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.
| 4101. |
The following tables has 3 columns and 4 rows. Based on each table, there are THREE questions. Each question has FOUR options (A), (B), (C), and (D). ONLY ONE of these four options is correct Q |
| Answer» Answer :A | |
| 4102. |
A particle of charge -q and mass m moves in a circle aroung a long wire of linear charge density lambda. If r = radius of the circular path and T = time period of the motion circular path. Then : |
|
Answer» `T = 2pir(m//2klambdaq)^(1//2)` |
|
| 4103. |
Distance in free space at which intensity of 5 eV neutron beam reduces to half will be nearly : (Take half - life of the neutron = 12.8 min) |
| Answer» Answer :D | |
| 4104. |
The following tables has 3 columns and 4 rows. Based on each table, there are THREE questions. Each question has FOUR options (A), (B), (C), and (D). ONLY ONE of these four options is correct QPick the correct option |
|
Answer» (IV) (HI) (P) |
|
| 4105. |
A physical energy of the dimension of length that can be formula cut of c,G and (e^(2))/(4 pi epsilon_(0)) is [c is velocity of light G is universal constant of gravilation e is change |
|
Answer» `C^2[Ge^2/(4πε_0)]^(1/2)` |
|
| 4106. |
An electron moving in a circular orbit of radius r makes n rotations per second per second. The magnetic moment of the orbital electron is |
| Answer» Answer :D | |
| 4107. |
Using Gauss's theorem derive an expression for the electric field at any point outside a charged spherical shell of radius R and of charge densitysigma C//m^(2) |
|
Answer» SOLUTION :Consider a uniformaly charged THIN spherical shell of radius R and having a CHARGE Q. To FIND electric field at a point outside the shell situated at a distance (` r gt R ) `from the centre of shell, consider a sphere of radius r as the Gaussian surface. All points on this surface are equivalent relative to given charged sheel and,thus electric ` oversetto E ` at all points of Gaussian surface has same magnitude `oversetto Eand hat n ` are parallel to each other. ` therefore ` Total electric flux over the Gaussian surface ` phi _in = int oversetto Ehat n ds=E . 4 pi r^(2) "" ...(i) ` According to Gauss.s theorem ` phi _in =(1)/( in _0)` ( charged enclosed) ` = (Q)/( in _0) ""...(II) ` Comparing (i) and ( ii),we get ` E. 4 pi r^(2)=(Q)/( in_0)or E =(Q)/( 4 pi in _0 r^(2)) ` Thus , for anypoint outside the shell, the effect is, as if whole charge Q is concentrated at the centre of teh spherical shell. If surface charge density of shell be ` sigma ` thenQ = ` 4 pi R ^(2) sigma ` and therefore ` E= ( 4 pi R ^(2) sigma)/( 4 pi in_0 r^(2)) =( sigma R^(2))/( in _0 r^(2)) ` 
|
|
| 4108. |
Three equal charges are placed at the three corners ABC of a square ABCD. If the force between the charges at A&B ("on" q_(1) and q_(2)) is F_(12) and that between q_(1) and q_(3) is F_(13) the ratio of magnitudes F_(12)//F_(13) is _____ |
|
Answer» |
|
| 4109. |
In davisson-Germer experiment ,intensity of electron beam scattered by….. |
|
Answer» photometer |
|
| 4110. |
Showthat the normal componentof electrostaticfield has a discontinuly form one sideof a charged. Surface to another given by (vec(E_(2)) - vec(E_(1))). hat(n) = (sigma)/(in_(0))where hat(n) is a unitvectornormal to the surfaceat a pointand sigma at a pointand sigma is the surface charge density at thatpoint. (The direction of hat(n) is from side 1 to side 2). Hence show that justy outsidea conductor, theelectricfield sigma hat(n)//in_(0). (b) Show that the tangential componetof electrostaticfieldis contionous from one side fo a charged surface to another. |
|
Answer» Solution :PROCEEDING as in Art, NORMAL of electricfield intensity due to thin infinitie plane sheetof charge, on left side (side 1) `vec(E)_(1) = - (sigma)/(2in_(0)) hat(n)` and on right side (side 2), `vec(E_(2)) = (sigma)/(2in_(0)) hat(n)` Discotinuityis the normalcomponentfrom ONE sideto the other is `vec(E_(2)) - vec(E_(1)) = (sigma)/(2 in_(0)) hat(n) + (sigma)/(2 in_(0)) hat(n) = (sigma)/(in_(0)) hat(n) or (vec(E_(2)) - vec(E_(1))) hat(n) = (sigma)/(in_(0)) hat(n). hat(n) = (sigma)/(in_(0))` Inside a closedconductor, `vec(E)_(1) = 0``:. E = vec(E_(2)) = (sigma)/(in_(0)) hat(n)` (b) To show that the tangentialcomponentof electrostaticfield is continousfrom one sideof a chargedsurfaceto another, we usethe FACTTHAT work done byelectrostaticfield on a closedloop is zero. |
|
| 4111. |
A wavefront is an imaginary surface |
|
Answer» phase is same for all POINTS |
|
| 4112. |
A full-wave rectifier is used to convert 50 Hz A.C into D.C, then the number of pulses per second present in the rectified voltage is |
|
Answer» 50 |
|
| 4113. |
Who wrote an influential pamphlet 'What is the third Estate'? |
|
Answer» Mirabeau |
|
| 4114. |
