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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.
| 1301. |
If the value of potential gradient on potentiometer wire is decreased, then the new null point will be obtained at |
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Answer» LOWER length |
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| 1302. |
In an experiment on photo electric effect, the slope of cut - off voltage versus frequency, incident light is found to be 4.12 xx 10^(-15) Vs. Calculate the of Planck's constant. |
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Answer» `6.6xx10^(-34)Js` `4.12xx10^(-15)=(hc)/(e)` `:. h=4.12xx10^(-15)xxe` `=4.12xx10^(-15)xx1.6xx10^(-19)` `=6.529xx10^(-34)Js`. |
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| 1303. |
Two metal balls with radii r_(1)=1 cm and r_(2)=2cm at a large distance R=100 cm from each other, are connected to a battery of emf epsi=3000 V Find the of interactiobetwenn the balls. What is the signification of the ward large? |
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Answer» |
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| 1304. |
A point sound source is situated in a medium of bulk modulus 1.6 xx 10^5 N//m^2. An observer standing at a distance 10 m from the source writes down the equation for the wave asy = A sin (15pi x - 6000 pi t). Here y and x are in metres and t is in second. The maximum pressure amplitude tolerable by the observer's ear is 24 pi Pa. Find : the displacement amplitude A of the waves received by the observer. |
| Answer» SOLUTION :`10 MU m` | |
| 1305. |
A point sound source is situated in a medium of bulk modulus 1.6 xx 10^5 N//m^2. An observer standing at a distance 10 m from the source writes down the equation for the wave asy = A sin (15pi x - 6000 pi t). Here y and x are in metres and t is in second. The maximum pressure amplitude tolerable by the observer's ear is 24 pi Pa. Find : the power of the sound source. |
| Answer» SOLUTION :`288 pi^2 WATT` | |
| 1306. |
A : Kepler's law cannotbe used for asteroidsand comets . R:Asteroidsand cometsdo not revolvearoundsun underits gravitational force . |
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Answer» If both Assertion & Reasonare true . Andthe REASONIS the correct explanationof THEASSERTION , then mark (1) |
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| 1307. |
A solid cube and a solid sphere have equal surface areas. Both are at the same temperature of 120^(@)C. Then : |
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Answer» both of them same temperature of `120^(@)C`. Then : Correct choice is (a). |
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| 1308. |
Determine the Wavelength of lightemitted from LED which is made up of GaAsP semiconductor whoseforbidden energy gap is 1.875 eV.Mention the colour of the lightemitted. |
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Answer» Solution :`E_g=(HC)/(lambda)` Therfore ,`lambda=(hc)/(E_g)=(6.6xx10^(-34)xx3xx10^(8))/(1.875xx1.6xx10^(-19))=660nm` The wavelength 660 nm CORRESPONDS to red COLOUR light. |
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| 1309. |
Zenor breakdown semiconductor |
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Answer» FORWARD BIAS only |
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| 1310. |
Listout theadvantagesof stationaryarmature- rotatingfieldsystemof ACgenerator . |
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Answer» Solution :(i) The current is DRAWN DIRECTLY from fixed terminals on the stator without the USE of BRUSH contacts. (ii) The insulation of STATIONARY armature winding is easier. (iii) The number of sliding contacts (slip rings) |
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| 1311. |
The magnetic dipole moment of steel wire of length L is m. It is bent from the middle and arranged as 60°. So the new magnetic dipole moment will be ...... |
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Answer» `(m)/(sqrt2)`  ANGLE between both is 120°. `THEREFORE` Resultant mangetic dipole moment, `= SQRT(( (m)/(2))^(2) = ((m)/(2))^(2) + 2 ((m)/(2) ) ((m)/(2)) xx COS 120^(@) )` `= sqrt((m^2 )/(4) + (m^2)/( 4) + 2((m^2)/( 4)) (-(1)/(2) ) ) =sqrt((m^2)/(4) ) = (m)/(2)` |
