InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
What contradiction is found by using ampere circuital law to obtain magnetic field during charging of capacitor ? |
Answer» Solution :In figure (a) parallel PLATE capacitor (C ) is shown in its charging state. During charging of capacitor at time t current i(t) flows which is obtained from battery. No current flow inside capacitor. A parallel plate capacitor C, as part of a circuit through which a time dependent current i(t) flows, (a) a loop of radius r, to determine magnetic FIELD at a point P on the loop , (b) a pot - shaped surface passing through the interior between the capacitor plates with the loop shown in (a) as its rim , (c ) a tiffinshaped surface with the circular loop as its rim and a flat circular bottom S between the capacitor plates. the circular bottom S between the capacitor plates. The arrows show uniform electric field between the capacitor plates. At a point P outside capacitor, consider loop of radius r whose centre is on the wire plane is perpendicular to wire. From symmetry it can be shown that at any point on the loop magnetic field is along circumference of the loop and magnitude is equal of at any point of the loop. By using ampere circuital law point P outside capacitor, `oint VEC(B).vec(d)l=mu_(0)i(t)` ....(1) where `Sigma I = i(t)` is total current. If `|vec(B)|=B`and `oint vec(d)l=2pir`, then `B(2pi r)=mu_(0)i(t)`...(2) Now consider another surface shown in figure (b) which do not touch current CARRYING conductor bottom of surface lies in region between plates. Also in figure (c ) consider oval shaped surface which is between plates of capacitor and radius equal to r. In figure (b) and (c ) current do not flow in closed surface considered hence closed surface having radius r by using ampere circuital law. In `int vec(B).vec(d)l=mu_(0)Sigma I` `int dl=2pi r` `Sigma I=mu_(0)i(t)=0` Thus, we get different value of magnetic field outside and inside which is contradictory. Thus, outside and inside which is contradictory. Thus, outside plates of capacitor we get non - zero magnetic field and inside (between) plates of capacitor magnetic field is zero. This contradiction is by using ampere circuital law hence it can be said that there is something missing in ampere circuital law. |
|
| 52. |
The fraction of atoms of radioactive element that decays in 6 days is (7)/(8). The fraction that decays in 10 days will be |
|
Answer» `(77)/(80)` |
|
| 53. |
A circuit consists of a resistance R connected to n similar cells. If the current in the circuit is the same whether the cells are connected in series or in parallel, then the internal resistancer of each cell is given by |
|
Answer» `R = R//n` |
|
| 54. |
A uniform non conducting ring has mass m and radius R. tow point charges q and q' are fixed on its circumference at a separation of sqrt(2)R the ring remains in equilibrium in air with its plane vertical in a region where exists a uniform vertically upward electric field E. Given E=(4mg)/(7q) (a) Find angle theta in equilibrium position (see figure). (b) The ring is given a small rotation in the plane of the figure and released. Will it perform oscillations? |
|
Answer» (B). Yes |
|
| 55. |
The time required for the light to pass through a glass slab (refractive index = 1.5) of thickness 4 mm is (c = 3 xx 10^(8)ms^(-1)"speed of light in free space") |
|
Answer» `10^(-11)` `(n_(g))/(n_(a)) = (c_(a))/(c_(g))` `(3)/(2) = (3xx10^(8))/(c_(g))` `c_(g) = 2 xx 10^(8)` We have, Time = `("DISTANCE")/("Speed")` `t = (4 xx 10^(-3))/(2 xx 10^(8))` or "" `t = 2 xx 10^(-11) s.` |
|
| 56. |
What do you understand by "sharpness of resonance" in a series LCR circuit ? Find expression for Q-factor of the circuit. |
|
