetic field that is externally applied and the electric field between its electrodes forms the basis of the operation of the magnetron.
Question 30 5 / -1
The magnetic field strength of a current carrying wire at a particular distance from the axis of the wire
Solution
Concept:
Magnetic field strength:
The space or region around the current carrying wire/moving electric charge or around the magnetic material in which force of magnetism can be experienced by other magnetic material is called as magnetic field by that material. The value of this magnetic field is called as magnetic field strength of that magnetic material. Biot-Savart Law :
The law who gives the magnetic field generated by a constant electric current is Biot-savart law. Let us take a current carrying wire of current I and we need to find magnetic field at a distance r from the wire then it is given by: \(dB = \;\frac{{{\mu _0}\;I}}{{4\pi }}\left( {\frac{{\overrightarrow {dl} \times \hat r}}{{{r^2}}}} \right)\) = magnetic field due to current carrying wire element dl at the point
Where μ0 = 4π × 10-7 T.m/A is the permeability of free space /vacuum, dl = small element of wire and r ̂ is the unit position vector of point where we need to find magnetic field.
We can get the net magnetic field at that point by integrating the above equation:
Net magnetic field = \(B = \;\frac{{{\mu _0}I}}{{4\pi }}\smallint \frac{{\overrightarrow {dl} \times \hat r}}{{{r^2}}} = \;\frac{{{\mu _0}I}}{{4\pi }}\smallint \frac{{dl\;Sin\theta }}{{{r^2}}}\)
Where θ is the angle between dl and r vector.
Explanation:
The magnetic field strength of a current-carrying wire at a particular distance (a) at point P is given by; \(B = \;\frac{{{\mu _0}I}}{{4\pi a}}\left( {Sin{\theta _2} + \;Sin{\theta _1}} \right)\) From the above expression of magnetic field strength B we can see that the magnetic field strength of a current-carrying wire at a particular distance from the axis depends on the current of the wire. Hence option 1 is correct. It doesn’t depend on radius of wire and temperature of the surrounding. So options 2 and 3 are not correct.
Question 31 5 / -1
In an LC oscillation circuit the capacitance of the capacitor is 40 μF and the self inductance of the inductor is 100 mH. Find the natural frequency of the circuit.
Solution
CONCEPT:
LC Oscillations:
We know that a capacitor and an inductor can store electrical and magnetic energy, respectively. When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations in mechanical systems .Let a capacitor and an inductor are connected as shown in the figure. Let a capacitor be charged Qo at t = 0 sec .The moment the circuit is completed, the charge on the capacitor starts decreasing, giving rise to a current in the circuit. The angular frequency of the oscillation is given as, \(⇒ ω_o=\frac{1}{\sqrt{LC}}\)
Where L = self-inductance and C = capacitance
The charge on the capacitor varies sinusoidally with time as, ⇒ Q = Qo cos(ωo t)
The current in the circuit at any time t is given as, ⇒ I = Io sin(ωo t)
Where Io = maximum current in the circuit
The relation between the maximum charge and the maximum current is given as, ⇒ Io = ωo Qo
CALCULATION:
Given C = 40 μF = 40×10-6 F and L = 100 mH = 10-1 H
Where L = self-inductance and C = capacitance
We know that the angular frequency of the LC oscillation is given as,
\(⇒ ω_o=\frac{1}{\sqrt{LC}}\)
\(⇒ ω_o=\frac{1}{\sqrt{10^{-1}×40×10^{-6}}}\)
⇒ ωo = 0.5 × 103 rad/sec
⇒ ωo = 0.5 × 103 rad/sec
⇒ ωo = 500 rad/sec
Hence, option 2 is correct.
Question 32 5 / -1
In photo electric effect, the maximum initial velocity of emitted electrons depends upon
Solution
Concept:
Photoelectric effect: When electromagnetic radiation hits a material, electrons are emitted, this effect is known as the Photoelectric effec t.Electrons emitted in this manner are called photo-electrons .
Max Kinetic energy of these photo-electrons is given byK = h ν - ϕ -- (1)
K is max kinetic energy, ν is frequency, ϕ is a work function.
Work Function: The minimum energy required to eject electrons from metal by photoelectric effect is called work function. ϕ = h ν 0 -- (2)
Where ν0 is the minimum frequency or threshold frequency required for the photoelectric effect to happen.
