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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In an electric field an electron is kept freely. If the electron is replaced by a proton, what will be the relationshipbetween the forces experienced by them? |
| Answer» SOLUTION :MAGNITUDE of FORCE will be same but DIRECTION will be REVERSED. | |
| 2. |
An elastic ball is dropped on a long inclined plane. If bounces, hits the plane again, bounces and so on. Let us Label the distance between the point of the first and second hit d_(12) and the distance between the points of second and the third hit is d_(23). find the ratio of d_(12)//d_(23). |
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Answer» `(2)/(1)` |
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| 3. |
A thin equiconvex lens of glass ( mu = 1.5) with radius of curvature 4m is placed on a horizontal plane mirror. When the space between the lens and mirror is filled with a liquid, an object held at a distance 6m vertically above the lens is found to coincide with its own image. What is the refractive index of the liquid. |
| Answer» SOLUTION :`MU =4//3` | |
| 4. |
A body of mass 2 kg is moving according to the equation for displacement at seconds as x(t) = pt^(2) + rt^(3). If p = 3 ms^(-1) , q = 4 ms^(-1) and r = 5 ms^(-1) the force acting after 2 sec is : |
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Answer» `136N` `(dx)/(dt)=p+2qt+3et^(2)` `a=(d^(2)x)/(dt^(2))=0+2q+6rt` ACCELERATION at `t = 2s = 0 + 2 xx 4 + 6 xx 5 xx 2` `= 68 ms^(-2)` `"Force" = ma = (2 xx 68) N` `= 136 N` HENCE .a. is the right choice |
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| 5. |
PhMgBr+(X)underset((2)H^(oplus))(to)Ph-overset(OH)overset(|)underset(Ph)underset(|)(C)-Ph,(x) is - |
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Answer» `PH-OVERSET(O)overset(||)(C)-H` |
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| 6. |
If the gravitional force between two objects were proportional to 1/R (and not as 1R^(2)) where R is separation between them , then a particle in circular orbit under such a force would have its orbital speed v proportional to |
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Answer» `R^(1)` `therefore (GMm)/(R )=(mv^(2))/(R )` `rArr v=sqrt(GM)`. i.e. v is independent of R. THUS correct CHOICE is (c ). |
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| 7. |
In x-ray spectra, the K_gamma lines in k series corresponds to |
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Answer» The VACANCY CREATED in K-shell being FILLED by an ELECTRON from L-shell |
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| 8. |
A body at higher temperature T K radiates heat at a rate which is proportional to |
| Answer» ANSWER :D | |
| 9. |
n equal resistors are first connected in series and then in parallel. The ratio of the equivalent resistance in two cases is |
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Answer» N |
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| 10. |
A ball is thrown with velocity 8 ms^(-1)making an angle 60° with the horizontal. Its velocity will be perpendicular to the direction of initial velocity of projection after a time of (g =10ms^(-2)) |
| Answer» SOLUTION :`(1.6)/(SQRT(3))s` | |
| 11. |
The kinetic energy of an electron, which is accelerated in the potential difference of 100 volts, is |
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Answer» 416.6 cal `= 1.602 xx 10^(-17) J` |
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| 12. |
A resistor of 100 ohm is connected in series with an inductor of 10H and a capacitor of 0.1 mu F All these elements are connected to a 220 volt, 50 Hz a.c. supply. Calculate the total impedance of the circuit. |
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Answer» |
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| 13. |
In an inductor, current rises to a steady value at a constant rate. Comment. |
| Answer» SOLUTION :No, CURRENT RISES only EXPONENTIALLY. | |
| 14. |
Locate the centre of mass of a system of two spherical masses of 5 kg and 10 kg kept with their centres 1 m apart. |
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Answer» 3/2 |
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| 15. |
Two identical samples of a gas expands so that volume is doubled. The first sample undergoes isothermal expansion while the second is expanded adiabatically. The final pressure: |
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Answer» in first sample is greater Thus, correct CHOICE is (a). |
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| 16. |
Thermal resistance of two metallic plates of same material and cross -section joined in series are 3 and 4 respectively. What is thermal resistance of the combination ? |
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Answer» `3//4` THUS correct CHOICE is (C). |
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| 17. |
What do the graphs represent? |
| Answer» SOLUTION :INTENSITY of scatteredelectrons VERSUS INCIDENT ANGLE. | |