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| 1312. |
Nuclear element having Z ge 83 are ____and continuously emit____disintegration ? |
| Answer» SOLUTION :UNSTABLE, RADIOACTIVE | |
| 1313. |
A piece of copper and another of germanium are cooled from room temerature to 80K. The resistance of |
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Answer» each of them INCREASES |
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| 1314. |
Ifwavelenght of light in air is 2400 xx 10_(-10) m, then what will be the wavelenght of light in glass (mu = 1.5)? |
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Answer» 1600 Å |
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| 1315. |
Electric field intensity at any point is the ……. Experienced by ……… placed at that point. |
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Answer» |
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| 1316. |
A body is projected at an angle such that K.E. at the highest point is reduced to half the energy at point of projection. The angle of projection is : |
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Answer» 30° `:. Ecos^(2)theta=1/2E` `costheta=1/sqrt(2)IMPLIES theta=45^@` |
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| 1317. |
For which angle of incidence reflected ray is completely polarised? |
| Answer» SOLUTION :POLARISING ANGLE. | |
| 1318. |
Statement-1 : A block of mass m starts moving on a rough horizontal surface with a velocity V. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to angle of 30^(@) with the horizontal and the same block is made to go up on the surface with the same initial velocity V. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. Statement-2 : The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. |
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Answer» statement-1 and 2 are true and statement-2 is a correct explanation for statement-1 |
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| 1319. |
A circular current coil with 100 turns and area 20 cm^(2)is placed in a uniform magnetic field in such a manner that magnetic field lines are parallel to the plane of the coil. Torque experienced by the coil is 0.1 Nm when a current of 1 A is passed through the coil. What is the magnitude of magnetic field intensity? |
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Answer» Solution :NUMBER of turns, ` N = 100` Area of coil , `A = 20 cm^(2) = 20 xx 10^(-4) m^(2)` Torque, ` tau = 0.1 ` Nm, Current , `I = 1 A` As the magnetic field lines are parallel to the plane of coil. So, angle between area VECTOR to the plane of coil and magnetic field i.e. ` theta = 90^(@)` As ` tau = NIAB sin theta` ` B = tau/(NIA sintheta)` ` B = (0.1)/(100 xx 1 xx 20 xx 10^(-4) sin 90^(@))` ` B = 0.00005 xx 10^(4)` ` B = 0.5 ` T |
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| 1320. |
A capacitor with plate separation d is charged to V volts. The battry is disconnected and a dielectric slab of thickness (d)/(2) and dlelectric constant '2' is inserted between the plates. The potential difference across its terminals becomes |
| Answer» ANSWER :D | |
| 1321. |
A layer of oil 3 thick is floating on a layer of coloured water 5 cm thick. The refractive index of the coloured wateris 5/3 and the apparent depth of the two liquids is 36/7 cm. Then the refractive index of the oil is |
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Answer» `7/4` |
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| 1322. |
In a Young's double-slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is : |
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Answer» 3 `d = 6.1 x`  width of CENTRAL MAXIMA for single slit `omega_0 = (2D lambda)/x` If n maxima is OBSERVED by Young.s double-slit in central maxima of single slit `n((lambda D)/(d)) = (2 D lambda)/x` `implies n = 6.1 xx 2 = 12.2 ~~ 12` |
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| 1323. |
A batteryof emf 10 v and internal resistane 3 Omega is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistane of the resistors ? What is the terminal voltage of the battery when the circuit is closed ? |
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Answer» SOLUTION :`E= I (R+r ) ` ` 10= 0.5 ( R+ 3 ) ` ` R= 17 Omega ` `V =E - IR = 10-0.5 xx 3= 8.5 V ` |
