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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.
| 1101. |
Electric current is the flow of charges along a defi nite direction and take place through metals as well as semiconductors.Give the sketch of graph with V along X-axis and I along Y-axis for a metal at room temperature. |
| Answer» SOLUTION :STRAIGHT LINE GRAPH | |
| 1102. |
Electric current is the flow of charges along a definite direction and take place through metals as well as semiconductors. Give the physical significance of the slope of the graph. If the above graph is drawn at 100°C, compared the nature of the graph with the graph at room temperature. |
| Answer» Solution :Slope gives conductance. When the temperature INCREASES, resistance ALSO increased and hence slope DECREASES | |
| 1103. |
Magnitude of displacement current through an amperian circuit is |
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Answer» `1+epsilon_(0)(dphi_(E))/(DT)` |
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| 1104. |
Rain drops are falling down ward vertically at 4 kmph. For a person moving forward at 3 kmph feels the rain at |
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Answer» 7kmph |
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| 1105. |
In forced vibration, the body vibrates with the |
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Answer» same FREQUENCY of EXTERNAL FORCE. |
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| 1106. |
On a new-moon day, a man on earth sees the annular eclipse of the sun? Diametre of the sun. How high would he have rise above the earth's surface so as just see the total eclipse of the sun? Diametre of the sun =1.4xx10^(9)m, diametre of the moon =1.7xx10^(6)m.Distances of the sun and the moon from the position of the man on earth are respectively,1.5xx10^(11)m. and 3.8xx10^(8)m. |
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| 1107. |
Three identical metallic uncheged spheres A, B, and C, each of radius a, are kept at the corners of an equilateral triangle of side d(d gt gt a) as shown in (Fig. 3.137). The fourth sphere (of radius a), which has a charge q, touches A and is then removed a position far away. B is earthed and then the earth connection is removed. C is then earthed. The charge on C is . |
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Answer» `(Q a)/(2 d)((2 d - a)/(2 d))` `(k q_2)/a + (KQ)/( 2d) + (k q_1)/d = 0` or `q_2 = (-q a)/(2 d)((d - a)/d)`  .
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| 1108. |
Modem is device which performs |
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Answer» MODULATION |
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| 1109. |
The resistance of a coil used in a platinum-resistance thermometer at 0^(@)C is 3.00 Omega and at 100^(@)C is 3.75 Omega. Its resistance at an unkown temperature is measured as 3.15 Omega. Calculate the unknown temperature. |
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| 1110. |
A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed is 5 S, then : |
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Answer» S=ft  For same change in velocity .f.t t=constant. `:. (1)/(2)ft_(1)^(2)+ft_(1)^(t)+(1)/(2)((f)/(2)).(2t_(1))^(2)=15 S` `S+ft_(1) +2S=15 S` `ft_(1)t=12 S` Now `S=(1)/(2)ft_(1)^(2)=(1)/(2)f((t^(2))/(36))=(ft^(2))/(72)` |
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| 1111. |
Which of the following field lines pattern could represent a magnetic field? |
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Answer»
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| 1112. |
Consider the following statement A and B and identify the correct answer A) Intensity of all bright diffraction fringes is not same B) The minima of a diffraction pattern are not perfectly dark. |
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Answer» A is false but B is TRUE |
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| 1113. |
Who is Ricky, Jonak, Monpi and Tinky? |
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Answer» Siblings |
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| 1114. |
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30^@ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil? |
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Answer» SOLUTION :It is given that each SIDE of square coil `l = 10 cm = 0.1 m,` hence area `A = l^2 = (0.1)^(2) = 0.01 m^2` `N = 20 , I = 12 A , theta = 30^@ and B = 0.80 T` `:.` Torque `TAU = N B A I sin theta = 20 xx 0.8 xx 0.01 xx 12 xx sin 30^@` `= 20 xx 0.8 xx 0.01 xx 12 xx 1/2 = 0.96 N m^(-1)`. |
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| 1115. |
A magnet vibrates 20 times in one minute. If its pole strength increased to 4 times, what the number of vibrations? |
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| 1116. |