Explanation:
Combining Equation (1) and (2) we get K = h ν - h ν 0
So, It is clear that the kinetic energy depends upon both frequency and threshold frequency. Kinetic energy is related to speed. So, we can say speed depends upon both frequency and threshold frequency. So, the correct option is both 2) and 3) above
Question 33 5 / -1
The direction of wave front of a wave with the direction of propagation of wave is:
Solution
CONCEPT:
Huygens’s principle:
Wavefront: The locus of all particles in a medium , vibrating in the same phase is called waveFront .The direction of propagation of light (ray of light) is perpendicular to the waveFront . Every point on the given wavefront acts as a source of a new disturbance called secondary wavelets which travel in all directions with the velocity of light in the medium. A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wavefront at that instant . This is called a secondary wavefront .
EXPLANATION :
From the above, it is clear that the direction of propagation of light (ray of light) is perpendicular to the waveFront . Therefore option 2 is correct.
Question 34 5 / -1
If the series arrangement of two similar cells of emf E and internal resistance r is connected to a load of resistance R, then find the current in the load resistance.
Solution
CONCEPT:
Cell
The cell converts chemical energy into electrical energy . Cells are of two types:Primary cell: This type of cell cannot be recharged.Secondary cell: This type of cell can be recharged. The electromotive force of a cell
It is the maximum potential difference between the two electrodes of the cell when the cell is in an open circuit. The potential difference of a cell
It is the maximum potential difference between the two electrodes of the cell when the cell is in a closed circuit. If a cell of emf E and internal resistance r is connected across a load resistance R, then the current in the load resistance is given as,
\(⇒ I=\frac{E}{R+r}\)
Cells in series:
A set of batteries are said to be connected in series when the positive terminal of one cell is connected to the negative terminal of the succeeding cell. The equivalent emf of the series arrangement is given as, ⇒ Eeq = E1 + E2 + ... + En
The equivalent internal of the series arrangement is given as,
⇒ req = r1 + r2 + ... + rn
CALCULATION:
Given E1 = E2 = E, r1 = r2 = r, Load resistance R = R
The equivalent emf of the series arrangement is given as,
⇒ Eeq = E1 + E2
⇒ Eeq = E + E
⇒ Eeq = 2E ----(1)
The equivalent internal resistance of the series arrangement is given as,
⇒ req = r1 + r2
⇒ req = r + r
⇒ req = 2r ----(2)
We know that if a cell of emf E and internal resistance r is connected across a load resistance R , then the current in the load resistance is given as,
\(⇒ I=\frac{E}{R+r}\) ----(3)
By equation 1, equation 2, and equation 3 if the series arrangement of two similar cells of emf E and internal resistance r is connected to a load of resistance R, then the current in the load resistance is given as,
\(⇒ I=\frac{E_{eq}}{R+r_{eq}}\)
\(⇒ I=\frac{2E}{R+2r}\)
Hence, option 3 is correct.
Question 35 5 / -1
If R1 = 2Ω, R2 = 4Ω, R3 = 6Ω, determine the electric current that flows in the circuit below.
Solution
CONCEPT:
Kirchhoff’s Voltage Law: The algebraic sum of voltage (or voltage drops) in any closed path of a network that is traversed in a single direction is zero.
Here, the assumed current I causes a + ve drop of voltage when flowing from +ve to – ve potential while – ve drop of voltage when a current flowing from – ve to + ve for the above circuit,
If we apply KVL,
−V + I R1 + I R2 = 0
\({\rm{I\;}} = \frac{{\rm{V}}}{{{{\rm{R}}_1} + {{\rm{R}}_2}}}\)
CALCULATION:
Given R1 = 2Ω, R2 = 4Ω, R3 = 6Ω and E1 = 10 V, E2 = 5 V
We assume the direction of the current (I) is the same as the direction of the clockwise rotation. So applying KVL in the loop. – IR1 + E1 – IR2 – IR3 – E2 = 0
Substituting the values in the equation, we get –2 I + 10 – 4 I – 6 I – 5 = 0 -12I + 5 = 0 I = -5/-12 I = 0.416 A = 0.42 A
The electric current that flows in the circuit is 0.42A. The electric current is signed positive which means that the direction of the electric current is the same as the direction of the clockwise rotation was assumed.
So option 1 is correct.
Question 36 5 / -1
Which of the following statements are true about image formation in a plane mirror?
