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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.
| 40101. |
A nuclide A undergoes alpha -decay and another nuclide beta^(-)undergoes B decay a) All the a-particle emitted by A will have almost the same speed b) The b-particle emitted by A may have widely different speeds c) All the b-particle emitted by B will have almost the same speed d) The b-particle emitted by B may have widely different speeds |
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Answer» a, B are TRUE |
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| 40102. |
Frequency of wave is 6xx10^15 Hz. The wave is: |
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Answer» RADIO active |
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| 40103. |
Draw typical output characteristics of an n-p-n transistor in CE configuration. Show how these characteristics can be used to determine output resistance. |
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Answer» Solution :Output characteristics is the plot between collector-emitter voltage `(V_(CE))` and the collector current `(I_(c))` at different constant values of base current `(I_(B))`. Output resistance is defined as the ratio of the change in collector-emitter voltage `(Delta V_(CE))` to the change in collector current `(DeltaI_(c))` at a constant base current `(I_(B))`.  Initially with the increase in `V_(CE)` the collector current increases almost linearly. This is because the junction is not reverse biased. When the SUPPLY is more than REQUIRED to reverse bias the base-collector junction, `I_(C)` increases very little with `V_(CE)`. The RECIPROCAL of slope of the linear part of the curve gives the value of output resistance i.e., `r_(0)=((Delta V_(CE))/(Delta I_(c)))I_(B)`. |
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| 40104. |
When the current in a coil changes from 2A to 10 A in 0.5sec, an e.m.f. of 8V is induced in the coil. The coefficient of self-inductance of the coil is .......... henry |
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Answer» 0.1 |
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| 40105. |
A cylindrical glass rod of radius 0.1 m and RIsqrt3 lies on a horizontal plane mirror. A horizontalray of light goind perpendicular to the axis of rod is incident on itDeviation suffered by ray in second rod is |
| Answer» ANSWER :D | |
| 40106. |
A cylindrical glass rod of radius 0.1 m and RIsqrt3 lies on a horizontal plane mirror. A horizontalray of light goind perpendicular to the axis of rod is incident on itAt what distance a second identical rod be placed on the mirror such that emergent ray from the second rod is in line with incident ray on 1st rod. |
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Answer» 21.5cm |
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| 40107. |
A cylindrical glass rod of radius 0.1 m and RIsqrt3 lies on a horizontal plane mirror. A horizontalray of light goind perpendicular to the axis of rod is incident on itAt what height from plane mirror should the ray be incident so that it emerges from the rod at a height 0.1 m above the mirror |
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Answer» `18(2)/(3) cm` |
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| 40108. |
A body of mass 10 kg is raised to height 10 m above the ground with an acceleration of 1.5 m//s^(2). The work done in this case taking g = 10 ms^(-2) is: |
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Answer» 1000 J F=m(g+a)  WORK DONE,`W=m(g+a)XXH` `10xx(10+1.5)xx10=1150 J` |
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| 40109. |
A solid homogenous sphere of mass M and radius r is moving on a rough horizontal surface, partly rolling and partly sliding. During this kind of motion of this sphere |
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Answer» total KINETIC energy is conserved Angular momentum is conserved only when external TORQUE is zero. Total kinetic energy and rotational kinetic energy about the centre of mass is not conserved. |
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| 40110. |
A metallic shell has a point charge 'q' kept inside its cavity. Which one of the following diagrams correctly represents electric lines of forces |
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Answer»
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| 40111. |
Consider the circuit shown in the figure where all the resistances are of magnitude 1 k.Omega. If the current in the extreme right resistance X is 1 mA, the potential difference between A and B is |