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| 1324. |
What we call the maximum value of electric current in either direction ? |
| Answer» SOLUTION :PEAK VALUE | |
| 1325. |
A pipe's lower end is immersed in water such that the length of air column from the top open end has a certain length 25 cm. The speed of sound in air is 350 m/s. The air column is found to resonate with a tuning fork of frequency 1750 Hz. By what minimum distance should the pipe be raised in order to make the air column resonate again with the same tuning fork? |
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Answer» 7cm |
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| 1326. |
When a metallic surface is illuminated with radiation of wavelength lambda, the stoppingpotentials is V. If the same surface is illuminated with radiation of wavelength 2 lambda, the stopping potential is ( V )/( 4). The threshold wavelength for the metallic surface is |
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Answer» `4lambda` `e((V)/(4)) = (hc)/(2lambda) - (hc)/(lambda_(0))""……(2)` `(1)+ (2)` `(4(eV))/(eV)` = `((1)/(lambda) - (1)/(lambda_(0)))/((1)/(2lambda) - (1)/(lambda_(0))) = (((lambda_(0) - lambda)/(lambda lambda_(0))))/(((lambda_(0) - 2 lambda)/(2 lambda lambda_(0))))` On solvingwe GET`lambda_(0) = 3 lambda` |
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| 1327. |
Which of the followingrelation hold good for refraction between a pairof media with i_1 and i_2 as angles incidenceand refraction v_1 and v_2 as velocitiesof light in the media ? |
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Answer» `upsilon_(2) SIN i_(1) = upsilon_(1) sin i_2` |
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| 1328. |
Radiant energy is incident on a body at the rate of 1000 joules per minute.If the reflection coefficient of the body is 0.2 and its transmission coefficient is 0.3, find the radiant energy (i) absorbed (ii) reflected (iii) transmitted by the body in 3 minutes. |
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Answer» Solution :a+ r+t=1a= 1 - r -t = 1-0.2 -0.3= 0.5 ` Let Q_(a) , Q_(r)and Q_(t)` bet the QUANTITIES of radiant energy absorbed, reflected and transmitted in 3 MINUTES by the body, RESPECTIVELY. (i) ` a = Q_(a)/Q "" therefore Q_(a) = aQ = 0.5 xx 3000 = 1500 J` (ii) ` r = Q_(r)/Q "" thereforeQ_(r) = rQ = 0.2 xx 3000= 600J` ` (III) t = (Q_t)/Q "" thereforeQ_(t_) = tQ = 0.3 xx 3000 = 900 J` |
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| 1329. |
ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with an angular velocity omega as figure. The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor ABDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of lambdaper unit length. Find the current in the rotating conductor, as it rotates by 180^@. |
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Answer» Solution :(i)Considering situation of OP wire after `t=I/8=pi/(4omega)`  Suppose at this instance wire OP intercept BD at Q. Suppose OQ is x. Area of `DeltaODQ=1/2xxODxxQD` `=1/2xxlxxltantheta` `[because tan theta = (QD)/(OD)=(QD)/l]` `=1/2l^2 tantheta` FLUX linked with `DeltaODQ,` `phi=AB=1/2l^2 tan thetaB=1/2 l^2 Btan(omegat)(because theta = omegat)` Induced emf , `epsilon=(d phi)/(dt)=d/(dt)(1/2 l^2 B tan (omegat))` `=epsilon=1/2 Bl^2 OMEGA^2 sec^2 omegat` Induced current, `I=epsilon/R =epsilon/(lambdax)` (Here only x length of wire OP is inside closed loop , HENCE resistance `R=lambdax`) Now `cos theta = l/x rArr x=l/(cos theta) =l/(cos omegat)` `therefore I=epsilon/(lambdal) cos omegat=1/(lambdal). 1/2 Bl^2 omega sec^2 omegat cos omegat` `therefore I=1/2(Blomega)/(lambdacos omegat)` (ii) For `t=T/8=pi/(4omega)` to `t=(3T)/8=(3pi)/(4omega)`  In this case wire OP intercept AB at Q suppose OQ =x Flux linked with closed loop OQBDO `phi=AB=B(l^2+l^2/(2tantheta))` `phi=Bl^2(1+1/2 cos omegat)(theta = omegat)` Induced emf , `epsilon=-(dphi)/(dt)` `=-d/(dt)Bl^2(1+1/2cot (omegat))` `=-Bl^2xx1/2[-"cosec"^2 (omegat)](omega)` `epsilon=1/2Bl^2omega "cosec"^2 (omegat)` Induced current `I=epsilon/R=epsilon/(lambdax)=(epsilonsinomegat)/(lambdal)` `(sin theta =l/x rArr x=l/(sin omegat))` `I=1/(lambdal)xx1/2Bl^2 omega"cosec"^2 (omegat). sin(omegat)` `I=1/2(Blomega)/(lambda sin (omegat))` (iii) Now `t=(3T)/8=(3omega)/(4pi)` to `t=T/2=pi/omega` Suppose wire PQ intercept AC at Q and OQ =x (see figure)  Flux linked with closed loop OQABDO is , `phi=BA=B(2l^2-l^2/(2tan (omegat)))` Induced emf, `epsilon=(dphi)/(dt)` `=d/(dt)[Bl^2(2-1/(2 tan (omegat)))]` `=(Bl^2)/2 omega "cosec"^2 (omegat)` Induced current , `I=epsilon/R = epsilon/(lambdax)=epsilon/(lambdal)sin(omegat)` `=1/(lambdal) sin (omegat). (Bl^2omega)/2 cos^2 (omegat)` `=1/2(Blomega)/(lambda sin (omegat))` |