A. The image is larger in size than the object.
B. The image is formed at the same distance as the object.
C. The image is laterally inverted.
D. The image is virtual
Solution
Concept:
Image formation by Plane mirror
The plane mirror always forms a virtual image of a real object. The virtual images can’t be formed on a screen. The size of the image formed by a plane mirror is equal to the size of the object. The distance between object and mirror is the same as the distance between the mirror and the image. The image is laterally inverted , which means the left-right inversion. For example, the left hand of the object will appear on the right side of the image. Explanation:
So,
Statement A is false, as image size is equal to object size. Statement B is true, as the distance between image and mirror is the same as the distance between object and mirror. Statement C is true, as the image is laterally inverted. Statement D is true, as the image is virtual. B,C and D are tue.
Question 37 5 / -1
When the angle of incidence on the surface of a transparent material is 60°, the reflected light is completely polarized. The velocity of the refracted ray inside the material is:
Solution
Reflection of Light:
When a ray of light reflects back in the same medium this phenomenon is called reflection of light. It is governed by laws of reflection. When reflection happens, the angle of incidence is equal to the angle of reflection. Refraction of Light:
When a ray of light passes from one transparent medium to another, it bends its path. This phenomenon is called refraction. Refractive index is defined as the ratio of the speed of light in air to speed of light in a given medium. \(\frac{c}{v} = \frac{n_2}{n_1}\) -- (1)
c is the speed of light in air, v is the speed of light in the medium
Snell's Law: When light travels from one medium 1 (refractive index n1 ) to medium 2 (refractive index n2 ) to another , then n1 Sin i = n2 Sin r
i is the angle of incidence, r is the angle of refraction.
Polarization:
A light wave oscillates in different phases that are both electric and magnetic fields. When polarized, it oscillates only in a single phase. When unpolarized light incidents on a transparent medium, a fraction of light rays are reflected and a fraction goes under refraction. The fraction of refracted light is more. The fraction of reflected light is less, and it gets polarized up to some extent. The polarization in the refracted ray is negligible. The angle of Polarization: The angle of incidence, when a light ray falls on a transparent medium, such that the reflected ray is completely polarized is called angle of polarization. The angle of Polarization is given by Brewster's law \(θ_p =tan ^{-1}(\frac{n_2}{n_1})\)
or
\(tanθ _{p} = \frac{n_2}{n_1}\) -- (2)
Calculation:
Given Angle of polarization θp = 60 °
Now using Equation (2)
\(tan 60 ^\circ = \frac{n_2}{n_1}\)
\(\implies √3 = \frac{n_2}{n_1}\) -- (3)
Putting (1) in (3)
\(\implies √3 = \frac{c}{v}\)
\(\implies v = \frac{c}{√3}\)
Speed of light c = 3 × 10 8 m/s
\(\implies v = \frac{3 × 10^8}{√3}\)
⇒ v = √ 3 × 10 8 m /s
So, speed of light in the medium is √ 3 × 10 8 m /s
Question 38 5 / -1
The drift velocity of electrons (v) in a current-carrying conductor is related to its electric field (E) as:
Solution
The correct answer is option 2) i.e. vd ∝ E
CONCEPT :
Drift velocity : The average velocity attained by the electrons under the influence of an electric field is termed the drift velocity. The free electrons are always in a constant random motion in the conductor. When the conductor is connected to a battery, an electric field is generated. When these free electrons are subjected to an electric field , they slowly drift in the direction of the electric field applied, while maintaining the random motion. At this point, the net velocity of the electrons is called its drift velocity . The drift velocity (vd ) of the electron in a conductor is given by:
\(v_d =\frac{I}{nAq}\)
Where I is the current flowing through the conductor, A is the area of cross-section of the conductor, q is the charge on the electron and n is the number of electrons.
Mobility is defined as the magnitude of drift velocity per unit electric field. Mobility, \(μ = \frac{v_d}{E}\)
EXPLANATION :
Mobility, \(μ = \frac{v_d}{E}\) ⇒ vd = μE
⇒ vd ∝ E
Question 39 5 / -1
Which characteristic of X-ray spectra does not change when the target material is changed?
Solution
The correct answer is option 4) i.e. cutoff wavelength
CONCEPT :
X-ray spectrum: The X-ray spectrum is defined as the energy distribution of X-ray radiation.X-rays are produced when an electrons incident on the target material inside an X-ray tube emits electromagnetic radiation . There are two types of X-rays produced - characteristic lines and continuous spectrum. Characteristic lines are produced due to the transition of an electron within the atom of the target material.A continuous spectrum is produced due to electrons colliding with the target and decelerating. Therefore, the continuous spectrum has a minimum wavelength .X- rays are emitted when the kinetic energy of the electrons is completely converted into photon energy. KE of electron = Photon energy
eV = \(\frac{hc}{λ}\)
Where λ is the minimum wavelength.