The pressure of 4 gm of hydrogen occupying 10 litre of volume at 10^@C (R=8.315 J/moleK, Molecular weight of H_2=2) |
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Answer» `3xx10^5 N/m^2` |
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| 1117. |
A block mass 3 kg slides down a frictionless inclined plane of length 6 m and height 4 m. if the block is released from rest at the top of the incline, what is its speed at the bottom? |
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Answer» 5m/s |
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| 1118. |
In Kundt's tube experiment the following observations were made : Length ofthe brass rod is 100 cm, average length of a loop in air is 10.3 cm and in carbon-di-oxide=8.0 cm. Calculate thevelocityofsound in brass and in CO^(2). Whatisthefrequency ofthe note ? |
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Answer» Solution :LET `lambda_(R), lambda_(a)` and`lambda_(g)` be the wavelengths in cm in therod, in air and in gas respectively. Then `(lambda_(r))/(2) = 100 or lambda_(r) = 200 cm` `(lambda_(a))/(2) = 10.3 `or `lambda_(a) = 20.6 cm ``(lambda_(g))/(2) = 8.0 or lambda_(g) = 16cm` Let `v_(r), v_(a)` and `(v_(g))` be the sound velocities in rod, air and gas respectively.Then `(v_(r))/(v_a) = (lambda_(r))/(lambda_a) or v_(r) = (lambda_(r))/(lambda_(a)) xx v_(a)` `v_(r) = (200)/(20.6) xx 350` = `3.4 xx 10^(3) m//s` `v_(g) = 350 xx (16)/(20.6) ` = `271.8 m//s` If n be frequency of the note, then we have `V'_(a) = nlambda_(a)` `n = (v_(a))/(lambda_(a)) = (350)/(0.206)` = 1699.02 HZ |
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| 1119. |
In polarisation by reflection of light angle between reflected light and refracted light is |
| Answer» Solution :In POLARISATION by reflection of light, angle between REFLECTED light and refracted light is `(pi)/(2)` | |
| 1120. |
Pick out the incorrect statement regarding reverse saturation current in a P-n junction diode |
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Answer» This current doubles for every `100^@`C RISE in temperature |
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| 1121. |
The dimensional formula for entropy is identical to that of |
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Answer» UNIVERSAL GAS constant |
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| 1122. |
What should be angle between twomirrors that what so ever may be angle of incidence, the incident and reflected rays from the two mirros are parallel to each other: |
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Answer» `(pi)/(3)`  `DELTA = (180 - 2i_(1)) + (180 - 2i_(2))` For RAYS to parallel, `delta = 180^(@)` `therefore 360^(@) - 2 (i_(1) + i_(2)) = 180^(@)` `therefore i_(1) + i_(2) = 90^(@)` Also in `DeltaABC , (90 - i_(1)) + (90 - i_(2)) + theta = 180^(@)` or `i_(1) + i_(2) = theta` `therefore "" theta = 90^(@) = (pi)/(2)` |
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| 1123. |
The resultant amplitude due to superposition of two light waves having amplitudes a_1 and a_2 and initial phases alpha_1 and alpha_2 is: |
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Answer» `R=sqrt(a_1^2+a_2^2+2a_1a_2)` |
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| 1124. |
Permeabilityin a magnetic cirucit is similar __________in anelectricciruit |
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Answer» `mu_(R) lt x lt 0` |
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| 1125. |
what Maxwell's equations imply ? |
| Answer» Solution :They predict that the TIME and space DEPENDANT electric and magnetic FIELDS propagate as transverse WAVES called electromagnetic waves which posses VELOCITY of light. | |
| 1126. |
Two parallel beams of light of wavelength lambda inclined to each other at angle theta (ltlt1) are incident on a plane at near normal incidence. The fringe width will be : |
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Answer» `(lambda)/(2theta)`  `R=beta sinalpha` `QR` is the difference between the light reaching at `Q` and `P` respectively  `PQ=beta sinalpha` for given CASE `alpha=(theta)/(2)`  For `PQ` to be one fringe. the path difference between the INTERFERING light beams will change by `lambda` while moving from `P` to `Q` `|"path difference at"P-"path difference at" Q|=lambda` `|(betasin'(theta)/(2)-(-betasin'(theta)/(2)))|=lamda rArr2beta sin'(theta)/(2)=lambda beta=(lambda)/(2SIN(theta//2))` for NEAR normal incidence `sintheta~thetabeta=(lambda)/(theta)`  `TAN theta//2=(d//2)/(D) (therefore tan theta//2~theta//2)` `therefore theta=(d)/(D)` `beta=(lambdaD)/(d)` `beta=(lambda)/(theta)` |
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| 1127. |
Dichroism is the property where |
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Answer» unequal ABSORPTION of ordinary and EXTRAORDINARY rays TAKES place |
