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.
| 3951. |
A man observes a coin placed at the bottom of a beaker which contains two immiscible liquids of refractive indices 1.2 and 1.4 as shown in the figure. Find the depth of the coin below the surface, as observed from above. |
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Answer» `d=3.6/1.2+7/1.4=8` |
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| 3952. |
A capacitor can be used to produce a desired electric field. We consldered the parallel plate arrangement as a basic type of capacitor. Similarly, an inductor (symbol) con be used to near the middle of a long solenoid ) as our basic type fo inductor. if we establish a current i in the winding ( or turns ) of an inductro ( a solenoid ) the current produces a magnetic flux phi_(B) through the central region of the inductor. the inductance of the inductor is then [ L = (N phi_(E))/(i) ]( inductance defined ) where N is the number of turns. the windings of the inductor are said to the linked by the shared flux, and the product N phi_(E ) is called the magnetic flux linkage . the inductance L is thus a measure of flux produced by the inductor per unit of current . Now, consider a special type of inductor whose radius of turn R as shown in the adjacent figure and total number of turns is N. This special type of solenoid produces a magnetic field |vec(B)| = mu l r^(2) uniformly along the axis of the solenoid as shown in figure. r is the distance from the axis of solenoid. The value of total flux linkages when current l flows through this special type ofinductor is given by |
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Answer» `(pi)/(2)mu_(0)"NIR"^(4)` |
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| 3953. |
A capacitor can be used to produce a desired electric field. We consldered the parallel plate arrangement as a basic type of capacitor. Similarly, an inductor (symbol) con be used to near the middle of a long solenoid ) as our basic type fo inductor. if we establish a current i in the winding ( or turns ) of an inductro ( a solenoid ) the current produces a magnetic flux phi_(B) through the central region of the inductor. the inductance of the inductor is then [ L = (N phi_(E))/(i) ]( inductance defined ) where N is the number of turns. the windings of the inductor are said to the linked by the shared flux, and the product N phi_(E ) is called the magnetic flux linkage . the inductance L is thus a measure of flux produced by the inductor per unit of current . Now, consider a special type of inductor whose radius of turn R as shown in the adjacent figure and total number of turns is N. This special type of solenoid produces a magnetic field |vec(B)| = mu l r^(2) uniformly along the axis of the solenoid as shown in figure. r is the distance from the axis of solenoid. Value of self inductance of this inductor is given by |
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Answer» `(4pi)/(3)mu_(0)NR^(3)` |
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| 3954. |
A capacitor can be used to produce a desired electric field. We consldered the parallel plate arrangement as a basic type of capacitor. Similarly, an inductor (symbol) con be used to near the middle of a long solenoid ) as our basic type fo inductor. if we establish a current i in the winding ( or turns ) of an inductro ( a solenoid ) the current produces a magnetic flux phi_(B) through the central region of the inductor. the inductance of the inductor is then [ L = (N phi_(E))/(i) ]( inductance defined ) where N is the number of turns. the windings of the inductor are said to the linked by the shared flux, and the product N phi_(E ) is called the magnetic flux linkage . the inductance L is thus a measure of flux produced by the inductor per unit of current . Now, consider a special type of inductor whose radius of turn R as shown in the adjacent figure and total number of turns is N. This special type of solenoid produces a magnetic field |vec(B)| = mu l r^(2) uniformly along the axis of the solenoid as shown in figure. r is the distance from the axis of solenoid. If an external magneitc field vec(B) perperdicular to the axis of the solenoid is applied, then total flux linkage will |
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Answer» CHANGE |
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| 3955. |
