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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.
| 5051. |
An electric dipole is kept in nonuiform electric field. It experiences |
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Answer» a FORCE and a TORQUE |
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| 5052. |
Draw a typical illuminated P-N junction solar cell |
Answer» SOLUTION :
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| 5053. |
The radio of resolving powers of an optical microscope for two wave lengths lambda_(1)= 4000 A^(0) and lambda_(2)= 6000 A^(0) is: |
| Answer» Answer :D | |
| 5054. |
As shown in the figure, a magnet is moved with some speed towards a coil at rest. Due to this induced electromotive force, induced current and induced charge in the coil are E,I and Q respectively. If the speed of the magnet is doubled, the incorrect statement is |
| Answer» Answer :D | |
| 5055. |
In the figure, an air lens of radii of curvature 10 cm (R_(1)=R_(2)=10 cm) is cut in a cylinder of glass (µ=2/3). The focal length and the nature of the lens is |
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Answer» 15cm,concave `1/f=(""_(g)µ_(a)-1)(1/R_(1)-1/R_(2))` `=(2/3-1)[1/10-(-1/10)]` `rArr`f=-15 cm |
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| 5056. |
The temperature coefficient of resistivity of a semiconductor is |
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Answer» ALWAYS POSTIVE. |
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| 5057. |
An atom in the state .^(2)P_(3//2) is located in the external magnetic field on induction B=1.0kG. In terms of the vector model find the angular precession velocity of the total angular momentum of that atom. |
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Answer» Solution :If `vec(M)` is the total angular momentum vector if he atom then there is magnetic moment `vec(MU)_(m)=g mu_(B)vec(M)//ħ` associated with it, here `g` is the Lande factor. In a magnetic field of INDUCTION `vec(B)`, an energy `H'= -g mu_(B)vec(M).vec(B)/ħ` is associated with it. This interaction term CORRESPONDS to a presession of the angular momentum vector beacuse if leads to an equation of motion of hte angular momentum vectro of the form `(dvec(M))/(dt)=vec(Omega)xxM` where `vec(Omega)=(g mu_(B)vec(B))/(ħ)` Using Gaussian unit expression of `mu_(B)=0.927xx10^(-20) erg//gauss, B=10^(3)gauss ħ= 1.054xx10^(-27)` erg sec and for the `.^(2)P_(3//2)` state `g=1+((3)/(2)xx(5)/(2)+(1)/(2)xx(3)/(2)-1xx2)/(2xx(3)/(2)xx(5)/(2))=1+(1)/(3)=(4)/(3)` and `Omega=1.17xx10^(10)rad//s` The same formula is VALID in `MKS` units also But `mu_(B)= 0.972xx10^(-23)A/m^(2),B=10^(-1)T` and `ħ=1.054xx10^(-34)` Joul e sec. The answer is the same. |
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| 5058. |
Energy stored in the choke coil is in the form of ____ |
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Answer» heat |
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| 5059. |
A doubly ionized lithium atom is hydrogen like with atomic number Z = 3 . Find the wavelength of the radiation required to excite the electron in Li^(2+) from the first to the third Bohr orbit. Given the ionization energy of hydrogen atom as 13.6 eV. |
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Answer» Solution :The energy of `n^(th)` orbit of a hydrogen-like ATOM is given as `E_(n) = - (13.6Z^(2))/(n^(2))` Thus for `Li^(2+)` atom, as Z= 3 the electron energies of the first and third Bohr orbits are Forn = 1 , `E_(1) ` = - 122.4eV, For n = 3 , `E_(3)`= - 13.6eV. Thus the LEVEL is, E = `E_(3) - E_(1) `= - 13.6 - (-122.4) = 108.8eV Therefore , the radiation needed to CAUSE this transition should have photons of this energy. hv = 108.8 eV The WAVELENGTH of this radiation is or `lambda = (hc)/(108.8eV) = 114.25`Å |
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| 5060. |
A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory? Explain the meaning of this statement. |