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Answer» 34V  Now ` i_1= 1 mA ` ( given) POTENTIAL differneceis SAMEFOR resistanceconnectedin parallel ` thereforei_1(1+1) = i_2(1) implies i_2 = 2 mA ` ` (i_1 +i_2) +i_2 (1)=i_3(1) implies i_3 = 5 mA ` `( i_1+i_2+i_3) 1+ i_3(1)= i_4(1)implies i_4= 13 mA ` ` (i_1+i_2+i_3+i_4) 1+ i_4 (1)=V_(AB)implies V_(AB )= 34 V.` |
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| 40112. |
Vehicles carrying inflammable material usually have metallic ropes touching the ground during motion, why? |
| Answer» SOLUTION :When the vehicle moves, its body gets CHARGED in addition to the CHARGE developed on tyres due to air friction. If the accumulation of the charge becomes large, it may prove hazadous to the vehicle carrying inflammable materials. If the METALIC ropes are touching the vehicle, the charge PRODUCED leaks to ground and the vehicle is saved from catching file. | |
| 40113. |
A hydrogen atom (ionization potential 13.6 eV) makes a transition from third excited state to first excited state. The energy of the photon emitted in the process is _____eV. |
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Answer» 1.89 |
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| 40114. |
The molar specific heat of oxygen at constant pressure, C_(p)7.03" cal/mol"^(@)C and R=8.31" Joules/mol"^(@)C. The amount of heat taken by 5 moles of oxygen when heated at constant volume from 10^(@)C" to "20^(@)C will be approximately: |
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Answer» 25 cal or `C_(v)=C_(p)-(R )/(J)` `=7.03 -(8.31)/(4.18)=5.04` CALS/mol `Q=5xxC_(v) xx Delta t` `=5xx5.04 xx10 ~~ 250` cals `therefore` Correct choice is (B). |
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| 40115. |
The radio waves of frequencies 300 MHz to 3000 MHz belong to: |
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Answer» SUPER HIGH FREQUENCY band |
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| 40116. |
Define bias voltage. |
| Answer» Solution :The INTERNAL repulsion of the depletion layer stops further diffusion of free ELECTRONS ACROSS the junction. This difference in potential across the depletion layer is called BARRIER potential. | |
| 40117. |
On reflection the _____as well as _____ remains unchanged. |
| Answer» SOLUTION :VELOCITY, FREQUENCY, WAVELENGTH | |
| 40118. |
Where did Akash work in the beginning ? |
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Answer» GOVERNMENT Hospital |
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| 40119. |
A person moving on a motor cycle in a ground takes a turn through 60^(@) on his left after every 50m. Then find the magnitude of displacement suffered by him after 9th turn |
| Answer» Answer :A | |
| 40120. |
A moving coil galvanometer can measure a current of 10^(-6)A. What is the resistance of the shunt to measure 1A? |
| Answer» SOLUTION :`(i_q)/(i) = (S)/(G + S) = N` | |
| 40121. |
A semi circular current loop is placed in an uniform magnetic field of 1 tesla as shown. If the radius of loop is lm, the magnetic force on the loop is |
| Answer» Answer :B | |
| 40122. |
The length of a compound microscope is 14cm and its magnifying power when final image is formed at near point is 25. If the focal length of eyepiece is 5cm. The distance of object from the objective and the focal length of objective lens are |
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Answer» `59/25 CM, 59/31 cm ` |
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| 40123. |
For same rating which have the maximum efficiency from following :- |
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Answer» Motor |
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| 40124. |
Does the photoelectric emission take place due to incidence of visible light and ultraviolet rays on the surfaces of different types of metal in our everyday experience? |
| Answer» Solution :Yes, the photoelectric emission does take place. But in absence of any positively charged COLLECTOR, the emitted photoelectrons accumulate on themetal surface and make a LAYER of negative charges. Within a very short SPAN of TIME, the emission of photoelectrons COMES to a stop due to the repulsion of the negatively charged layer. Hence the photoelectric emission stops after a while. | |
| 40125. |
What is the effect on the interference fringes in a Young's double-slit experiment due to each of the following operations : the screen is moved away from the plane of the slits, |