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| 1330. |
(A): The induced current flows so as to oppose the cause producing it. (R): Lenz.s law is based on energy conservation. |
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Answer» Both A and R are true and R is the CORRECT EXPLANATION |
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| 1331. |
A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B = 0.2T. The normal to the plane of the coil makes an angle of 60^(@) with the direction of the magnetic field. The magnetic flux linked with the coil is |
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| 1332. |
A neutron of kinetic energy 65 eV collides inelastically with a singly ionised helium atom at rest. It is scattered at an angle 90^@ with respect to original direction. If the energy of scattered neutron is 6.36 eV, find the frequency of emitted radiation from the helium atom after suffering collision. |
| Answer» SOLUTION :`9.85xx10^15` HZ | |
| 1333. |
When tuning fork is vibrating the vibration of two tuning prongs : |
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Answer» differ in phase by `pi`/4 correct CHOICE is (b). |
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| 1334. |
A car of mass 1000 kg is moving with a speed of 40 ms^(-1)on a circular path of radius 400m. If its speed is increasing at the rate of 3ms^(-2) the total force acting on the car is |
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Answer» 3000N |
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| 1335. |
Which diode is reverse biased ? |
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Answer»
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| 1336. |
The energy density at a point in a medium of dielectric constant 8 is 26.55xx10^6J//m Calculate the field intensity at that point. |
| Answer» SOLUTION :k=8 ENERGY DENSITY `=26.55xx10^-6J//m^3epsilon_0=8.85xx10^-12F//m.` Energy density `1/2epsilon_0kE^2E=SQRT((2("energydensity"))/(epsilon_0k)=sqrt((2xx26.55xx10^-6)/(8.85xx10^-12xx8))=sqrt(3/4xx10^6)sqrt3/2xx10^3=8.66xx10^2N//C.` | |
| 1337. |
The semi circular ring of radius R and mass m is free to rotate about the point O in shown figure. |
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Answer» TENSION in the string is `(mg)/(2)` (B) `Mgxx(R)/(2)=(2MR^(2))alpha` (D) Centre of mass tangential acceleration only. |
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| 1338. |
Threshold frequency for photosensitive material is v_(0). If photons of frequency 2v_(0) fall on this surface, the electrons come out with a maximum speed of 4xx10^(6)ms^(-1). When photons of frequency 5v_(0) fall on the same surface, the maximum speed of the ejected electrons will be |
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Answer» `2xx10^(7)ms^(-1)` `((v_(1))_(max))/((v_(2))_(max))=SQRT((v_(1)-v_(0))/(v_(2)-v_(0)))=sqrt((2v_(0)-v_(0))/(5v_(0)-v_(0)))=1/2 implies (v_(2))_(max)=2(v_(1))_(max)=2xx4xx10^(6)=8xx10^(6)ms^(-1)`. |
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| 1339. |
What physical quantities may X and Y represent? (Y represents the first mentioned quantity) |
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Answer» Pressure versus temperature of a GIVEN gas (CONSTANT VOLUME) |
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| 1340. |
Draw a block diagram of a simple modulator for obtaining amplitude modulated signal. A carrier waves of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75% ? |
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Answer» SOLUTION :Block DIAGRAM of a simple MODULATOR for OBTAINING amplitude modulated signal: `A_(c)= 12 V," " mu= 75% = 0.75` `therefore"" mu = (A_(M))/(A_(c))` or, `"" A_(m)= muA_(c) = 0.75 xx 12 V = 9 V` `therefore` The peak voltage of the modulating signal should be 9 V. 