Time taken by the sunlight to pass through a window of thickness 4 mm, whose R.I. is 1.5 is: |
|
Answer» `2XX10^10 sec` |
|
| 4115. |
What is forbidden energy gap ? |
| Answer» SOLUTION :The separation between highest energy level of VALENCE band and LOWEST energy level of conduction band is CALLED as FORBIDDEN energy gap. | |
| 4116. |
The potential difference of a cell in an open circuit is 6 V, which falls to 4 V when a current of 2 A is drawn from the cell. Calculate the emf and the internal resistance of the cell. |
|
Answer» |
|
| 4117. |
A potentiometer wire 10m long and having 20Omega resistance is connected in series with a battery and a resistance R = 2160Omega. A daniel cell of emf 1.08 v is balanced across the P.D. of the resistance R. The emf of thermo couple is balanced across 300 cm of the wire. The emf of thermo couple is |
| Answer» Answer :C | |
| 4118. |
For obtaining diffraction pattern , aperture of the slitshould be of the order of |
|
Answer» `LAMBDA ` |
|
| 4119. |
In a semiconducting material, the mobilities of electrons and holes are mu_(e) and mu_(h) respectively. Then |
|
Answer» `mu_(e) lt mu_(h)` |
|
| 4120. |
Which of following are not electromagnetic waves ? |
|
Answer» X - rays |
|
| 4121. |
When a source of frequency f_(0) moves away from a stationary observer with a certain velocity, an apparent frequency f' is observed. When it moves with the same speed towards the observer, the observed frequecy is 1.2 f'. If the velocityof sound is v, then the acutual frequency f_(0) is |
|
Answer» `12/11 F'` |
|
| 4122. |
What are thesallent features of corpuscular theory of light? |
|
Answer» Solution :(i) Light is emitted as TINY, massless (negligibly small mass) and perfectly ELASTIC particles called corpuscles. (II) As the corpuscles are very small, the source of light does not suffer appreciable loss of mass even if it emits light for a long time. (iii) On account of high speed, they are unaffected by the force of gravity and their path is a straight line in a medium of uniform refractive index. (IV) The energy of light is the kinetic energy of these corpuscles. (v) When these corpuscles impinge on the RETINA of the eye, the vision is produced. The different size of the corpuscles is the reason for different colours of light. (vi) When the corpuscles approach a surface between two media, they are eitherattracted or repelled. (vii) The reflection of light is due to the repulsion of the corpuscles by the medium and refraction of light is due to the attraction of the corpuscles by the medium. |
|
| 4123. |
The ratio of velocities of a proton and an alpha particle is 4 : 1. The ratio of their De Broglie wave lengths will be |
|
Answer» `4:1` |
|
| 4124. |
When used as a switch, a transistor is operated in |
|
Answer» the cut-off REGION and ACTIVE region |
|
| 4125. |
A charged particle is released from rest in a region of steady and uniform electric and magnetic fields, which are parallel to each other . What will be the nature of the path followed by the charged particle ? Explain your answer: |
| Answer» Solution :It will FOLLOW a straight line path, in the direction or OPPOSITE to the direction of the ELECTRIC field . Magnetic Lorentz FORCE will be not act on this particle because its motion is in the same direction or opposite direction with the magnetic field (since electric and magnetic fields are PARALLEL) | |
| 4126. |
What does a polaroid consist of ? Using polaroid show that light waves are transverse in nature ? |
|
Answer» Solution :A POLAROID consists of long chain molecules ALIGNED in a particular direction. If an unpolarised (ordinary) light wave is incident on a polaroid then the electric vectors ASSOCIATED with the propagating wave along the direction of aligned molecules get absorbed. thus, the emergent light will have electric vectors oscillating along a direction perpendicular to the direction of aligned molecules. it clearly SHOWS that light waves are TRANSVERSE in nature. |
|
| 4127. |
When the poet chose one of the paths, what did he hope to do? |
|
Answer» COME BACK and TRY the other path |
|
| 4128. |
क्या त्वरण ऋणात्मक हो सकता है |
|
Answer» हा |
|
| 4129. |
Given below a circuit diagram of an AM demodulator. For good demodulation of AM signal of carrier frequency f, the value of RC should be: |
|
Answer» `RC=(1)/(f)`  For good demodulation of `AM` signal the value of `RC` (which is a time-constant) is chosen such that `(1)/(F) lt lt R C` or `R C gt gt (1)/(f)` where `f` is the frequency of the carrier signal. |
|
| 4130. |
A circular current loop of magnetic moment M is in an arbitary orientation is an external magnetic field vec B. The work done to rotate the loop by 30^@ about an axis perpendicular to its plane is…. |
|