| Answer» Solution :(c) When external magnetic FIELD applied on FERROMAGNETIC sample with net magnetic field zero, is increased gradually and then decreased to its initial value, the DOMAINS in that sample do not return COMPLETELY to their original configuration but retains some magnetic field called residual or remanent magnetic field as its "memory.. of past. From this remanence value, we can come to know about no. of hysterisis cycles repeated by given ferromagnetic sample. Such memory given ferromagnetic sample. Such memory magnetic storage of information on tapes of CASSETTES and on the disks of computers. The no. of bits in the information signal can be related to no. of hysterisis cycles repeated and that is how, ferromagnetic sample is used as a memory storage device. | |
| 5061. |
How we obtain the image in a simple microscope? |
| Answer» SOLUTION :The IMAGE is obtained at the DISTANCE of distinct VISION | |
| 5062. |
A body starting from rest moves along a straight line with constant acceleration. The variation of speed (v) and distance (x) is given by graph represented in: |
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Answer»
For UNIFORM accelerated motion, VELOCITY time graph is parabola with arms streching alongs distance AXIS Hence correct choice is (b). |
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| 5063. |
In the given v -1 graph, the distance travelled by the body in 5 second will be |
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Answer» 20 m `=(1)/(2)xx2xx20+3xx20+(1)/(2)xx1xx20+(1)/(2)xx1xx20` =100 m |
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| 5064. |
A current I flows along a straight conductor with round cross-section. Find the flux of the Poynting vector across the lateral surface of the conductor's segment with resistance R. |
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Answer» Solution :Suppose the radius of the conductor is `R_(0)`. Then the conduction current density is `j_(c) = (I)/(piR_(0)^(2)) = sigmaE` or `E = (I)/(piR_(0)^(2)sigma) = (rhoI)/(piR_(0)^(2))` where `rho = (1)/(sigma)` is the resistivity. Inside the conductor there is a MAGNETIC field given by `H.2pi R_(0) = I` or `H = (I)/(2piR_(0))` at the edge LTBRGT `:.` Energy flowing in per second in a section of length `l`is `EH xx2piR_(0)l = (rhoI^(2)l)/(piR_(0)^(2))` But the resistance `R = (rhol)/(piR_(0)^(2))` Thus the energy flowing into the conductor `= I^(2)R`. |
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| 5065. |
The de-Broglie wave length of a ball of mass 0.12 kg is 2.76xx10^-34m. Calculate the speed of ball. |
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Answer» SOLUTION :`lambda=H/(MV)` `THEREFORE( h/(mlambda))=((6.6xx10^-34)/(0.12xx2.76xx10^-34))=19.9ms^-1 ` |
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| 5066. |
The maximum peak to peak voltage of an AM wave is 24mV and the minimum peak to peak voltage is 8mV. The modulation factor is not equal to |
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Answer» 0.1 |
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| 5067. |
Two point charges q_1 and q_2 are placed in an external uniform electric field as shown in figure. The potential at the location of q_1 and q_2 are V_1 and V_2, i.e., V_1 and V_2 are potentials at location of q_1 and q_2 due to external unspecified charges only. Then electric potential energy for this configuration of two charged particle is |
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Answer» `(q_1V_1+q_2V_2)/2` `W_(2)=q_(2)V_(2)+(q_(1)q_(2))/(4piepsilon_(0)r)` So total work done is `U`, `q_(1)V_(1)=q_(2)V_(2)+(q_(1)q_(2))/(4piepsilon_(0)r)` |
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| 5068. |
A charge oil drop is moving with a velcoity v_(1) As it acquire charge it moves up with the velcity V_(2) in the same electric field . It fall freely with a velocity 'V' in the absence before and after acquiring additional charge is |
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Answer» `(V_(1)+V)/(V_(2)-V_(1))` `Eq_(2)-mg=6pinaV_(2)TO2` |
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| 5069. |
The region of the atmosphere above troposphere is known as |
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Answer» Lithosphere |
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| 5070. |
A refrigerator abstracts 1600 calories of heat from the ice trays. The coefficiency of performance is 4. What is the work done by the motor of the refrigerator in heat-units i.e. calories ? |
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Answer» 300 cal |