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| 1129. |
How is the internal resistance of a battery affected by the separation of its electrodes and their surface area ? |
| Answer» Solution :If we INCREASE the separation between ELECTRONS then the INTERNAL resistance increases. If we increase the SURFACE AREA of electrode then also the internal resistance of cell decreases because more number of ions can access electrode simultaneously. | |
| 1130. |
What is the frequency of a photon whose energy is 66.3 eV? |
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Answer» Solution :Here E = 66.3 e V = `66.3 XX 1.6 xx 10^(-19) J` Frequency v = `E/h = (66.3 xx 1.6 xx 10^(-19))/(6.63 xx 10^(-34)) = 1.6 xx 10^16 HZ` |
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| 1131. |
The maximum kinetic energyof photoelectrons |
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Answer» a. depends on collector plate |
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| 1132. |
On a velocity time graph of a body, the ratio of average acceleration during time interval OA and AB is: |
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Answer» 1 |
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| 1133. |
The resultant of two forces each of magnitude P acting at a point is sqrt2p. The angle between two vectors is |
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Answer» `45^@` |
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| 1134. |
An oscillator is producing FM waves of frequency 2 kHz with a variation of 10 kHz. What is the modulation index? |
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Answer» 0.67 |
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| 1135. |
Impedance in L-C-R series circuit is a ……... number. Its value is given by …….. |
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Answer» REAL number , `R+j(Z_L +Z_C)` `=R+jomegaL-j/(omegaC)` `=R+j (OMEGAL -1/(omegaC))` `thereforeZ=R+j (X_L -X_C)` which is complex number |
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| 1136. |
If the proton is moving at 90° to the uniform mag netic field what will be the change in kinetic en ergy of the proton? Give reason. |
| Answer» SOLUTION :Zero. Since the FORCE is perpendicular to the DI RECTION of VELOCITY work done is zero. | |
| 1137. |
In Young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4, while the lower slit is covered by another glass plate having the same thickness as the first one but having the refractive index 1.7. Interference pattern is observed using light of wavelength 5400 Å . It is found that the point P on the screen where central maximum (n = 0) fall before the glass plate were inserted, now has % the original intensity. It is further observed that what used to be the fifth maximum earlier lies below the point P while the sixth minima lies above P. Calculate the thickness of glass plate. (Absorption of light by glass plate may be neglected). |
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| 1138. |
STATEMENT-1 Superposition takes place only between those waves emitted by coherent sources. STATEMENT-2 All coherent sources emit energy in proper order. |
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Answer» Statement-1 is true Statemetnt-2 is True,Statement -2 is a correct EXPLANATION for statement -1. |
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| 1139. |
Eightequal chargedareplaced at the eight corners of a cubewhich is aon each side .Calculate the force experienced by each charge q. |
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| 1140. |
Define the term 'rms value of the current'. How is it related to the peak value ? |
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Answer» Solution :Incomplete cycle, the square root of average value of square of ALTERNATING current is CALLED the root MEAN square (rms) value of alternating current. `Irms = (I_(0))/(sqrt(2))` `= 0.707 I_(0)`,where `I_(0)` is the PEAK value of current. |
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| 1141. |
A ray of unpolarised light in incident on the surface of a glass plate of refractive index 1.54 at the polarising angle. If tan 57^@C = 1.54, then the angle of refraction in glass is : |
| Answer» ANSWER :C | |
| 1142. |
Name two factors on which the resistivity of a given material depends ? A carbon resistor has a value of62 kOmegawitha tolerance of 5%. Give the colour code for the resistor. |
| Answer» Solution :Temperature, MATERIAL Blue, Red, ORANGE, Gold. | |
| 1143. |
In the process of nuclear fission of 1 gram uranium , the mass lost is 0.92 milligram. The efficiency of power house can by it is 10 %,Toobtain 400 megawatt power from the power house , how much uranium will be required per hour?(c= 3xx10^(8)ms^(-1)) |