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| 1341. |
Light of wavelengthlambda_(0)in air entersa medium of refractiveindex n . If twopointsA and bin thismedium lie along the pathof this light at a distancex , thenphase difference phi_(0) between thesetwopoints is |
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Answer» `phi_(0) =1/n ((2PI)/(lambda_(0)))x` |
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| 1342. |
The dimensional formula of capacitance ls ...... Take Q as the dimension formula of charge. |
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Answer» `M^(1)L^(-2)Q^(-2)` `:. [C] = ([Q^(2)])/([J])` `= (Q^(2))/(M^(1)L^(2)T^(-2))` `=M^(-1)L^(-2) T^(2)Q^(2)` |
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| 1343. |
Temperature gradient in a rod 0.5 m long is 80^(@)Cm^(-1) The temperature of the hotter end is 60^(@)C. The temperature of cooler end will be : |
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Answer» `40^(@)C` `rArr(60-T_(2))/(0*5)=80""rArrT_(2)=+20^(@)C`. THUS correct choice is (d). |
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| 1344. |
A 5 muF capacitor is charged to a potential of 100 V and then disconnected from the power supply. Now it is connected in parallel to another capacitor of 20 muF. The common potential of the capacitors will be _____________ . |
| Answer» SOLUTION :`20 V , V = (C_1 V_1)/(C_1 + C_2)= ((5muF) xx 100V)/((5 + 20 )muF) = 20V` | |
| 1345. |
The wavelength of matter wave is called de Broglie wavelength An alpha particle, a proton and an electron having de-Broglie wavelength lambda_alpha,lambda_p and lambda_e respectively are moving with the same momentum then |
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Answer» `lambda_alphagtlambda_pgtlambda_e` |
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| 1346. |
How old was Margie when she first learned the punch code for her homework and tests? |
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Answer» 11 |
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| 1347. |
The mutual inductance for a pair of 2 coils is 6 mH. Calculate the induced emf developed in one coil when a steady current of 5 A flows through the other coil. |
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Answer» |
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| 1348. |
In Young's double slit experiment the point source is placed slightly off the central axis as shown in figure. The plane of slits (S_(1), S_(2)) and screen are separated by a distance of 2m. The separation between the slits is 10 mm. The position of the slit above the axis is 1mm. if the wavelength lambda = 4000 Å in the above arrangement The optical path difference between the two waves arriving at P is |
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Answer» 0.08 mm |
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| 1349. |
Draw symbol and truth table for AND gate. |
Answer» SOLUTION :AND GATE is a logic gate having two (or ,more) inputs and ONE output which is AVAILABLE only when all inputs are available. 
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| 1350. |
The K.E. of the most energetic electrons emitted from a metallic surface is doubled when the wavelength, lambda of the incident radiation is reduced from 400 nm to 310 nm. The work function of the metal is |
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Answer» 0.9 eV and `2XX(1)/(2)mv^(2)=(hc)/(lambda)-phi` DIVIDING (ii) by,(i) we get `2=(hc//lambda.-phi)/(hc//lambda-phi)` or `(2hc)/(lambda)-2phi=(hc)/(lambda.)-phi` `or phi=(2hc)/(lambda)-(hc)/(lambda.)` `=(2xx6.63xx10^(-34)xx3xx10^(8))/(4xx10^(-7)xx1.6xx10^(-19))=2.2eV` |
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