Minimum wavelength indicates the most energetic photon emitted due to a highly energetic electron, that collides with the target and decelerates to stop immediately after a collision.
EXPLANATION :
When the target material has changed the intensity of the X-ray spectra varies. The wavelength of characteristic lines varies. The minimum wavelength or the cutoff wavelength remains unchanged.
Question 40 5 / -1
The magnetic flux threading a coil changes from 12 × 10-3 Wb to 6.0 × 10-3 Wb in 0.01s. Calculate the induced emf.
Solution
Concept :
Magnetic flux ( ϕB ): It is a measure of the number of magnetic lines of force passing through some surface held in the magnetic field
Electromagnetic Induction (Induced emf): Faraday, in 1831, discovered that whenever the number of magnetic lines of force, or magnetic flux, passing through a circuit changes, an e.m.f. is produced in the circuit.
If the circuit is closed, a current flows through it. The e.m.f and the current so produced are called 'induced e.m.f .' and induced current and last only while the magnetic flux is changing. This phenomenon is known as 'electromagnetic induction'. Calculation :
By Faraday's Law , the Induced emf is given by:
\(e = -\frac{\Delta (N ϕ_B)}{\Delta t}\)
Here NϕB is the flux linked with the whole coil. putting the given values, we have \(e = -\frac{ (6.0 \times 10^{-3}Wb) - (12 \times 10^{-3}) Wb }{ 0.01 s}\)
e = 0.6 Wbs-1 = 0.6 V. [Wb = Vs]
Hence option 1 is the answer.
Question 41 5 / -1
If a free electron is placed in the path of a plane electromagnetic wave, it will start moving
Solution
CONCEPT.
Lorentz force: It is defined as the force exerted on a charged particle q moving with velocity v through an electric field E and magnetic field B . The entire electromagnetic f orce on the charged particle is called the Lorentz force and is given by ⇒ FL = qE + qvBsinθ
Electromagnetic wave
Those waves which do not need a material medium for their propagation and can travel even through a vacuum, are called electromagnetic waves. Electromagnetic waves are produced when an electric field comes in contact with the magnetic field.
EXPLANATION:
When a free electron is placed in the path of a plane electromagnetic wave, it will experience a force due to the electric field of the electromagnetic wave and starts moving opposite to the direction of the electric field . Since the electric field and the magnetic field of the electromagnetic waves are mutually perpendicular to each other. So when the electron will move opposite to the direction of the electric field , by default it will move perpendicular to the magnetic field . We know that when a charged particle moves perpendicular to the magnetic field, the magnetic field tries to change the direction of the charged particle and will rotate the electron in the plane parallel to the magnetic field . So the electron will finally move opposite to the direction of the electric field by tracing a helical path. Hence, option 4 is correct.
Question 42 5 / -1
An electric current is passed through a circuit. The circuit has two wires of the same material that are connected in parallel. If the lengths and radii of the wires are in the ratio 4 : 3 and 2 : 3 respectively, then the ratio of the currents passing through the wire will be-
Solution
The correct answer is option 3) i.e. 1 : 3
CONCEPT :
Resistance: The hindrance to the flow of current offered by a material is called electrical resistance.The SI unit of electrical resistance is the ohm (Ω). The resistance of a conducting wire is given by the equation: \(R = ρ\frac{l}{A}\)
Where R is the resistance, l is the length of the wire and A is the cross-sectional area of the material.
CALCULATION :
Let I1 and I2 be the current through the resistors R1 and R2 .
Since the two resistors are of the same material, their resistivity is the same. ⇒ ρ1 = ρ2
The ratio of resistances = \(\frac{R_1}{R_2} = \frac{ρ_1\frac{l_1}{A_1}}{ρ_2\frac{l_2}{A_2}}\)
\(\Rightarrow \frac{R_1}{R_2} = {\frac{l_1}{l_2}}\times {\frac{A_2}{A_1}}\)
\(\Rightarrow \frac{R_1}{R_2} = {\frac{l_1}{l_2}}\times {\frac{\pi r_2 ^2}{\pi r_1 ^2}} ={\frac{l_1}{l_2}}\times {(\frac{r_2 }{ r_1})^2} \)
\(\Rightarrow \frac{R_1}{R_2} = {\frac{4}{3}}\times {(\frac{3}{2})^2} \)
\(\Rightarrow \frac{R_1}{R_2} = \frac{3}{1}\)
For resistances connected in parallel , the voltage drop across their ends will be the same .