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Answer» Solution :(a) Angular separation of the fringes remains CONSTANT `(= lambda//d)`. The actual separation of the fringes INCREASES in proportion to the distance of the screen from the plane of the two slits. (b) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below. (c) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below. (d) Let s be the size of the source and S its distance from the plane of the two slits. For interference fringes to be seen, the condition `s//S lt lambda//d` should be satisfied, otherwise, interference patterns produced by DIFFERENT parts of the source overlap and no fringes are seen. Thus, as S decreases (i.e., the source slit is brought closer), the interference PATTERN gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed. (e) Same as in (d). As the source slit width increases, fringe pattern gets less and less sharp. When the source slit is so wide that the condition`s//S lt lambda//d` is not satisfied, the interference pattern disappears. (f ) The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is white. For a point P for which `S_(2)P-S_(1)P=lambda_(b)//2`,where `lambda_(b)(~~4000Å)` represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear RED in colour. Slightly farther away where `S_(2)Q-S_(1)Q=lambda_(b)=lambda_(r)//2" where "lambda_(r)(~~8000Å)` is the wavelength for the red colour, the fringe will be predominantly blue. Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen. |
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| 40126. |
Two masses M_(1) and M_(2) are attached to the ends of string which passes over the pulley attached to the top of a double inclined plane. The angles of inclination of the inclined planes are alpha and beta. Take g=10ms^(-2). If M_(1)=M_(2) and alpha=beta, what is the acceleration of the system? |
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Answer» ZERO |
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| 40127. |
The circuit is completed at t=0, then heat loss in R=4Omega resistance, till the capacitor gets fully charged. |
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Answer» `64 MU J` `q(t)=(3CE)/2(1-e^(1/(2R_(1)C)))` `I=(dq)/(dt)=(3E)/(4R_(1))e^(-1/(2R_(1)C))` HEAT loss in `R_(1)=intI^(2)R_(1)dt` `=576muJ` so, heat loss in `R=4Omega` is `=576xx2/3=384muJ` 
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| 40128. |
What is the magnitude of the electric field at the point (3hat(i)-2hat(j)+4hat(j))m if the electric potential is given by V = 2xyz^(2), where V is in volt and x, y, z are in meters? |
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Answer» |
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| 40129. |
Find the derivative of given functions w.r.t the corresponding independent varible. y = (cot x)/(1 + cot x) |
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Answer» |
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| 40130. |
A current carrying circular loop is freely suspended by a long thread. The plane of the loop will point in the direction |
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Answer» WHEREVER LEFT free |
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| 40131. |
An electrical device which offers a low resistance to the current in one direction but a high resistance to the current in opposite direction is |
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Answer» CURRENT AMPLIFIER |
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| 40132. |
Under which of the following conditions E value of the cell, for the cell reaction given is maximum? Zn(s) +Cu^(2+) (aq) hArrCu(s)+Zn^(2+) (aq) |
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Answer» `C_1=0.1M,C_2=0.01M` `E=E^@-(2.303RT)/(nF)logQ` `impliesE= E^@-(2.30 3RT)/(nF)log ((ZN^(2+))/(Cu^(2+)))` `E_(CELL)^@=E_c^@-E_A^@=0.34-(-0.76)V=1.1V` `E=1.1-0.059/nlog""C_2/C_1(Zn^(2+)=C_2 Cu^(2+)=C_1)` By analysing from the above equation, E VALUE of the cell will be maximum when , `log""C_2/C_1` WOULD come out to be minimum, when `log""C_2/C_1` values would be minimum then, `log""0.01/0.1=log10^-1` (minimum). Thus option (a) is CORRECT. |
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| 40133. |
Draw the logic circuit corresponding to the Boolean expression. Y=AB+barBC. |
Answer» Solution :It is an expression SHOWING sum of TWO products. Thus we require two AND gates and an OR gate. To GENERATE `barB` we also require a NOT gate. Their CIRCUIT can be drawn as follows. 