Answer» Solution :We KNOW that current amplitude in a series LCR circuit is maximum when `omega = omega_(0) = 1/sqrt(LC)`. The maximum value of current at resonance is `(I_(m))_(rms) = V_(m)/R`For any other value of `omega` , the current amplitude is less. Choose values of o for which current amplitude is `1/sqrt(2) (I_(m))_(max)` . In Fig. 7.34, these values of WARE represented by POINTS B and C having angular frequencies `omega_(1)` and `omega_(2)`such that `omega_(1) = omega_(2)`and `omega_(2) = omega_(0) + Deltaomega`.The quantity `omega_(0)/(2.Deltaomega)`is regarded as a measure of sharpness of resonance and is also known as the Q-factor of the circuit. Smaller the value of Aw, sharper is the resonance. It can be easily calcualted that `Deltaomega = R/(2L)`  `THEREFORE` Q-factor representing the sharpness of resonance `Q = (omega_(0))/(2Deltaomega) = (omega_(0)L)/R` As, `omega_(0) =1/sqrt(LC)`, hence, `Q = 1/sqrt(LC).1/R = 1/Rsqrt(L/C)` |
|
| 57. |
A transformer cannot be used to step up dc voltage. Why ? |
| Answer» SOLUTION :A dc voltage is applied across the primary ,then the MAGNETIC flux LINKED with the coil does not VARY with time end hence no emf is induced across the secondary. | |
| 58. |
A stretched string is vibrating in the second overtone then the number of nodes and antinondes between the ends of the string are respectively : |
|
Answer» 3 and 2 
|
|
| 59. |
The empty space between the shaded baals and holow balls and hollow balls as shown in the diagram is called |
|
Answer» hexagonal VOID |
|
| 60. |
What are the factors on which the rate of thermionic emission depends? |
| Answer» SOLUTION :The rate of thermionicemission DEPENDS on the temperature and work function of the metal. HIGHER the temperature, greater will be thermionic EMISSION. For metals with lower work function, thermionicemission OCCURS at low temperature. | |
| 61. |
Positive charge inside 2^(He) atom is………….. |
|
Answer» `1.6 xx 10^(-19)` C |
|
| 62. |
(a) State Gauss's law for magnetism. Explain its significance. (b) Write the four important properties of the magnetic field lines due to a bar magnet. |
|
Answer» Solution :(a) According to Gauss. law for magnetism , "the NET magnetic flux through may closed surface is zero" i.e., `phi_B = oint vecB .""vecd s = 0` It signifies that in magnetism isolated monopoles do not exist. There are no sources/sinks of magnetic field `vecB`. The SIMPLEST magnetic element is a dipole or a current loop. (b) Four important properties of the magnetic field lines DUE to a bar magnet are given below: (1) Magnetic field lines are a visual and intuitive realisation of the magnetic field. A magnetic field line is a smooth curve in a magnetic field, tangent to which at any point gives the direction of magnetic field at that point. (2) In free SPACE around a magnetic dipole the magnetic field lines start from N-pole and end at S-pole. However, inside the magnet they travel from S-pole to N-pole. Thus, magnetic field lines of a magnet or a solenoid form continuous closed curves. (3) The magnetic field lines do not intersect one another. It is so since the direction of the magnetic field would not be unique at the point of intersection. (4) The larger the number of magnetic field lines crossing per unit normal area in a given region, the stronger is the magnetic field `vecB` there. |
|
| 63. |
Ultraviolet light of wavelength 280 nm is used in an experiment on photoelectric effect with lithium ( varphi = 2.5 eV) cathode. Find (a) the maximum kinetic energy of the photoelectrons and (b) the stopping potential. |
|
Answer» Solution : (a) The maximum kinetic energy is ` K_max = (hc/lambda)- varphi. ` `= (1242 eV nm/ 280 nm)- 2.5 eV. ` ` = 4.4 eV - 2.5 eV = 1.9eV. ` (B) STOPPING potential V is given by ` eV= K_max` ` or, V=(K_max)/E = 1.9 eV/e = 1.9 V.` |
|
| 64. |
An alpha- particle and a photon are accelerated from rest through the same potential difference V. Find the ratio of de Broglie wavelength associated with them. |
|
Answer» Solution :`K.E. = ( 1)/( 2 ) m upsilon^(2) = QV` or `upsilon= sqrt((2qV)/(m))` de Broglie WAVELENGTH,` LAMBDA = ( h )/( mv)` For same POTENTIAL difference, `( lambda_(ALPHA))/( lambda_(p))= sqrt((m_(p)q_(p))/( m_(alpha)q_(alpha)))=sqrt((m_(p)e)/( 4m_(p) 2e)) = (1)/( 2sqrt( 2))` |
|
| 65. |
Which one of the series of hydrogen spectrum is in the visible region ? |
|
Answer» Lyman series |
|
| 66. |