Answer» MB `=mB(COS0^@-COS30^@)` `=mB(1-sqrt3/2)` `=((2-sqrt3)/2)mB` |
|
| 4131. |
Energy of the lowest level of H-atom is-13.6 eV. The energy of the emitted photons in transition from fourth to second energy state is of: |
|
Answer» 2.55eV |
|
| 4132. |
A charged particle with some initial velocity is projected in a region, where non-zero non varying uniform and/or magnetic fields are present. In list-I, information about existence of electric and/or magnetic field and direction of initial velocity of charged particle are given in List-II, the probable path of charged particle is mentioned. Match the entries of List-I with entries of List-II (consider gravity free space) {:("List-I","List-II"),(P. vec(E ) = 0. vec(B) = 0"and initial velocity is at an unknown angle with" vec(B),1. "Parabola"),(Q. E != 0. vec(B) = 0 "and initial velocity is at known angle with" vec(E ),2."circular"),(R. vec(E ) != 0. vec(B) != vec(E ) || vec(B ) "and initial velocity is perpendicular to" vec(E ),3. "Helical path with non uniform pitch"),(S. vec(E ) != 0. vec(B) = 0. vec(E ) "perpendicular to" vec(B) "and initial velocity is perpendicular to both" vec(E ) "and" vec(B), 4. "cycloid"):} |
|
Answer» 2,1,3,4 |
|
| 4133. |
सिन्धु नदी का उद्गम होता है - |
|
Answer» हिन्दुकुश पर्वतमाला से |
|
| 4134. |
A transistor is used in a common emitter mode as an amplifier. Then |
|
Answer» the base EMITTER JUNCTION is reverse biased |
|
| 4135. |
Read the paragraph and the figure carefully and answer the questions. The figure shows a constant deviation prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass, it is equivalent to two 30^(@)-60^(@)-90^(@) prisms and one 45^(@)-45-90^(@) prism. The angle theta_(1) is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. For this prism, mu=2 sin theta_(1) The ratio of (theta_(1))/(theta_(2)) is |
|
Answer» `1` |
|
| 4136. |
Assertion (A) : Lines of Lyman series of hydrogen spectrum lie in ultraviolet region but lines of Balmer series lie in visible light region.Reason (R) : Subsequent to the emission of a line of Balmer series we must obtain the first line of Lyman series of hydrogen atom. |
|
Answer» If both assertion and reason are true and the reason is the correct EXPLANATION of the assertion. |
|
| 4137. |
Read the paragraph and the figure carefully and answer the questions. The figure shows a constant deviation prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass, it is equivalent to two 30^(@)-60^(@)-90^(@) prisms and one 45^(@)-45-90^(@) prism. The angle theta_(1) is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. For this prism, mu=2 sin theta_(1) The total deviation of the incident ray when it emerges out of the prism is |
|
Answer» `90^(@)` |
|
| 4138. |
Four spheres of radius r each of mass m placed with their centres on the four corners of the square of side 'a'. The M.I. of the system about an axis along one of the sides of square is : |
|
Answer» `(4)/(5)mr^(2)+2ma^(2)`  `I_(A)=(2)/(5)mr^(2),I_(B)=(2)/(5)mr^(2)` `I_(C)=(2)/(5)mr^(2)+Ma^(2)` `I_(D)=(2)/(5)mr^(2)+Ma^(2)` `therefore I=I_(A)+I_(B)+I_(C)+I_(D)` `=(2)/(5)mr^(2)+(2)/(5)mr^(2)+(2)/(5)mr^(2)+ma^(2)+(2)/(5)mr^(2)+ma^(2)` `=(8)/(5)mr^(2)+2ma^(2)` |
|
| 4139. |
A particle travrersed along a straight line for firsthalf time with velocity V_(0).For the remaining part,hald of the distance is travered with velocity V_(1)and other half distance with velocity V_(2).Find the mean velocity of the particle for the total journey . |
|
Answer» `(2V_(0)(v_(1)+v_(2)))/(v_(1)+v_(2)+2v_(0))` |
|
| 4140. |
Read the paragraph and the figure carefully and answer the questions. The figure shows a constant deviation prism ABCD. The incident ray is PQ and the emergent ray is ST. Although it is made up of one piece of glass, it is equivalent to two 30^(@)-60^(@)-90^(@) prisms and one 45^(@)-45-90^(@) prism. The angle theta_(1) is the angle of incidence on face AB. The path of the ray inside the prism is indicated in the figure. For this prism, mu=2 sin theta_(1) The angle of refraction on the face AB is |
|
Answer» `30^(@)` |
|
| 4141. |
A train travels at 108 kmph towards east. Earth's magnetic field is B=0.4xx10^(-4) Tesla and acts downwards at 60^(@) to the horizontal. Calculate the induced e.m.f. between the ends of a horizontal axis PQ of the train. Given PQ = 2m. Also find which end of PQ is at a higher potential? |
|
Answer» Solution :Induced e.m.f. along PQ is due to the vertical component of B `B_(v)=0.4xx10^(-4)sin60=sqrt(3)xx0.2xx10^(-4)T` `PQ=l=2m` `v=108xx(5)/(18)m//s=30m//s` Induced e.m.f., `varepsilon=B_(v)lv=sqrt(3)xx0.2xx10^(-4)xx2xx30` `=12sqrt(3)xx10^(-4)V=20.784xx10^(-4)V=2.078mV` Using Fleming.s Right HAND Rule, we can see that `varepsilon` ACTS from Q to P. HENCE .P. is at a higher potential. 