| 19. |
In an LCR circuit if resistance increases, quality factor |
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Answer» INCREASES |
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| 20. |
Two coherent sources are 0.15 mm apart and fringes are observed at 1 m away with monochromatic light of wavelength 6000Å.Find (a) the fringe width in air, (b) the fringe width in a liquid of refractive index 5/2. |
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Answer» Solution :(a) Fringe width `beta = (lamda D)/(d) = (6000 XX 10^(-10) xx 1)/(0.15 xx 10^(-3)) , i.e., beta = 4mm` (B) Fringe width in a medium `beta = beta/mu = (4)/(5//2) = 1.6mm` |
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| 21. |
Using Kirchhoffs rules in the given circuit, determine (i) the voltage drop across the unknown resistor R and (ii) current I in the arm EF. |
Answer» Solution :Taking upper PORTION as mesh using Kirchhoffs second law portion  `1 xx 2 -I xx 3 = 3 -5` or ` 2 - 3 I =-2` or ` 3 I =4` or `I = 4/3 A` So total current through R, ` I . =1 + 4/3 + 7/5 A` `therefore` VOLTAGE drop across R, `5-3 xx 4/3 =1` Volt. |
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| 22. |
Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify ? |
| Answer» Solution :Copper (a DIAMAGNETIC material) has negative SUSCEPTIBILITY but ALUMINIUM (a paramagnetic material) has positive susceptibility. Negative susceptibility signifies that the material is repelled in a magnetic FIELD and magnetic field LINES are expelled from the material. | |
| 23. |
Light of frequency 7.21xx10^(14)Hz is incident on a metal surface. Electrons with a maximum speed of 6.0xx10^(5)m//s are ejected from the surface. What is the threshold frequency for photoemission of electrons ? |
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Answer» Solution :Here `v=7.21xx10^(14)Hz and v_(max)=6.0xx10^(5)m//s` Using the relation`H(v-v_(0))=(1)/(2)mv_(max)^(2),` we have `v-v_(0)=(mv_(max)^(2))/(2h)=(9.11xx10^(-31)xx(6.0xx10^(5))^(2))/(2xx6.63xx10^(-34))=2.48xx10^(14)` `v_(0)=v-2.48xx10^(14)=7.21xx10^(14)-2.48xx10^(14)=4.73xx10^(14)Hz`. |
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| 24. |
Define ionisation potential. |
| Answer» Solution :The minimum accelerated POTENTIAL which would PROVIDE an electron SUFFICIENT energy to ESCAPE from the OUTERMOST orbit. | |
| 25. |
While doing an experiment with potentiometer Fig. it was found that the deflection is one sided and (i)the deflection decreased while moving from one end A of the wire to the end B, (ii) the deflection increased, while the jockey was moved towards the end B. (a) Which terminal + or -ve of the cell E_(1), is connected at X ub case (i) and how is E_(1) is related to E ? (b) Which terminal of the cell E_(1) is connected at X in case (ii) ? |
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Answer» Solution :In the given potentiometer experiment, the deflection in the galvanometer will be one sided if the emf of cell `E_(1)` is greater than the emf of battery `E` because than emf `E_(1)` is greater than fall of POTENTIAL across the potentiometer wire due to battery `E`. (a) The deflection in galvanometr will decrease while moving from one end `A` of the poteniomter wire to the end `B`, if POSITIVE of the cell `E_(1)` is is connected to `X` and negative of the cell to `Y`, because in this situation, the current in galvanometer decrease with the INCREASE in length of potentiometer wire. (b) The deflection in galvanometer will increase while moving from end `A` of the poteniomter wire to the end `B` if - ve of the cell `E_(1)` is connected to `X` and `+` ve to `Y`. In this situation, the current in galvanometer increases with the increase in the length of potentiometer wire. 
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| 26. |
Which of the following gates corresponds to the truth table given below ? |
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Answer» NAND |
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| 28. |
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16A, coil Y has 25 turns and carries a current of 18 A. The sense of the current in X in antoclockwise, and clockwise in Y, for an observer looking at hte coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre. |
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Answer» Solution :For coil X : `R_1= 16 cm = 0.16 m , N_1 = 20, I_1 = 16A` `:.` Magnetic field at the centre of coil X `B_1 = (mu_0 N_1 I_1)/(2 R_1) = (4 pi xx 10^(-7) xx 20 xx 16)/(2 xx 0.16) = 4pi xx 10^(-4) T` As per right HAND palm rule, `B_1` is DIRECTED horizontally towards the east. For coil `Y: R_2 = 10 cm = 0.10 m , N_2 = 25, l_2 = 18 A` `:. B_2 = (mu_0 N_2 I_2)/(2 R_2) = (4 pi xx 10^(-7) xx 25 xx 18)/(2 xx 0.10) = 9 pi xx 10^(-4) T` As per right hand plam rule, `B_2` is directed horizontally towards WEST. `:.` Net magnetic field B at the centre point in westward direction and is given as : `B = B_2 - B_1 = (9 pi xx 10^(-4) - 4 pi xx 10^(-4)) T = 5 pi xx 10^(-4) T ~= 1.6 xx 10^(-3) T` towards west. 