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| 5071. |
A crate rests on the flatbed of a truck that is initially traveling at 15 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding? |
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Answer» `0.20` |
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| 5072. |
A tunning fork produces 4 beats/s both with 50 and 40 cm of a stretched wire of sonometer. The frequency of fork is : |
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Answer» 36 Hz Then `v.= ((V)/(V - U_(s)) )` V `rArr 1000 = ((350)/(350 - 50)) v rArr v = (1000 xx 300 )/(350 )` When the source is moving away from the OBSERVER then ` v.. = ((V)/(V + U_(s)) ) ` V ` rArr"" v.. = ((350)/(350 + 50) ) xx (1000 xx 300)/(350)` v.. = 750 Hz correct choice is (d). ` v_(2) = (1)/(2l_(2) sqrt((T)/(m))` `therefore (v_(1))/(v_(2)) = (l_(2))/(l_(1)) = (40)/(50) = (4)/(5) " " rArr ""V_(1) = (4)/(5) v_(2)` LET the unknonw frequency of TUNNING fork = v `therefore "" v- v_(1) = 4 ` `v_(2) - v = 4 ` Adding `"" v_(2) - v_(1)= 8 ` `v_(2) - (4)/(5) v_(2)= 8 ` `(1)/(5) v_(2) = 8"" rArr v_(2) = 40` Hz. `therefore "" 40 -v =4"" rArr "" v = 36 ` Hz. Hence the correct choice is (a) . |
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| 5073. |
A conducting rim of radius R rolls without slipping on a horizontal surface . The speed of the ring 's centre of mass is v_(0) . A uniform magnetic field B perpendicular to the ring 's plane is swithched on. The emf induced across point of contact and the topmost point of the rim is 2B v_(0)R. |
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Answer» |
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| 5074. |
Define beta_("dc") ofa transistor. |
| Answer» Solution :A small change in base CURRENT produces a LARGE variation in the COLLECTOR current. This property of an amplifier is KNOWN as amplification | |
| 5075. |
Out of microwaves, ultraviolet and infrared rays , Which radiation is most effective for emission of electrons from a metallic surface ? |
| Answer» Solution :ULTRAVIOLET rays are most effective for photoelectric EMISSION as they have highest FREQUENCY and hence most ENERGETIC. | |
| 5076. |
Two parallel horizontal conductors are suspended by light vertical threads 75.0 cm long. Each conductor has a mass of 40.0gm per metre, and when there is no current they are 0.5 cm a part. Equal magnitude current in the tow wires in a separation of 1.5 cm . Find the values and directions of currents. |
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Answer» Solution :The situation is shown in figure, Here, we have ` T cos theta = mg (i)` `T sin theta = F = (mu_0)/(4 PI) L (2i_1 i_2)/(d) or ` `T sin theta = (mu_0)/(4pi) l (2i^2)/(d)....(ii)` from eqs. (i) and (ii) `tan theta = (mu_0)/(4pi) l . (2i^2)/(d). (1)/(mg)....(iii)`  where `theta` is small, `tan theta ~~ sin theta` From figure `sin theta = (0.5 xx 10^(-2))/(75 xx 10^(-2))` `m = 40.0 xx 10^(-3) l KG` where l = length of conductor in meter Substituting in eq. (iii). we get `(0.5 xx 10^(-2))/(75 xx 10^(-2)) = 10^(-7)l (2i^2)/((1.5 xx 10^(-2))) xx(1)/((40xx 10^(-3))lxx 9.8)` Solving , we get `i = 14` amp. For REPULSION , the currents are in opposite direction. |
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| 5077. |
Do you agree with the statement that an induced emf has no direction of its own? Explain. |
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Answer» Solution :YES, it is TRUE that induced EMF has no DIRECTION of its own. Because if the magnetic flux in a circuit increases, the direction of induced emf will be such as to decrease the flux. If the magnetic flux in a circuit decreases then the direction of induced CURRENT will be such as to increase the flux. |
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| 5078. |
Of the following, NAND gate is |
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Answer»
`beta_(1) = (20/21)/(1-(20/21)) = 20` `beta_(2) = (100/101)/(1-(100/101)) = 100` The value of `beta` RANGES between 20-100 |
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| 5079. |
Mention the range of mass number for which the Binding energy curve is almost horizontal. |
| Answer» SOLUTION :For A =30 to 120 (A is MASS NUMBER) | |
| 5080. |
