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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.
| 7001. |
A thin symmettrical double convex lens of refractive index mu_(2)-1.5 is placed between a medium of refractiv eindex mu_(1)=1.4 to the left and another medium of refractive index mu_(3)=1.6 to the right.Then the system behaves as |
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Answer» a convex LENS For parallel beam of light, `(-mu_(1))/(u)+(mu_(2))/(V)=(mu_(2)-1)/(R )` `rArr -(1.4)/(0oo)+(1.5)/(V)=(1.5-1.4)/(R )rArrV=15R` For second refraction surface, `-(mu_(2))/(1.5r)+(mu_(3))/(V)=(mu_(2))/®` `rArr -(1.5)/(15R)+(1.6)/(V)=(1.6-1.5)/(R)rArrV'=oo` Hence ,the combination behaves as glass plate |
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| 7002. |
The valence band and conduction band of a solid overlap at low temperature, the solid may be |
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Answer» a metal |
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| 7004. |
if three sources of sound of equal intensities and having frequencies 300 Hz , 301 Hz and 302 Hz are sounded together, the number of beats heard is |
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Answer» 1 |
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| 7005. |
Depict the equipotential surfaces due to an isolated point charge . |
Answer» SOLUTION :
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| 7006. |
Ratio of radius of gyration of a thin uniform circular disc about the axis passing through its centre and normal to its plane to radius of gyration of the same disc about its diameter is |
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Answer» 0.084027777777778 |
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| 7007. |
Theminimumnumberof gatesrequiredto realizethisexpressionZ =DABC+ DA bar(BC ) is |
| Answer» Answer :A | |
| 7008. |
A mixture of diatomic gases is obtained by mixing m_1 and m_2 masses of two gases, with velocity of sound in them c_1 and c_2respectively. Determine the velocity of sound in the mixture of gases. (Temperature of the gas remains constant) |
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Answer» `C= SQRT( (m_1c_2^2 - m_2c_1^2)/(m_2 + m_1))` |
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| 7009. |
A proton entersa magnetic field of 1.5 T with a velocity of 2 xx 10^(7) ms^(-1) at an angle of 30^@ with the field. The force on the proton will be |
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Answer» `2.4 xx 10^(-12) N` `:.` Force on PROTON `F = q v B sin theta = 1.6 xx 10^(-19) xx 2 xx 10^(7) xx 1.5 xx sin 30^@ = 2.4 xx 10^(-12) N`. |
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| 7010. |
A certain force F increase the length of a given wire by 1mm. What force is required to increase its length by 2mm? |
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Answer» a)3F |
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| 7011. |
Prove the laws of reflection by using Huygens' principle. Or, In Young's double slit experiment, what is the effect on the interference pattern if, (i)The distance between the two slits is halved. (ii) The distance between the screen and the plane of slits is doubled. (iii) one of the slits is covered with translucent paper. |
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Answer» Solution :Fringe width, `y = (D lambda)/(2d)` (i) Distance 2d between the slits is halved, then y is DOUBLED. (ii) Distance D between the screen and the PLANE of the slits is doubled, then again y is doubled. (iii) Due to increase in the optical path for RAYS passing through the covered slit, the fringe PATTERN will be displaced SIDEWAYS. |
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| 7012. |
When a nucleus in an atom undergoes a radioactive decay the electronic energy levels of the atom |
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Answer» do not change for any type of radioactivity |
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| 7013. |
Mirage is formed due to ______ . |
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Answer» reflection of light |
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| 7014. |
A charge of 2C is placed on the x-axis at 1 m from the origin along -ve X-axis. Infinite number of charges each of magnitude 2C are placed on X-axis at lm, 2m, 4m, ..... from origin along +ve X-axis. The first charge is positive and alternate charges are of opposite in nature. The electric field intensity at the origin |
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Answer» `(1)/( 1- PI in_0)` ALONG+ vex-axis |
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| 7015. |
Define critical angle with reference to total internal reflection. Calculate the critical angle for glass-air surface, if a ray of light which is incident in air on the glass surface is deviated through 15°, when angle of incidence is 45°. |
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Answer» Solution :The critical angle for a pair of media may be defined as the angle of INCIDENCE in the denser medium for which the angle of REFRACTION in the rarer medium is 90° and beyond which the LIGHT is totally internally reflected back in the denser medium itself. In present problem angle of incidence in air i = 45°. As the ray deviates through 15° on entering into glass and glass is optically denser, HENCE, `r = 45^(@) 15^(@) = 30^(@)` `therefore` Refractive index of glass `n_(ga) = (sin i)/(sin r) =(sin 45^(@))/(sin 30^(@)) = (1//sqrt(2))/(1//2) = sqrt(2)` `therefore`Critical angle for glass-air interface is given by `sin i_( c) = 1/(n_(ga)) = 1/sqrt(2) rArr i_( c) = sin^(-1) (1/sqrt(2)) = 45^(@)` |