A capacitor can be used to produce a desired electric field. We consldered the parallel plate arrangement as a basic type of capacitor. Similarly, an inductor (symbol) con be used to near the middle of a long solenoid ) as our basic type fo inductor. if we establish a current i in the winding ( or turns ) of an inductro ( a solenoid ) the current produces a magnetic flux phi_(B) through the central region of the inductor. the inductance of the inductor is then [ L = (N phi_(E))/(i) ]( inductance defined ) where N is the number of turns. the windings of the inductor are said to the linked by the shared flux, and the product N phi_(E ) is called the magnetic flux linkage . the inductance L is thus a measure of flux produced by the inductor per unit of current . Now, consider a special type of inductor whose radius of turn R as shown in the adjacent figure and total number of turns is N. This special type of solenoid produces a magnetic field |vec(B)| = mu l r^(2) uniformly along the axis of the solenoid as shown in figure. r is the distance from the axis of solenoid. If it is given that total flux linkage through the inductor is KIL^(2) (where L is the inductance). then its inductance is given by |
| Answer» ANSWER :C | |
| 3956. |
A capacitor can be used to produce a desired electric field. We consldered the parallel plate arrangement as a basic type of capacitor. Similarly, an inductor (symbol) con be used to near the middle of a long solenoid ) as our basic type fo inductor. if we establish a current i in the winding ( or turns ) of an inductro ( a solenoid ) the current produces a magnetic flux phi_(B) through the central region of the inductor. the inductance of the inductor is then [ L = (N phi_(E))/(i) ]( inductance defined ) where N is the number of turns. the windings of the inductor are said to the linked by the shared flux, and the product N phi_(E ) is called the magnetic flux linkage . the inductance L is thus a measure of flux produced by the inductor per unit of current . Now, consider a special type of inductor whose radius of turn R as shown in the adjacent figure and total number of turns is N. This special type of solenoid produces a magnetic field |vec(B)| = mu l r^(2) uniformly along the axis of the solenoid as shown in figure. r is the distance from the axis of solenoid. Adjust the value of K such that inductance become twice of the previous value. |
| Answer» ANSWER :D | |
| 3957. |
A uniform string of length L and mass M is fixed at both ends while it is subject to a tension T. It can vibrate at frequencies (v) given by the formula where n= 1,2,3.......... |
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Answer» `V = n/2 sqrt(T/(ML))` `v=n/(2L)sqrt(T/MU)`, where `mu = M/L` `therefore v = n/(2L) sqrt((TL)/(M)) = n/2 sqrt(T/(ML))` |
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| 3958. |
Which of the following circular rods, (given radius r and length l) each made of the same material and whose ends are maintained at the same temperature diff. will conduct most heat ? |
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Answer» `R=2r_(0),L=2l_(0)` Since `A=pir^(2)` and `dx=l` `:.(dQ)/(dt)prop(r^(2))/(l)` Now `(r^(2))/(l)` will be maximum when `r=2r_(0)` and `l=l_(0)` Coorect choice is (c ). |
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| 3959. |
A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has length 4.8 m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave-pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of the wave-pulse. CalculateA) the time taken by the wave -pulse to reach the other end R of the wire andB) the amplitude of the reflected and transmitted wave pulses after the incident wave pulse cross thejoint Q. |
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Answer» 0.14 s, 1.5 CM, 2 cm |
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| 3961. |
Which plotis the adsorptionisobarforchemisorptionwherex istheamountof gasadsorbedon mass m (at constant pressure at temperature T ? |
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Answer»
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| 3962. |
Recovering information from a carrier is known as |
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Answer» Modulaion |
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| 3963. |
A pointergalvanometerwitha scaleof 30divisionshas aresistanceof 12Omegafullscaledeflectionis obtained fora currentof 3mA .Calculatethe currentsensitivityof thegalvanometer.Howwill youconvert thisgalvanometerintoavoltageof range0 to18 V ? |
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Answer» Solution :GIVEN`: theta= 30 ,G = 12 Omega,I =3 mA,V=18V,S_1 =? ,R=?` CURRENT sensitivity `S_1 =theta/1 = (30 )/(3 XX 10^(-3))` ` S_I = 10 xx 10^3 ` `S_I =10^4d I v //A ` `R=(V)/(I_g) -G ` `= ( 18)/( 3 xx 10^(-3) ) -12` `R=600-12 = 5988Omega ` Byconnecting`R=5988Omega `in serieswith THEGALVANOMETER . |