One end of a copper rod of length 1.0m and area of cross-section 10^(-3)m^(2)is immersed in boiling water and the other ens in ice. If the coefficient of thermal conductivity of copper is 92 cal/m-s- ""^(@)C and the latent heat of ice is 8xx10^(4) cal/kg, then the amount of ice which will melt in one munite is : |
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Answer» `8xx10^(-3)` kg where L is latent HEAT and `t=1` minute. `(KA(T_(1)-T_(2)))/(l)=mL` `m=(92xx10^(-3)xx(100-0)60)/(1xx8xx10^(4))` `=6*9xx10^(-3)kg` Correct CHOICE is (d). |
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| 5081. |
Applications such as electrostatic point spraying and powder coating. Are based on the property of ................ Between charged bodies. |
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Answer» |
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| 5082. |
If A={2,3,4,5} and B={4,5,7), then n(AnnB)=? |
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Answer» 2 |
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| 5083. |
Figure shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 cm.The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.a.Calculate the capacitance and the rate of change of potential difference between the plates.b.Obtain the displacement current across the plates.c.Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor ? |
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Answer» Solution :a.`R=12xx10^(-2)m, d=5xx10^(-3)m, I=0.15A` `A=PI r^(2)=pi XX (12xx10^(-2))^(2)m^(2)` `C=(epsilon_(0)A)/(d)=(8.85xx10^(-12)xx pi xx(12xx10^(-2))^(2))/(5xx10^(-3))=80.1xx10^(-12)F=80.1 pF` `q=CV` `I=(dq)/(dt)=C.(dV)/(dt)` `(dV)/(dt)=(I)/(C )=(0.15)/(80.1xx10^(-12))=1.875xx10^(9)Vs^(-1)` b.Displacement current = Conduction current`"" THEREFORE I_(D)=0.15 A` c.Yes |
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| 5084. |
Is there any benefit to adding trivalent or pentavalent impurity to the copper ? |
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Answer» YES Adding impurity in COPPER REDUCES its conductivity so that the flow decreases. |
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| 5085. |
In Millikan's oil drop experiment, an oil drop with density 1.5 g cm^(-3)and a radius of 10 xx 10^(-7)m is supported by an external electric field. Calculate the required electric field strength if charge on the drop is 15e. |
| Answer» SOLUTION :`25.64xx10^3` N/C | |
| 5086. |
For a body in a uniformly accelerated motion, the distance of the body from a reference point at time t is given by x=al+ bt^(2) + C where a,b, c are constants. The dimensions of 'c' are the same as those of (a) x (b) at (c) bt^(2) (d) a^(2) //b |
| Answer» ANSWER :D | |
| 5087. |
Define mass defect and nuclear binding energy. For a nucleus ""_(Z)^(A)X, write the value of mass defect and nuclear binding energy. |
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Answer» If `Deltam` is the mass defect, then Binding energy, `B.E.=Deltam C^(2)` (in joule) `=Zm_(p)+(A-Z)M_(n)-M_("nucleus")` Ifmass defect is in a.m.u. then `B.E.=(Deltamxx931.5)` M EV Expression for binding energy. The mass defect `(Deltam)` OFA nucleus containing Z protons and (A-Z) neutrons is given by `Deltam=ZM_(p)+(A-Z)M_(n)-M_("nucleus")` ...(1) where m is mass of the nucleus. So `M_("nucleus")=(ZM_(p)+(A-Z)M_(n))-DELTA m` ...(2) Since mass of atom `M(""_(Z)X^(A))` is given by `M=M_("nucleus")+ZM_(e)` Using eq (2), we get `M=(ZM_(p)+(A-Z)M_(n))-Deltam+ZM_(e)` `:.` Mass defect, `Deltam=Z(M_(p)+M_(e))+(A-Z)M_(n)-M` `=ZM_(H)+(A-Z)M_(n)-M [ :.M_(p)+M_(e)=M_(H)]` `:.B.E.=Delta mc^(2)` or `B.E.=[ZM_(H)+(A-Z)M_(n)-M]c^(2)` |
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| 5088. |
Shown a circular coil of N turns and radius a, connectedto a battery of emf epsilon through arheostat. The rheostat has a totallength L and resistance R. The resistance of the coil is r. Asmall circular loop of radius a' and resistance r' is placed coaxially with the coil. The centre of the loop is at a distance x from the centre of hte coil. In the beginning, the sliding contact of the rehostat is at the left end and then on wards it is moved towards right at a constant speed v. Find the emf induced int he small circular loop at the instant (a) the contact begins to slide and (b) it has slid thrugh half the length of the rheostat. |