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| 40134. |
Why are NAND andNOR gates known as digital building blocks or universal gates ? |
| Answer» Solution :The repeated USE of NAND or NOR gates alone can give all other gates LIKE OR, AND, and NOT gates, HI digital circuits the NAND or NOR gates SERVE as building blocks and hence they are named so. | |
| 40135. |
A piece of copper and another of germanium are cooled from room temperature to 80K. The resistance of ………………. . |
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Answer» each of them increase |
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| 40136. |
A sphere of density rho and radius a has a concentric of radius b as shown in the (a) Sketch the gravitational force exerted by the spehre on the particle of mass m located at a distance r from the centre of the sphere as a function of r in the range 0 le r le oo (b) Sketch the corresponding curve for the potential energy u (r) of the system , |
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Answer» |
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| 40137. |
What is the nature of symmetry of the dipole field ? |
| Answer» Solution :The dipole field has a cylindrical SYMMETRY. The axis of CYLINDER passes through the dipole axis. | |
| 40138. |
Wavelength of an electron having energy 10 keV ……..Å |
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Answer» 0.12 `=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx10^(4)xx1.6xx10^(-19)))` `=0.12xx10^(-10)m` `therefore lambda=0.12 Å` |
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| 40139. |
Find the Q value of the reaction H^1+Li^7 to He^4 + He^4. Determine whether the reaction is exothermic or endothermic? The atomic mass of H^1, He^4 and Li^7 are 1.007825 u, 4.002603 u and 7.016004 u respectively. |
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Answer» Solution :The total mass of the initial PARTICLES `m_(1)= 1.00 78 25 + 7.016004 = 8.0233829 u and` the total mass of the final particles `m_(f) = 2XX 4 .002603 = 8.005206 u ` DIFFERENCE between initial and final mass of particles `Deltam=m_(i) -m_(f) = 8.023829 - 8.005206 = 0.018623` u This mass is converted into energy and the reaction is exothermic . The Q value is POSITIVE and given by `Q = (Deltam)c^2 = 0.018623 xx 931.5 = 17.35 MeV` |
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| 40140. |
What is missing in the following nuclear reaction""_(1)H^(2)+""_(1)H^(2) rarr""_(2)He^(3)+? |
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Answer» POSITRON |
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| 40141. |
During beta-decay of stationary nucleus , an electron is observed with akinetic energy 1.0 MeV . From this ,what can be concluded about the Q-value of the decay ? |
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Answer» Q=1.0 MEV |
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| 40142. |
A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side of the principal axis, the intensity of light |
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Answer» FIRST DECREASES and then increases |
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| 40143. |
In J.J. Thomson e/m experiment, a beam of electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if |
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Answer» B is increased by 208 times If m is increased by 208 times then B should be increased `SQRT(298)=14.4` times. |
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| 40144. |
Stationary waves of frequency 200 are formed in air. If velovity of the wave is 360 m/s the shortest distance between antinodes will be : |
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Answer» 1.8 m `lambda= (u)/(v) = (360)/(200) = 1.8 `m distance bet. TWO conseutive ANTINODES `lambda// 2 = (18)/(2) = 0.9` m. Hence correct choice is (d). |
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| 40145. |
A dipcircle is initally in the magnetic meridianif it is now ratated throiugh an angle thetain the horizontalplane then the tangent of the new angle of dip is increased in the ratio |
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Answer» `(TAN PHI)/(tan Phi)=(1)/(SIN theta)` `therefore tan Phi =(V)/(H)` then tan `Phi=(V)/(H)=(V)/(H cos theta)` Then `(tan Phi)/(tan Phi)=(1)/(cos theta)` |
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| 40146. |
यदि A={1, 2, 3} तथा B={3, 4, 1} हो तो AnnB होगा- |
| Answer» Answer :B | |
| 40147. |
Three pure inductance each of 3 H are connected as shown in figure. The equivalent inductance of this connection between points A and B is ...... |
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Answer» 1 H  `therefore` The equivalent inductance , `1/L=1/L_1+1/L_2+1/L_3` `therefore 1/L=1/3 +1/3 +1/3` `therefore 1/L=1 rArr L=1H` |
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| 40148. |
When a diamagnetic substance is brough near the north or the south pole of a bar magnet, it is |
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Answer» ATTRACTED by the poles |
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| 40149. |
What we name the conductors which do not obey ohm's law? |
| Answer» SOLUTION :NON OHMIC EX. DIODE | |
| 40150. |
The electric current in a circular coil of two turns produced a magnetic induction of 0.2T at its centre. The coil is unwound and rewound in to a coil of four turns. The magnetic induction at the centre of the coil now is, in tesla (if the same current flows in the coil) |
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Answer» 0.2 |
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