Potential energy of a bar mangnet of magnetic moment M placed in a magnetic field of induction B such that it makes an angle theta with the direction of B is (take theta =90^(@) as datum) |
|
Answer» `-M B sin theta ` |
|
| 67. |
In the given circuit |
|
Answer» A is at HIGH POTENTIAL |
|
| 68. |
A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits areseparated by a distance d and the screen is placed parallel to the plane of the slits. If the incident beam makes an angle theta = sin^(-1)((lambda)/(2d)) with the normal to the plane of the slits, there will be .....at the centre P_(0) of the pattern. |
|
Answer» 1/4the MAXIMUM intensity |
|
| 69. |
If the mass of the proton were just a little closer to the mass of the neutron, the following reaction would be possible even at low collision energies : e^(-) + p to n+v Why would this situation be catastrophic ? By what percentage would the proton mass have to increased to make this reaction possible ? |
|
Answer» |
|
| 70. |
In a certain clock, a pendulum of length L_(1), has a period T_(1) = 0.95s. The length of the pendulum is adjusted is adusted to a new L_(2) such that T_(2) = 1.0s. What is the ratio L_(2)//L_(1)? |
|
Answer» `0.90` |
|
| 71. |
The capacity of a parallel plate capacitor formed by the plates of same area A is 0.02muFwith air as dielectric. Now one plate is replaced by a plate of area 2A and dielectric (k=2) is introduced between the plates, the capacity is |
|
Answer» `0.04muF` |
|
| 72. |
Compare the emissivities of a caesium-coated tungsten cathode at 1000 K and that of a pure tungsten cathode at 2700 K. Assume the constant B in the Richardson-Dushmann equation to be the same for both cathodes. |
|
Answer» `x=(i_(1))/(i_(2))=((T_(1))/(T_(2)))^(2)EXP((A_(2))/(kT_(2))-(A_(1))/(kT_(1)))` |
|
| 73. |
For a normal eye the least distance of distinct vision is …….. Fill in the blank. |
|
Answer» |
|
| 74. |
…………………. signals are continuous variations of voltage or current |
| Answer» SOLUTION :ANALOG SIGNALS. | |
| 75. |
In the following potentiometer circuit AB is a uniform wire of length 1 in and resistance R the potential gradient along the wire and balance length AO (= l). |
|
Answer» SOLUTION :Current flowing in the potentiometer wire `I=E/(R_"total ")=20/(15+10)=2/25A` `:.` Potential DIFFERENCE across the wire `=2/25xx10=20/25=0.8A` `:.` Potential gradient `k=(V_(AB))/(I_(AB))=0.8/1.0=0.8V//m` Now, current flowing in the circuit containing experimental cell, `=15/(1.2+0.3)=1A` Potential difference across LENGTH `AO – 0.3 xx 1 = 0.3 V` Length`AO=(0.3)/(0.8)m=0.3/0xx100cm = 37.5cm ` |
|
| 76. |
Which one among the following happens when a swing nses to a certain height from its rest position ? |
|
Answer» Its potential energy DECREASES while KINETIC energy INCREASES |
|
| 77. |
The shortest wavelength of X-rays, emiited from a X-ray tube, depend upon |
|
Answer» CURRENT in the tube |
|
| 78. |
A stationary electron placed in a magnetic field experiences a mechanical push equal to |
|
Answer» cvB |
|
| 79. |
Refrigerator is an apparatus which takes heat from a cold body, work is done on it and the work done together with the heat absorbed is rejected to the source. An ideal refrigerator can be regarded as Carnot's ideal heat engine working in the reverse direction. The coefficient of performance of refrigerator is defined as beta = ("Heat extracted from cold reservoir")/("work done on working substance") = (Q_(2))/(W) = (Q_(2))/(Q_(1)- Q_(2)) = (T_(2))/(T_(1) - T_(2)) A Carnot's refrigerator takes heat from water at 0^(@)C and discards it to a room temperature at 27^(@)C . 1Kg of water at 0^(@)C is to be changed into ice at 0^(@)C. (L_(ice) = 80"kcal//kg") What is the coefficient of performance of the machine? |
|
Answer» `11.1` |
|
| 80. |
If half the space between two concentric conducting spheres be filled with dielectric of dielectric constant k and the rest is filled with air. Show that the capacitance of the capacitor thus formed will be same as if the whole part is filled with the dielectricof dielectric constant 1/2(1+k): |
|
Answer» |
|
| 81. |