|
|
| 4143. |
If mu_e and mu_b are electron and hold mobility, E be the applied electric field, the current density for intrinsic conductor is equal to : |
|
Answer» `n_ie(m_e+m_n)E` |
|
| 4144. |
A solenoid of length 3.14 cm has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid ? |
|
Answer» Solution :For a current carrying long SOLENOID, magnitude of magnetic field produced inside the solenoid PARALLEL to its axis is, `B=mu_(0)NI` `=mu_(0)(N/l)I` `=(4pixx10^(-7))(500/0.5)(5)` `=20xx3.14xx10^(-4)` `thereforeB=6.28xx10^(-3)T` |
|
| 4145. |
Two point charges + 3 muC and8 muC repel each other with a force of 40 N. If a charge of -5muC is added to each of them , then the force between them will become : |
|
Answer» `-10N` |
|
| 4146. |
Identify the work statement |
|
Answer» The electrical POTENTIAL energy of a SYSTEM of TWO PROTONS shall increase if the separation between the two is decreased |
|
| 4147. |
A concave mirror gives an image three times as large as its object placed at a distance of 20 cm from it. For the image to be real, the focal length should be : |
|
Answer» 10 cm u = - 20 cm, f = ? `therefore v = - 3u` By mirror formula, `(1)/(v) + (1)/(u) = (1)/(f)` `therefore - (1)/(3u) - (1)/(u) = (-1)/(f) RARR f = 15 cm` |
|
| 4148. |
Show that wave theory cannot explain fundamental chatacteristics of phtoelectric effect. |
|
Answer» Solution :Light is electromagnetic wave made up of electric and magnetic field.Phenomena like iterference,differaction and polarization can be explain SATISFACTORILY by wave picture of light. According to wave picture of light when light is incident on surface of metal,free electron will absorb the raidant energy continuosly. When intensity of incident radiation increases,amplitude of electric and magnetic field will also increase.Hence ,more enegy will be absorbed by electron. According to this maximum kinetic energy of electron on surface should increase with increase in intensity of light. A sufficient intense beam of any frequency should be able to impart enough energy to electron,so that they exceed minimum energy NEEDED to escape from the metal surface.A threshold frequency therefore should not exist. Above expectation of were theory are CONTRADICTORY to experimental result obtained in photoelectric effect. According to wave theory absorption of energy by electron take place continously over the entire wavefront of the radiation.As very large no.of electrons absorb energy ,ENRGY absorbed per electron will be very less. Calculation estimate that it can take hours (ir more time) for an electron to come out from metal surface. This CONCLUSION is also in contradiction to experimental result in which electron emission start in `10^(-9)` s. Thus ,wave theory (wave picture) is unable to explain basic feature of photoelectric effect. |
|
| 4149. |
An electron , initiallty at rest is released from a large distance from a proton (fixed in space ). The de-broglie wavelength of the electron when it is at distance r from proton is lambda the ratio of its kinetic energy at this distance with the kinetic energy of the electron is ground stae of hydrogen atom (Bohr model) with the radius of its orbit being r, is alpha Pick the correct options (s) |
|
Answer» `lambdapropsqrt(R)` |
|
| 4150. |
In L-C-R, A.C. series circuit, L = 9H, R = 10 Omega and C = 100 muF. Hence Q factor of the circuit is …… |
|
Answer» 25 `therefore Q=300/10`=30 |
|