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Answer» Solution :Magneticfield due to the coil (1) at the CENTER of (2) is ` B=( mu_0 Nia^2)/(2(a^2+x^2)^3//2)` Flux linked with the second, `phi=B.A'=( mu_0 Nia^2)/(2(a^2+x^2)^3//2) pi a'^2 ` `E.m.f.induced ` `=(d phi)/(dt)=( mu_0 Na^2 a'^2 pi)/(2(a^2+x^2)^3//2) (di)/(dt)` ` ( mu_0 N pi a^2 a'^2)/(2(a^2+x^2)^3//2) (d)/(dt) (e)/((R/L)x+r)` ` ( mu_0 N pia^2 a'^2)/(2(a^2+x^2)^3//2)e.(-1.(R)/(L).v)/((R)/(L)x+r)^2)` `(a) for x= L` `e=( mu_0 N pi a^2a'^2)/RvE)/(2L(a^2+x^2)^3//2 (R+r)^2)` ` (b) e=( mu_0 N pi a^2a'^2)/(2L(a^2+x^2)^3//2) (ERV)/(L((R/L)+r)^2)` `[for x = l/2,R/L, x=R/2]` |
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| 5089. |
A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth's magnetic field H_E at a place. If H_E= 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel ? Note that 1G=10^(-4) T. |
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Answer» Solution :Here all the SPOKES are identical and connected in parallel and so same INDUCED emf is obtained across all the spokes. Hence, we should find induced emf across any one spoke. Let it be `epsilon`. . Here, `epsilon=1/2 B omegaR^2 "" OMEGA=(120xx2pi)/60=4pi"rad"/"SEC"` `=1/2(0.4xx10^(-4))(4pi)(0.5)^2` `=0.2xx10^(-4)xx4xx3.14xx0.25` `therefore epsilon=6.28xx10^(-5)` V |
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| 5090. |
Calculate the effective capacitance combination given below: |
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Answer» |
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| 5091. |
In the given circuit, the potential difference across the 6muFcapacitor in steady state is |
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Answer» 1V |
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| 5092. |
The speed of sound in air is v. The velocity of a source of sound whose frequency appears to be doubled to a stationary observer is |
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Answer» `(v)/(4)`, TOWARDS the OBSERVER |
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| 5093. |
Assertion: Two particles of same charge projected with different velocity normal to electric field experience same force Reason:A charged particle experiences a force in electric field which is independent of velocity |
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Answer» Both Assertion and REASON are TRUE and Reason is the correct explanation of Assertion |
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| 5094. |
Calculate the momentum of an electron whose wavelength is 2 A^(@) . |
| Answer» SOLUTION :`3.313xx10^(-24)"KG MS"^(-1)` | |
| 5095. |
A physical quantity P is related to four observables a, b, c and d as follows P=(a^(3)b^(2))/((sqrt(c)d)). The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%. respectively. What is the percentage error in the quantity P? If the value of P calculated using the abovc rclation turns out to be 3.763, to what value should you round off the resultr? |
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Answer» 13%, 3.763 |
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| 5096. |
At a distance of 10 cm from a long straight wire carrying current the magnetic field is 0.04 T. At the distance of 20 cm, the magnetic field will be |
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Answer» `0.01 T` |
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| 5097. |
Matter waves are __ waves. They are not the electromagnetic waves or mechanical waves. |
| Answer» SOLUTION :PROBABILITY | |
| 5098. |
An electric heater consists of a nichrome coil and runs under 220 V, consuming 1 kW power. Part of its coil burned out and it was reconnected after cutting off the burn portion. The power it will consume now is - |
| Answer» Answer :A | |
| 5099. |
She has been honoured with titles like: |
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Answer» Vanamitra |
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| 5100. |
Y(x,t) = 0.8/[(4x+ 5t)^(2) +5] represents a moving pulse where x and y are in metre and t is in second. Then |
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Answer» PULSE is moving in POSITIVE x-direction |
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