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| 7016. |
Draw a circuit diagram to show biasing of a p-n-p transistor. Explain the transistor action. |
Answer» Solution :Action of p-n-p transistor. In a p-n-p transistor, when the emitter is forward biased, the holes in the emitter and the electrons in the base begin to move towards the junction, holes being attracted by the negative terminal and the electrons by the positive terminal of the battery. On reaching the base-emitter junction a small fraction of total NUMBER of holes with electrons to get neutralized. As the base layer is very thin the collector is kept at high negative POTENTIAL , almost all the holes are attracted by the collector, producing a hole-current between the emitter and collector.  The emitter current `I_(e)` is the SUM of base current and the collector current. i.e. `I_(e)=I_(b)+I_(C)` So `I_(c) lt I_(e)` Working of p-n-p In working of p-n-p, as each hole reaches the collector, an electron is emitted from the negative terminal of battery and the hole is neutralized. Simultaneously, a covalent bond is broken near the emitter, the electron so produced enter the positive terminal of the forward bias battery and the hole starts its JOURNEY towards the emitter base junction. Holes are the current carriers in p-n-p transistor, however the current in the outer circuit is always due to flow of electrons. |
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| 7017. |
A motor car weighing 1000 kg moves up an incline rising 1 in 49 at the rate of 54 km/hr. Find the power at which the engine is working . |
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Answer» 5 KW |
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| 7018. |
In a certain regionof space electric fieldis along the z direction throughuot the magnitude of electricis however not constant10^(-5) NC^(-1)per meter what are the forcetorque experienced by a system having a total dipole equal to 10^(-7)cm in the negative z direction |
| Answer» SOLUTION :Only (C ) is right the rest cannot REPRESENT electrostaicfield LINES (a) is wrong becausefield lines must be normal to a conlductor (b) is wrong because field lines cannot startwrong becauseelectrostaticfield lines cannotform closed loops | |
| 7019. |
A plane mirror 'M' and a concave mirror 'X' arc kept at a seperation of 40 cm with their reflecting faces facing each other as shown in figure. An object AB is kept perpendicular to the principal axis in position (1). Considering successive reflections first at mirror 'X' and then at 'M', a real image is formed infront of 'M at a normal distance 8 cm form it. If the object is moved to new position (2), the real image is formed at 20 cm from 'M' with the reflections as described earlier.Distance of position (2) of the object from the pole of concave mirror is |
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Answer» 11cm |
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| 7020. |
The ammonia molecule has a permanent electric dipole moment of 1.47 D where D is the debye unit and 1D=3.34 xx 10^(-30)Cm. Calculate the electric potential due to the ammonia molecule at a point distant 30 nm away along the dipole axis. |
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Answer» Solution :Data supplied, Dipole moment (p) `=1.47 xx 3.34 xx 10^(-30) C.m, r=30 xx 10^(-9) m =3 xx 10^(-8)m` Being ALONG the axis `theta=0, therefore cos theta=1` Potential `V=(1)/(4epsi_(0)). (p cos theta)/(r^(2))=(9 xx 10^(9) xx 1.47 xx 3.34 xx 10^(-30) xx 1)/(9 xx 10^(-16))=4.91 xx 10^(-5)V =49.1muV` |
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| 7021. |
Repeat above question if A is connected with D and B is connected with C. |
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Answer» Solution :Let potential of B and C is zero and common potential on capacitors is V, then at A and D it will be V 2V+3V=0 `rArr V="2 VOLT"` Now CHARGE on each plate is shown in the FIGURE. HEAT produced = `=400+150-1/2 xx 5 xx 4` `=550-10=540 muJ` |
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| 7022. |
A plane mirror 'M' and a concave mirror 'X' arc kept at a seperation of 40 cm with their reflecting faces facing each other as shown in figure. An object AB is kept perpendicular to the principal axis in position (1). Considering successive reflections first at mirror 'X' and then at 'M', a real image is formed infront of 'M at a normal distance 8 cm form it. If the object is moved to new position (2), the real image is formed at 20 cm from 'M' with the reflections as described earlier.Keeping the object in position (1), if the plane mirror is replaced with a convex mirror 'Y' of focal length 20 cm facing the mirror X after successive reflections first at 'X' and then at 'Y', the final image will be |
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Answer» at distance of `80/3`cm from the pole of mirror X and in front of its REFLECTING SURFACE. |
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| 7023. |
points A and B are situated along the extended axis of a 2cm long bar magnet at a distance x and 2x cm respectivly from the pole mearer to the points. The ratio of the magnetic fields at A and B will : |
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Answer» 4 : 1 exactly |