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1)/(2)QE, where Q is the charge on the capacitor and E is the magnitude of electric field between the plates. Explain the origin of the factor (1)/(2). |
|
Answer» Solution :Let the application of force F increases the distance between the plates by `trianglex`. Work done by the external force `=F*triangle x` This increases the energy. Energy DENSITY = u So, INCREASE in energy `= u(alpha triangle x) ""[alpha = " area of the plate "]` `:. F* trianglex = u alpha "" triangle x " or, " F = u alpha` `:. F = (1)/(2)in_0 E^2* alpha ""[:. u = (1)/(2) in_0E^2]` `:. F = (1)/(2) in_0 alpha E*E = (1)/(2) QE` The factor `(1)/(2)` arises in this relation due to the fact that average `(E )/(2) " of "0` and E is taken. |
|
| 82. |
A flint glass prism and a crown glass prism are to be combined in such a way that the deviation of the mean ray is zero. The refractive index of flint and crown glasses for the mean ray are 1620 and 1.518 respectively. If the refracting angle of the flint prism is 6.0^@, what would be the refracting angle of the crown prism ? |
|
Answer» Refractive index of flint glass `=mu_f=1.620` Refractive index of crown glass `=m_e=1.518` Refractive ANGLE of flint proms =`A_f=6^@` for ZERO net deviation of mean by `(mu_1-1)A_f=(mu_e-1)A_c` `rarr A_c=(mu_f-1)/(mu_c-1)` `A_f=(1.620-1)/(1.518-1)(6.0)^@=7.2^0` |
|
| 83. |
If the degrees of freedom of a gas are f, then the ratio of its specific heats (C_(p))/(C_(v)) is given by : |
|
Answer» <P>`1-(2)/(f)` `C_(P) =C_(v)+R =(1)/(2) f R+R =((f)/(2)+1)R` `therefore gamma =(C_(P))/(C_(v))=((f)/(2)+1)(R)/((f)/(2) cdot R)=1+(2)/(f)` Thus, CORRECT choice is (b). |
|
| 84. |
A cubical block rests on a plane rough surface with coefficient of static friction (1)/(sqrt(3)). AT what angle the plane be inclined so that the block just slides? |
|
Answer» `60^(@)` |
|
| 85. |
What is the forbidden energy gap (in joule) for a Germanium crystal ? |
|
Answer» `1.6xx10^-19J` |
|
| 86. |
A charged particle is moving with velocity vecv in a uniform magnetic field vecB. The magnetic force acting on it will be maximum when _____ . |
|
Answer» `vecvandvecB` are in same direction. `vecF=q(vecvxxvecB)` `thereforeF=qvBsintheta` Where angle between `vecvandvecB` is `THETA`. When `vecv_|_vecB`, then `theta=90^(@)` `thereforeF=qvBsin90^(@)` `thereforeF=qvB` is maximum force. |
|
| 87. |
The displacement of a particle is given by x = (t - 2)^(2) where x is in meters and t in seconds. The distance covered by the particle in first 4 seconds is |
| Answer» ANSWER :B | |
| 88. |
Assertion : Phase difference between two points separated by a distance equal to lambda//2 on same wavefront is pi. Reason : Path difference (Delta x) and phase difference (delta) are related as delta = (2 pi)/(lambda) Delta x. |
|
Answer» If both assertion and REASON are CORRECT and reason is a correct EXPLANATION of the assertion. |
|
| 89. |
Two wires A and B are of lengths 40 cm and 30 cm. A is bent into a circle of radius r and B into an arc of radius r. A current i, is passed through A and i, through B. To have the same magnetic induction at the centre, the ratio of i_1:i_2 is |
| Answer» ANSWER :A | |
| 90. |
100 gm of ice at 0^@c is mixed with 100gm of water at 100^@C . What will be the final temperature of the mixture |
|
Answer» a)`10^@C` |
|
| 91. |
Which of the following is the unit of magnetic induction ? |
|
Answer» T |
|
| 92. |
An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3KHz. Could this wave be demodulated by a diode detector which has the values of R and C as |
|
Answer» `R=1kOmega,C=0.01muE` `(1)/(v_(c))=(1)/(20)xx10^(-6)=0.05xx10^(-6)=0.5xx10^(-7)s` Band width =`20_(m)=3kHz=3xx10^(3)Hz` `v_(m)=1.5xx10^(3)Hz` `(1)/(v_(m))=(1)/(1.5)xx10^(-3)=0.7xx10^(-3)s`. Diode detector can demodulate when `(1)/(v_(c))ltltRCltlt(1)/(v_(m))` (i)RC=`1kOmegaxx0.01muF=10^(3)XX(0.01xx10^(-6))=10^(-5)s`. it can be demodulated. (ii) RC=`10kOmegaxx1xx10^(-8)=10^(-4)s`It can be demodulated (iii) RC`=10kOmegaxx1pF=10^(4)xx10^(-12)=10^(-8)s` It cannot be demodulated. |
|
| 93. |