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| 3964. |
Where did they decide to reach to save themselves and the ship? |
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Answer» Australia |
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| 3965. |
Which of the following represent correct order of conductivity in solids ? |
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Answer» K(metals ) GTGT k(INSULATORS ) lt k (SEMICONDUCTORS) |
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| 3966. |
Assertion: If a conductur is given charge then no excess inner charge appears Reason : Electric field inside conductor is zero |
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Answer» Both ASSERTION and REASON are true and Reason is the correct EXPLANATION of Assertion |
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| 3967. |
A 3310 Å photon liberates an electron from a material with energy 3 xx 10^(-19) J while another 5000 Å photon ejects an electron with energy 0.972 xx 10^(-19)J from the same material. Determine the value of Planck's constant and the threshold wavelength of the material. |
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Answer» SOLUTION :The energy of ejected electron is GIVEN by `E = (hc)/(lambda) - (hc)/(lambda_(0))` `3 XX 10^(-19) = hc [(1)/(3310 xx 10^(-10)) - (1)/(lambda_(0))]""…(1)` `9.72 xx 10^(-20) = hc [(1)/(5000 xx 10^(-10)) - (1)/(lambda_(0))]""...(2)` Subtracting (2) from (1), we GET `(3-0.972) xx 10^(-19) = (hc)/(10^(-10)) [(1)/(3310) - (1)/(5000)]` `2.028 xx 10^(-19) = (h xx 3 xx 10^(8))/(10^(-10)) [(1690)/(3310 xx 5000)]` `h = (2.028 xx 10^(-19) xx 10^(-10) xx 3310 xx 5000)/(3 xx 10^(8) xx 1690)` `h = 6.62 xx 10^(-34)` Js Now `W_(0) = (hc)/(lambda) - E` `= ((6.62 xx 10^(-34) xx 3 xx 10^(8))/(3310 xx 10^(-10))) - 3 xx 10^(-19) = (6-3) xx 10^(-19)` `W_(0) = 3 xx 10^(-19)` J Threshold wavelength, `lambda_(0) = (hc)/(W_(0))` `= (6.62 xx 10^(-34) xx 3 xx 10^(8))/(3 xx 10^(-19)) = 6.62 xx 10^(-7)`m `lambda_(0) = 6620 xx 10^(-10) m` |
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| 3968. |
Statement I : Common base amplifiers give voltage gain with phase change . Statement II : When reverse bias in P - N junction is increased , the width of depletion layer increases. |
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Answer» Statement I is TRUE , statement II is FALSE. |
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| 3969. |
The magnitude of the vector product of two vectors is 4. The magnitude of their scalar product is 4sqrt3 . The angle between the two vectors is : |
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Answer» `30^(@)` |
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| 3970. |
One enetimetre on the main scale of Vernier callipers is divided into ten equal parts. If 10 divisions of Vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is : |
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Answer» 0.01 CM |
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| 3971. |
In the circuit shown in the figure. L is ideal inductor and E is ideal cell. Swictch S is Closed at t = 0, Then: |
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Answer» after a LONG time INTERVAL potential DIFFERENCE across CAPACITOR and inductor will be equal. |
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| 3972. |
Resonance curve of a resonant circuit is graphical representation between |
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Answer» FREQUENCY and current |
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| 3973. |
Hydrogen atom does not emit X-rays because: |
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Answer» its energy level are too far apari |
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| 3974. |
State Ampere's circuital law and express it mathematically. Give the sign convention involved in the relation. |
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Answer» Solution :Ampere.s circuital law states that for an open SURFACE with a boundary, the INTEGRAL of the product of tangential component of magnetic field for an element and the length of element is equal to `mu_0` times the total current PASSING through the surface. Mathematically, `oint B_t dl = oint vecB . vec(dl) = mu_0 I` Where I is the current through the surface. the integral is taken over the closed loop coinciding with the boundary C of the loop. As per sign convention followed, let the fingers of the right hand be CURLED in the boundary is traversed in the loop integral `oint vecB. vec(dl)`, then the DIRECTION of the given thumb gives the sense in which the current I is regarded as positive.