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| 7024. |
Monochromatic radiation is incident on hydrogen (H) sample which is in ground state. If the hydrogen atoms emit radiation of ten different wavelengths after absorbing the incident radiation, then the wavlength of the incident radiation is (Let hc=1242eV-nm) |
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Answer» 84.4 NM Since , monochromatic radiation is incident on hydrogen SAMPLE which is in ground state. Hence , the energy of ground state is given by ` |E| = 136 eV ` ` (hc)/(lambda) = 136 "" (becauseE = (hc)/(lambda)) ` ` lambda = (hc)/(136) = (1242)/(136) nm ` = 91.3 nm= 95.1 nm Hence , the wavelength of incient radiation is 95.1 nm . |
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| 7025. |
For ideal diamagnetic substance, magnetic susceptibility is …..... . |
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Answer» Solution :For diamagnetic substance `-1 le chi_(m) LT 0` But for ideal diamagnetic substance, `chi_(m)=-1` (Which is SEMICONDUCTOR) |
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| 7026. |
What is the direction of induced electric field in a dielectric medium ? |
| Answer» Solution :The INDUCED ELECTRIC field is OPPOSITE in direction to that of the APPLIED electric field. | |
| 7027. |
+ q, + q, - q and - q are kept at vertices of square ABCD respectively. Find the force acting on Q at centre of square. Side of square is a. |
| Answer» SOLUTION :`(4sqrt(2)kQq)/a^(2) to `PARALLEL to AD or BC | |
| 7028. |
A proton and an electron are placed 1.6 cm apart in free space. Find the magnitude of electrostatic force between them. The nature of this force. |
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Answer» `9 xx 10^(-25)` repulsion |
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| 7029. |
A plane mirror 'M' and a concave mirror 'X' arc kept at a seperation of 40 cm with their reflecting faces facing each other as shown in figure. An object AB is kept perpendicular to the principal axis in position (1). Considering successive reflections first at mirror 'X' and then at 'M', a real image is formed infront of 'M at a normal distance 8 cm form it. If the object is moved to new position (2), the real image is formed at 20 cm from 'M' with the reflections as described earlier.Focal length of mirror 'X' is |
| Answer» Answer :B | |
| 7030. |
When a compact disc is illuminated by a source of white light, coloured lanes are observed. This is due to |
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Answer» dispersion |
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| 7031. |
The figure shows the initial position of a point source of light s, a detector D and lens L. Now at t = 0 all three starts moving toward right with different velocity as shown in figure. The times at which detector receives the maximum light. |
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Answer» `0.56s` and `4.45s` |
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| 7032. |
The electric field is a region (relative permittivity = 1) varies with distance "r" from origin according to relation E = Ar^2hatr, where A is a positive constant. The volume charge density in the region is proportional to |
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Answer» R |
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| 7033. |
(a)If alpha-decays of ._(92)U^(238) is energetically allowed (i.e., the decay products have a total mass less than tha mass of ._(92)U^(238)), what prevents ._(92)U^(238) form decaying all at once? Why is its half life so large? (b) The alpha-particle faces a Coulomb barrier. A neutron being unchanged faces no such barrier. Why does the nucleus ._(92)U^(238) not decay spontaneously, by emitting a neutron? |
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Answer» Solution :(a) As explained in theory, `alpha` -decay is caused by the quantum mechanical tunnelling of an alpha particle through a repulsive Coulomb barrier. The rate of tunneling would depend upon the height and width of the barrier. The decay cannot be all at once. And that is the reason why half life of `._(92)U^(238)` against `alpha` -decay is large, (b) The possible nuclear reaction is `._(92)U^(238)to._(92)U^(237)+._(0)n^(1)` The DATA shows that `m(._(92)U^(237))+m(n)` is GREATER than `m(._(92)U^(238))`. Therefore, the decay is not allowed energetically. Rather, some EXTERNAL energy has to be supplied to separated a neutron form `._(92)U^(238)`. |
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| 7034. |
The spheres of radius r and R having charge q and Q respectively on them . When they are joined with conducting ,wire, the energy of this system does not dissipated then ........... |
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Answer» QR= QR |
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| 7035. |
A: Particle theory fails to explain the velocity of light in air and water. R: According to particle theory, light in a dense medium is moving faster than in rare medium. |
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Answer» Both ASSERTION and reason are true and the reason is CORRECT EXPLANATION of the assertion. |
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| 7036. |
A rod of length 1 is rotated about its one end, perpendicular to a magnetic field of induction B. What is the e.m.f. induced in the rod? |