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Answer» Solution :Power to be OBTAINED from power house = 400 mega watt `:.` Energy obtained per HOUR = 400 mega watt `xx` 1hour = `(400xx10^(6) "watt") xx 3600 ` second `=144xx10^(10)` joule Here only 10% of input is utilized . In order to OBTAIN `144xx10^10` joule of useful energy, the output energy from the power house `(10E)/(100)=144xx10^(10)J` `E = 144 xx10^(11)` joule Let , this energy is obtained from a mass - loss of `Deltam` kg Then `(Deltam)c^2 = 144xx10^(11) ` joule `(Deltam) = (144xx10^(11))/(3xx10^(18))^2 = 16xx10^(-5)kg ` = 0.16 gm Since 0.92 milli gram `(=0.92 xx10^(-3)" gm")` mass is lost in 1 gm uranium , hence for a mass loss of 0.16 gm , the uranium required is `=(1xx0.16)/(0.92xx10^(-3))=174 gm` Thus to run the power house , 174 gm uranium is required per hour. |
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| 1144. |
There is a detective submarine installed inside sea water post 26/11 incident to detect terrorists. It is moving with constant speed V0 along a straight line and it sends a wave which travels with speed V_w = 1100 m/sec in water. Initially waves are getting reflected from a fixed Island and the frequency detected by the submarine is found to be 20% more than the original frequency. When a terrorist ship moving towards the submarine with constant speed V comes in between the submarine and the island. Frequency of waves reflected from the ship is 80% more than the original frequency, (density of water is 10^(3) kg//m^(3)) Velocity V_(0) is: |
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Answer» 50 m/sec |
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| 1145. |
There is a detective submarine installed inside sea water post 26/11 incident to detect terrorists. It is moving with constant speed V0 along a straight line and it sends a wave which travels with speed V_w = 1100 m/sec in water. Initially waves are getting reflected from a fixed Island and the frequency detected by the submarine is found to be 20% more than the original frequency. When a terrorist ship moving towards the submarine with constant speed V comes in between the submarine and the island. Frequency of waves reflected from the ship is 80% more than the original frequency, (density of water is 10^(3) kg//m^(3)) Speed of enemy ship V_(s) is: |
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Answer» 220 m/sec |
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| 1146. |
'Pitch' is a characteristic of sound that depends upon its |
| Answer» Answer :b | |
| 1147. |
There is a detective submarine installed inside sea water post 26/11 incident to detect terrorists. It is moving with constant speed V0 along a straight line and it sends a wave which travels with speed V_w = 1100 m/sec in water. Initially waves are getting reflected from a fixed Island and the frequency detected by the submarine is found to be 20% more than the original frequency. When a terrorist ship moving towards the submarine with constant speed V comes in between the submarine and the island. Frequency of waves reflected from the ship is 80% more than the original frequency, (density of water is 10^(3) kg//m^(3)) Bulk modulus of sea water is: |
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Answer» `10^(9) N//m^(2)` |
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| 1148. |
An alternating current is given by |
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Answer» `(7)/(sqrt2)A` |
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| 1149. |
For a glass prism (mu = sqrt(3)) , the angle of minimum deviation is equal to the angle of the prism . Find the angle of the prism . |
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Answer» SOLUTION :REFRACTIVE INDEX of material of prism. `mu=(SIN""((A+D_m)/(2)))/(sin(A/2))` but `D_m = A` `thereforemu= sin A /("sin"A/2)` `= (2 sin""A/2 cos"" A/2)/(sin ""A/2) [ because sin theta = 2 sin"" theta/2cos ""theta/2]` `thereforesqrt(3) = 2 cos""A/2` `therefore cos""A/2 = sqrt(3)/(2)` `therefore A/2 = 30^@` `thereforeA = 60^@` |
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| 1150. |
A particle starts from rest and travels a distance 's' with uniform acceleration, then it travels a distance '2s' with uniform speed and finally it travels a distance '38' with uniform retardation and comes to rest. If whole motion is in a straight line, the ratio of maximum velocity to average velocity is |
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Answer» Solution :Here s= area of v-t graph LET v be MAXIMUM velocity, then `s=(1)/(2) ut_(1)` `:. t_1=(2S)/(v)` Also `2s=vxxt_(2)` `t_2=(2s)/(v)` and`3s=(1)/(2)vt_(3)` `t_(3)=(6s)/(v)` Then `v_(av)=(6s)/((2s)/(v)+(2s)/(v)+(6s)/(v))=(6sxxv)/(10 s)=(3)/(5)v` Now `v_(max)=v` `:. (v_(max))/(v_(av))=(v)/(3//5v)=(5)/(3)` 
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