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| 3975. |
The ceiling of a long hall is 20 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms^(-1) can go without hitting the ceiling of the hall (g=10 ms^(-2)) |
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Answer» Solution :Here `H=20 m, u=40 ms^(-1)`. Suppose the BALL is thrown at an ANGLE `theta` with the HORIZONTAL. Now `H=(u^(2) sin^(2) theta)/(2g)rArr 20=((40)^(2) sin^(2) theta)/(2XX10)` or `sin theta=0.5 "" or theta=30^(@)` Now `R=(u^(2)sin 2 theta)/(g)=((40)^(2)xx sin 120^(@))/(10)` `=((40)^(2)xx0.866)/(10)=138.56 cm` |
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| 3976. |
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands seperated by energy gap respectively equal to (Eg)_C,(Eg)_(Si)and (Eq)_(Ge).Which of the following statement is true |
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Answer» `(Eg)_(SI)LT(Eg)_(Ge)lt(Eg)_c` |
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| 3977. |
There is friction between the block A and B while ground is smooth. Magnitude of force F is increasing linearly with time, a_(1) and a_(2) are accelerations of block A and B. Which graph best represents acceleration a_(1) and a_(2) of block A and B respectively with time? |
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Answer»
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| 3978. |
In the arrangement shown in the figure when the switch S_2 is open, the galvanometer shows no deflection for l=(L)/(2). When the switch S_2is closed, the alvanometer shows no deflection for l=(5)/(12 ) L. The internal esistance of 6V cell, and the emf E of the other battery are respectively internal resistance of cell of emf E is negligible) |
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Answer» `3 OMEGA , 8V ` |
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| 3979. |
A plane longitudinal wave of angular frequency omega = 1000 rad/s is travelling along positive x-direction in a homogeneous gaseous medium of density d = 1 kg m^(-3).Intensity of wave is I = 10^(-10) W.m^(-2) and maximum pressure change is (DeltaP)_(m) = 2 xx 10^(-4) Nm^(-2) Assuming at x = 0, initial phase of medium particles to be zero: Velocity of the wave is |
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Answer» `500 MS^(-1)` |
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| 3980. |
A plane longitudinal wave of angular frequency omega = 1000 rad/s is travelling along positive x-direction in a homogeneous gaseous medium of density d = 1 kg m^(-3).Intensity of wave is I = 10^(-10) W.m^(-2) and maximum pressure change is (DeltaP)_(m) = 2 xx 10^(-4) Nm^(-2) Assuming at x = 0, initial phase of medium particles to be zero: The equation of the travelling wave is given by |
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Answer» `y = 10^(-9) sin(1000t - 5X)m` |
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| 3981. |
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is, |
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Answer» 8 cm Actual WIDTH is the SUM of real depth from 2 sides Thickness of slab = `d_(1) n + d_(2) n` `= (5 xx 1.5) +(3 xx 1.5)`= 12 cm |
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| 3982. |
Consider a usual set-up of Young's double slit experiment with slits of equal slits of equal intensity as shown in the figure. Take O as origin and the y axis as indicated. If average intensity between y_(1)=-(lambdaD)/(4d) and y_(2)=+(lambdaD)/(4d) equals n times the intensity of maxima, then n equals ( take average over phase difference ) |
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Answer» `(1)/(2)(1+(2)/(PI))` |
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| 3983. |
The mensicus of a liquid contained in one of the limbs of a narrow U-tube is held in an electromagnetic with the mensicus in line with the field . The liquid is seen to rise . This indicates that the liquid is |
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Answer» ferromagnetic |
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| 3984. |
A prism (mu=1.5) has the refractive angle of 30^@. The deviation of monochromatic ray incident normally on its one surface will be : (Given sin48^@36'=0.75) |
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Answer» `18^@36'` |
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| 3985. |
In a semiconductor, the current flow is due to A) Electrons B) Holes C) Negatively charged ions D) Positively charged ions |
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Answer» Only A is TRUE |
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| 3986. |
Is there difference between magnetic lines of force and electric lines of force ? If your answer is yes. What is that difference ? |