From Ohm's law , V = IR, and since V is constant, \(R \propto \frac{1}{I}\)
\(\Rightarrow \frac{R_1}{R_2} \propto \frac{I_2}{I_1} \)
\(\Rightarrow \frac{I_2}{I_1} = \frac{1}{3}\)
Question 43 5 / -1
Which of the following statement is/are incorrect?
Solution
CONCEPT :
Meter bridge : Meter bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance .In the case of a meter bridge , the resistance wire AC is 100 cm long. Varying the position of tapping point B, the bridge is balanced. If in the balanced position of bridge AB = l, BC (100 – l) so that \(\frac{Q}{P} = \frac{{\left( {100 - l} \right)}}{l}\)
Also, \(\frac{P}{Q} = \frac{R}{S} \Rightarrow S = \frac{{\left( {100 - l} \right)}}{l}R\)
EXPLANATION :
From above it clear that the meter bridge works on the principle of Wheatstone bridge. Therefore, option 1 is correct. The Meter Bridge is instruments based on the principle of Wheatstone bridge and is used to measure an unknown resistance . Therefore option 2 is correct. A potentiometer is a device which used to measure the internal resistance of the battery . Therefore option 3 is incorrect. The meter bridge is sensitive when the resistance of all four arms of the bridge are off the same order . Two parts of the meter bridge wire form two arms with small values of resistance . So other resistances to be compared should also be comparable and hence small . Therefore, option 4 is correct.
Question 44 5 / -1
Which of the following produces magnetic field?
Solution
Concept:
Force:
It is a push or pull upon an object resulting from the object's interaction with another object. It is an effort that changes the state of an object at rest or at motion . Whenever any charged particle moves in a magnetic field, it experience can force by a magnetic filed, called Lorentz force. Electrical dipole:
It is defined as a couple of opposite charges q and –q separated by a distance 2a. the direction of electric dipoles in space is always from negative charge -q to positive charge q.
Diamagnetic substance:
Diamagnetic materials are substances that are usually repelled by a magnetic field. These materials can freely magnetize when placed in the magnetic field. Magnetization is in the direction opposite to that of the magnetic field. Time varying field:
There are two types of static fields namely electrostatic field and magnetostatic field . These can not vary with respect to time. These static fields are not dependent on each other. The time-varying electric and magnetic fields are dependent on each other. According to Modified Ampere's Maxwell law , the variations of electric field causes magnetic field. Explanation:
Force is an effort that changes the state of an object at rest or at motion. Electric dipole is the charge separation by some distance. As charge remains in rest, no magnetic field will produce.Diamagnetic material can freely magnetize when placed in the magnetic field but these can not produce their own magnetic field. Time varying electric field can produce magnetic field , as net charge is changing with respect to time.Hence, from the given option, only time varying electric fields can produce magnetic fields.
Question 45 5 / -1
Uniformly charged spherical shell having the radius R. Electric field at a distance r (r < R) from the center?
Solution
CONCEPT :
Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e. \({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\) But we know that electrical flux through a closed surface is \(\oint \vec E \cdot \overrightarrow {ds} \) \(∴\oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\)
Where, E = electric field, q = charge enclosed in the surface and εo = permittivity of free space.
EXPLANATION:
From Gauss’s law
\(\Rightarrow \phi = \frac{q}{{{\epsilon_o}}}\)
\(\Rightarrow \phi = \mathop \smallint \limits_s^\; E \times ds\)
\(\Rightarrow \phi = E \times 4\pi {r^2}\)
As the charge inside a spherical shell is zero , the Gaussian Surface encloses no charge . The Gauss's theorem gives
\(\Rightarrow E \times 4\pi {r^2} = \frac{q}{{{\epsilon_o}}} = 0\) ∴ E = 0 for r < R
Question 46 5 / -1
Total internal reflection takes place when the angle of incidence is..
Solution
CONCEPT :
Total Internal Reflection (TIR):
When a ray of light goes from denser to the rarer medium it bends away from the normal and as the angle of incidence in denser medium increases . The angle of refraction in the rarer medium also increases and at a certain angle, the angle of refraction becomes 900 , this angle of incidence is called critical angle (C) When the Angle of incidence exceeds the critical angle then the light ray comes back i nto the same medium after reflection from the interface. This phenomenon is called Total internal reflection (TIR) Conditions for TIR:
The ray must trave l from denser medium to rarer medium The angle of incidence I must be greater than the critical angle C. EXPLANATION :
From the above discussion, it is clear that Total Internal Reflection will take place when the ray of light travels from denser medium to rarer medium and the angle of incidence must be greater than the critical angle . Hence, option (3) is correct.