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Answer» SOLUTION :Induced e.m.f.(e) = `(dphi)/dt` = `B.(DA)/dt` But dA = `pil^2` and dt = T and T = `2pi/w` e = `B.pil^2//(2pi/omega) = `1/2Bl^2omega` |
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| 7037. |
भारत किस गोलार्ध में स्थित है? |
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Answer» उत्तरी गोलार्ध में |
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| 7038. |
A crown glass prism of refracting angle 8° is combined with a flint glass prism to obtain deviation without dispersion. If the refractive indices for red and violet rays for crown glass are 1.514 and 1.524 and for the flint glass are 1.645 and 1.665 respectively, find the angles flint glass prism and net deviation. |
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Answer» Solution :The condition for deviation with out dispersion is ` ( mu_A - mu_R) A= (mu_V - MU _R )A` ` thereforeA.= ( ( 1.524-1.514)xx 8^0)/( (1.665- 1.645 )) =( 0.08^0)/(0.02 )= 4^0` Mean REFRACTIVE INDEX for crown glass `mu= ( 1.514 + 1.524)/( 2) = 1.519` Mean refractive index for flint glass `mu+ (1.645 + 1.665 ) /( 2)= 1.655` ` therefore ` The netdeviation`( DELTA - delta .) = ( mu-1) A - ( mu.-1) A.` `=0.159 xx 8^0 - 0.655xx 4^0` ` = 1.53^@` |
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| 7039. |
भारत का क्षेत्रफल कितना वर्ग किलोमीटर है? |
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Answer» 32.8 हजार वर्ग कि मी. |
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| 7040. |
Two nuclides have mass number 64 and 125 respectively. Ratio of their radii is ............... . |
| Answer» SOLUTION :`4:5 [R_(1)/R_(2)=[A_(1)/A_(2)]^(1/3)=[64/125]^(1/3)=4/5]` | |
| 7041. |
Which sample A or B, shown in Fig. 13.03, has shorter mean life ? |
| Answer» Solution :SAMPLE B has shorter MEAN life because its decay rate is FASTER than that of sample A. | |
| 7042. |
What should we keep the ponds covered with in the breeding season? |
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Answer» Sand |
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| 7043. |
Find an expression for the equivalent capacitance of a combination of three capacitors in series . |
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Answer» Solution :Consider a SERIES combination of three capacitors of CAPACITANCES `C_(1) , C_(2)` and `C_(3)` , respectively . Obviously charge on each plate of each CAPACITOR has same magnitude and charges on two plates of a capacitor has opposite signs . If potential differences across the three capacitors be `V_1 , V_2` and `V_3` respectively then from the relation . Q = V C , we have `V_(1) = (Q)/(C_(1)) , V_(2) = (Q)/(C_(2)) ` and `V_(3) = (Q)/(C_(3))` The total potential difference across the whole circuit `V = V_(1) + V_(2) + V_(3) = (Q)/(C_(1)) + (Q)/(C_(2)) + (Q)/(C_(3))` If C be the resultant capacitance in series arrangement then `Q = C V ` or `V = (Q)/(C)` `THEREFORE(Q)/(C) = (Q)/(C_(1)) + (Q)/(C_(2)) + (Q)/(C_(3)) implies (1)/(C) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3))` 
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| 7044. |
Seed dormancy allows the plants to |
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Answer» OVERCOME UNFAVORABLE CLIMACTIC conditions |
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| 7045. |
A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r ) produced by the shell in the range 0 le r le oo, where r is the distance from the centre of the shell? |
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Answer»
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| 7046. |
A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected? |
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Answer» moment of inertia For sphere `(2)/(5)MR^(2)omega` = constant or `r^(2)omega` = constant `THEREFORE` when r decreases `omega` increases `therefore` L = constant |
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| 7047. |
An electron falling freely under gravity enters a region of uniform horizontal magnetic field pointing north to south. The particle will be deflected towards |
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Answer» east |
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| 7048. |
A system of two charges separated by a certain distance apart stores electrical potential energy. If the distance between them is increased, the potential energy of the system, |
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Answer» MAY increase or decrease |
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| 7049. |
A bob of massm attached to an inextensible string of length l is suspended from a vertical support. The bob length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed omega rad/s about the vertical. About the point of suspension : |
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Answer» ANGULAR momentum changes in direction but not in MAGNITUDE 
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| 7050. |
Show how a current loop acts as a magnetic dipole . Arrive at an expression for the magnetic dipole moment. |
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Answer» Solution :Consider a plane loop of WIRE carrying a CURRENT .I. as shown in figure . Looking at the FACE, the current is clockwise and so it has a south POLARITY . Thus the current loop behaves as a magnetic dipole . If A is the AREA of the plane of the loop, then dipole moment M = IA For n such turns , M = nIA S.I unit of M is `A.m^2` 
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