| Answer» SOLUTION :Yes. Even in CONSTANT field conditions , MAGNETIC lines of FORCE are CLOSED. | |
| 3987. |
A truth, which is self-evident is a/an |
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Answer» Axiom |
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| 3988. |
In Young's double slit experiment the two slits are d distance apart. Interence pattern is observed on a screen at a distance D from the slits. A dark fringe is observed on the screen directly opposite to one of the slits. The wavelength is |
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Answer» `(D^(2))/(2D)` |
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| 3989. |
If the mass of earth is eighty times the mass of a planet and diameter of the planet is one fourth that of earth, then acceleration due to gravity on the planet would be |
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Answer» <P>7.8 m/`s^(2)` `g_(e) = (GM_(e))/(R_(e)^(2)) ""`.... (i) mass of planet is `M_(p) = (M_(e))/(80) ` & radius `R_(p) = (R_(e))/(4)` So `g_(p) = (GM_(P))/(R_(P)^(2)) ""`.... (ii) From (i) & (ii), we GET `g_(P) = g_(e)" "(M_(p))/(R_(p)^(2)) xx (R_(e)^(2))/(M_(e)) = (g_(e))/(5) = 2 m//sec^(2) ("as g = 10m"//sec^(2))` |
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| 3990. |
Unpolarised light of intensity 32Wm^-2 passes through three polarizers such that transmission axes of the first and second polarizer makes and angle 30^@ with each other and the transmission axis of the last polarizer is crossed with that of the first. The intensity of final emerging light will be |
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Answer» (a) `32Wm^-2` Angle between `P_2` and `P_3=theta=90^@-30^@=60^@` The INTENSITY of LIGHT TRANSMITTED by `P_1` is `I_1=I_0/2=32/2=16W/m^2` According to Malus law the intensity of light transmitted by `P_2` is `I_2=I_1cos^2 30^@=16(sqrt3/2)^2=12W/m^2` Similarly intensity of light transmitted by `P_3` is `I_3=I_2cos^2 theta=12 cos^2 60^@=12(1/2)^2=3W/m^2` |
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| 3991. |
A pencil of light rays falls on a plane mirror and forms a real image, so the incident rays are |
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Answer» a) parallel |
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| 3992. |
Who is the writer of the letter? |
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Answer» WILLIAM Hamlet |
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| 3993. |
In the figure, potential difference between A & Bis |
Answer» Solution :`V_(AB) = (15)/( 5 + 15) XX 20 = 15V`
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| 3994. |
In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is 4A then that in the secondary is |
| Answer» Answer :B | |
| 3995. |
The energy of a photon of characteristic X-ray from a Cooling tubes comes from |
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Answer» the K.E of the ELECTRON of the target |
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| 3996. |
In an A.C. circuit in 1 second current reduces to zero value 120 times. Hence the frequency of A.C. current is ...... Hz. |
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Answer» 50 In 1s if the value of current becomes 120 times zero then its frequency would be f = 60HZ. |
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| 3997. |
A beta active radioactive source is in the form of a conducting sphere of radius a. it is surrounded by a concentric conducting shell of radius b(gta). The shell is grounded. beta particles are emitted with kinetic energy ranging from E_(1) to E_(2)(gtE_(1)) (a) Find the maximum potential that will the acquired by the sphere of radius a. (b) Find the total charge that will flow through the grounding wire AB. (c) Find the final maximum charge on the outer sphre |
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Answer» |
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| 3998. |
An air bubble at the bottom of a lake 16 m deep has avolumeof 1.10 cm^(3). If thetemperatureat the bottom is 5.5^(@) and at thetop is is 17.0^(@)C, whatis thevolumeof the bubblejustbefor it reachesthe surface ? |
| Answer» SOLUTION :`2.92 CM^(2)` | |
| 3999. |
Fig shows a p-n junction diode. a. What does V_(B) denote ? b. Name the region AB. |
| Answer» SOLUTION :a. POTENTIAL barrierb. DEPLETION REGION | |
| 4000. |
In Young's Double slit experiment, first maxima is observed at a fixed point P on the screen. Now the screen is continuously moved away from the plane of slits. The ratio of intensity at point P to the intensity at point O (centre of the screen) |
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Answer» REMAINS constant |
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