Question 47 5 / -1
The image formed by a plane mirror is
Solution
The correct answer is virtual and erect .
Key Points
A plane mirror always forms a virtual image (behind the mirror) . The image and object are the same distance from a flat mirror , the image size is the same as the object size , and the image is upright . The image is erect because the light rays remain parallel before and after reflection meaning they never pass through a focal point . Important Points
Plane Mirror
These are flat mirrors that reflect images in their normal proportions, reversed from left to right. Use looking glass, solar cookers, periscope, kaleidoscope, etc. Convex Mirror
Convex mirrors are also spherical mirrors. They bulge out and distort the reflected image, making it smaller. The focal length of a convex mirror is Positive. The image formed by a convex mirror is always virtual and erect. When an object is placed at infinity, the virtual image is formed at focus and the size of the image is smaller. Use Vehicle mirrors, Magnifying glasses, Street light reflectors, etc. Concave Mirror
Concave mirrors are spherical mirrors that curve inward like a spoon. They create the illusion of largeness. For a diverging or concave mirror, the focal length is always negative. Concave mirrors form both real and virtual images. Concave Mirror used for Shaving mirrors, Ophthalmoscope, Astronomical telescopes, Solar furnaces, etc.
Question 48 5 / -1
What is the linear magnification m, for the image formed at the near point D, by a simple microscope equal to? (f is focal length)
Solution
CONCEPT :
Simple microscope: It is an optical instrument used to see very small objects.Its magnifying power or magnification is given by
\(Magnifying\;power\;\left( m \right) = \frac{{Angle\;subtended\;by\;the\;image\;at\;the\;least\;distance\;of\;distant\;vision\;\left( \beta \right)}}{{Angle\;subtended\;by\;the\;object\;at\;the\;least\;distance\;of\;distant\;vision\;\left( \alpha \right)}}\)
\(m = 1 + \frac{D}{f}\)
Where D = Least distance of distant vision and f = focal length of a convex lens
EXPLANATION :
The magnifying power of a simple microscope is given:
\(m = 1 + \frac{D}{f}\)
So option 1 is correct.
Question 49 5 / -1
An electric dipole is placed in the uniform electric field. The torque experienced by the dipole will be maximum when the angle between the dipole and the electric field is:
Solution
CONCEPT :
Electric dipole
When two equal and opposite charges are placed at a very small distance to each other then this arrangement is called an electric dipole . The electric dipole moment is defined as the product of the magnitude of one charge and the distance between the charges in an electric dipole. ⇒ P = q × 2r
Where 2r = distance between the two charges
Electric dipole in a uniform external field:
We know that when a charge q is placed in electric field E, it experiences a force F, the force is given as, ⇒ F = qE -----(1)
So when an electric dipole is placed in the electric field according to the diagram, The force on the +q and the -q charge due to the electric field is given as,
⇒ F = qE
The net force on the electric dipole will be zero. The torque on the electric dipole is given as, ⇒ τ = pE.sinθ
\(⇒ \vec{τ} = \vec{p}\times\vec{E}\)
Where θ = angle between the dipole and the electric field
This torque will tend to align the dipole with the electric field.
EXPLANATION:
When an electric dipole of dipole moment p is placed in an electric field E at an angle θ, the torque experienced by the electric dipole is given as, ⇒ τ = pE sinθ -----(1)
By equation 1 it is clear that the torque will be maximum when the value of sinθ is maximum. We know that the value of sinθ is maximum at θ = 90° . Hence, option 3 is correct.
Question 50 5 / -1
When a current I flows through the coil of area A of a moving coil Galvanometer a torque equal to ________ acts on it.
Solution
CONCEPT:
Moving coil Galvanometer: It is a device that is used to measure and detect the small amount of electric charge i.e. electric current. It is the most sensitive device and used in many electrical instruments . It is the major basic component of Ammeter and Voltmeter . Principle: Moving coil galvanometer is based on the fact that when a current-carrying rectangular coil is placed in a magnetic field it experiences a torque .
EXPLANATION:
From above it is clear that when a current-carrying rectangular coil is placed in a magnetic field it experiences a torque . ∴ The moment of the deflecting couple (τ) = nBIA
Where
n = number of turns,
B = magnetic field,
I = current
A = area of coil