State the working of a.c. generator with the help of a labelled diagram.The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity omega. Deduce the expression for the alternating e.m.f. generated in the coil. What is the source of energy generation in this device? |
|
Answer» Solution :A.C. Generator: See Q.30 (or) 2008, O.D. Set-1 The ESSENTIAL parts of an a.c. generator are shown in the figure. Initially the armaturecoil ABCD is horizontal. As the coil is rotasted clockwise , the arm AB moves up and CD moves down. By Flemming's right hand rule, the induced current flows along ABCD, In second half rotation, the arm CD moves up and AB moves down. The induced current flows in the opposite direction i.e. along DCBA. Thus an alternating current flows in the CIRCUIT The magnetic flux linked with the coil at any instant is `""phi="NA A "cos omegat` `"Induced emf will be" ""E=-(dphi)/(dt)=-(d)/(dt)("NBA cos "omegat)=NBA omega sin omegat` `or ""E=E_(0) sin omegat` Where, `E_(0)=NBA omega`= peak value of induced emf. (b) To calculate the magnitude of e.m.f induced Suppose N=number of turns in the coil. A=area of enclosed by each turn of coil. `overset(rightarrow)B`=strength of magnetic field. `theta=` ANGLE which NORMAL to the coil makes with `overset(rightarrow)B` at any instant t,  `therefore` Magnetic flux linked with the coil in this position. `""phi=N(overset(rightarrow)B.overset(rightarrow)A)` `""=NBA cos theta` `""=NBA cos omega"".....(i)` Where `omega` be the angular velocity of the coil. At this instant t, if e in the e.m.f induced in the coil, then `e=(-dphi)/(dt)=(-d)/(dt)(NAB cos omegat)=-NAB (d)/(dt)(cos omegat)` `""=-NAB(-sin omegat)omega` `""e=NAB" "omega sin omegat......(ii)` The induced, e.m.f. will be maximum, when `""sin omegat="maximum "=1` `therefore""e_("max")=e_(v)=NAB_(omega)xx1.......(iii)` Put in (ii)`""e=e_(0) sin omegat` (c ) Mechanical energy is converted into electrical energy. ' |
|
| 94. |
The range of ammeter can be………………………but can not be……………….. . |
|
Answer» |
|
| 95. |
Light of wavelength 5900 xx 10^(-10)m falls normally on a slit of width 11.8 xx 10^(-7)m. The resulting diffraction pattern is received on a screen. The angular position of the first minimum is. |
|
Answer» `30^(@)` |
|
| 96. |
The thermal radiations are similar to : |
|
Answer» X-rays THUS correct CHOICE is (a). |
|
| 97. |
Number of turns in the primary coil and the secondary coil of an ideal transformer are 500 and 2500 respectively. If current in the secondary coil is 0.2 A, calculate current in the primary coil. Also calculate transformer ratio. If voltage in the secondary coil is 750 V, calculate voltage in the primary coil. |
|
Answer» 150 V `I_1`=? , `I_2`= 0.2 A `V_1`= ? , `V_2`= 750 V Transformation ratio `r=N_2/N_1` `=2500/500` `THEREFORE` r=5 For current in PRIMARY coil, `N_1/N_2=I_2/I_1` `therefore I_1=I_2 xx N_2/N_1` = 0.2 x 5 =1.0 A For voltage in primary coil, `N_2/N_1=V_2/V_1` `therefore V_1=V_2 xx N_1/N_2` `= 750xx500/2500` `therefore V_1`= 150 V |
|
| 98. |
Find the greatest possible angle through which a deuteron is scattered as result of elastic collision with an initially stationary proton ? Take m_(1) & m_(2) as masses of a proton & a deuterium. |
|
Answer» |
|
| 99. |
The figure shows the graphical variation of the rectance of a capacitor with frequency of acsource. (a) Find the capacitance of the capacitor. (b) An ideal inductor has the same reactance at 100 Hz frequency as the capacitor has at the same frequency. Find the value of inductance of the inductor. (c ) Draw the graph showing the variation of the rectance of this inductor with frequency. |
|
Answer» SOLUTION :(a) As per graph for a frequency v = 100 Hz the capacitive reactance `X_(C )=6Omega`. `because X_(C )=(1)/(C omega)=(1)/(C.2pi v)` `RARR C=(1)/(2pi v. X_(C ))=(1)/(2xx3.14xx100xx6)=2.65xx10^(-4)F or 265 muF` (b) As for v = 100 Hz, `X_(L)=X_(C )=6Omega` `rArr L.2pi xx 100=6 rArr L=(6)/(2pi xx 100)=9.55xx10^(-3)H or 9.55 MH` (c )`X_(L)-v` graph is shown here: 
|
|
| 100. |
Transistor input characteristics curves are the graphs drawn with |
|
Answer» Collector current `I_C` on y-axis and the collector EMITTER voltage `V_(CE)` on X-axis for a CONSTANT base current |
|