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| 29. |
The mechanical stress of a charged conductor is directly proportional to |
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Answer» SQUARE of SURFACE CHARGE density |
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| 30. |
What are the different type of continous charge distribution ? |
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Answer» Solution :LINE CHARGE density :`LAMBDA=("charge")/("length")` AREA charge density : `sigma=("charge")/("area")` volume charge density : `rho=("charge")/("volume")` |
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| 31. |
On the basis of Eq,(6.4g) find the number of free electrons in a metal at T=0 as a function of de Broglie wavelengths. |
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Answer» Solution :Using the formula of the previous section `dn(v)=(m^(3))/(PI^(2)ħ^(3))v^(2)dv` We put `MV=(2piħ)/(LAMBDA)`, where `lambda=` Broglie WAVELENGTH `mdv(2piħ)/(lambda^(2))d lambda` Taking account of the fact that `lambda` DECREASES when `v` increase we write `db(lambda)= -dn(v)=((2pi)^(3)dlambda)/(pi^(2)lambda^(4)).=(8pi)/(lambda^(4))d lambda` |
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| 32. |
A body of volume V and dinsity p is raised through height h in a liqiud of density sigmaltp. The change in potential energy of the body is : |
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Answer» V PGH |
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| 33. |
The blades of an aeroplane propeller are rotating at the rate of 600 revolutions per minute. Its angular velocity is: |
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Answer» 10 `PI` rad/s |
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| 34. |
Two satellites are in the parking orbits around the earth. The mass of one is 10 times that of the other The ratio of their speeds of revolution is, |
| Answer» Answer :C | |
| 35. |
A radioactive element forms its own isotope after 3 consecutive disintegrations.The particles emitted are |
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Answer» 3 `BETA`- particles An isotope should have same Z. `alpha`- particle = `._(2)He^(4)`, `beta`- particle = `._(-1)beta^(0)` The EMISSION of one `alpha` particle and the emission of TWO `beta` particles maintain the Z same. Hence for isotope formation `2beta` particles and one `alpha` particle are emitted. |
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| 36. |
A shell of mass 0.01 kg fired by a gun of mass 10 kg. If the muzzle speed of the shell is 50 ms^(-1), what is the recoil speed of the gun? |
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Answer» 0.01 m/s |
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| 37. |
In the figure, the ball B is released from rest when the spring is in natural length. For the block A of mass 2m to leave contact with the ground at some instant the minimum mass of B must be |
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Answer» 2m `2mV = sqrt(m^(2)v^(2)+m^(2)v^(2))` `2mV=sqrt(2m^(2)v^(2))` `V=(v)/(sqrt(2))`] Energy released in explosion`=2XX(1)/(2)mv^(2)+(1)/(2)xx2mxx(v^(2))/(2)=(3)/(2)mv^(2)` |
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| 38. |
Two inputs of NAND gate are shortened. This gate is equivalent to |
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Answer» OR |
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| 39. |
Suppose an electron is attracted towards the origin by a force (k)/(r ) where k is a constant and r is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the n^(th) orbital of the electron is found to be r_(n) and the kinetic energy of the electron to be T_(n). Then which of the following is true? |
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Answer» `T_(n)prop(1)/(n),r_(n)propn^(2)` `:.(mv_(n)^(2))/(r_(n))=(K)/(r_(n))` `:.v_(n)^(2)=(K)/(m)` `T_(n)=(1)/(2)m((K)/(m))` `:.T_(n)=(K)/(2)` Now KINETIC energy `=(1)/(2)mv_(n)^(2)` Here, there is no n TERM, so the kinetic energy is independent then n. From Bohr.s hypothesis, the angular momentum `mv_(n)r_(n)=(nh)/(2pi)` `:.r_(n)=(nh)/(2pimv_(n))` `:.r_(n)=(nh)/(2pimxsqrt((k)/(m))):.r_(n)=(nh)/(2pisqrt(mk))` `:.r_(n)propn` |
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| 40. |
Let S_(1) and S_(2) be the two slits in Young's double-slit experiment. If central maxima is observed at P and angle angle S_(1)PS_(2) = theta, then the fringe width for the light of wavelength lambda will be (assume theta to be a small angle) |
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Answer» `LAMBDA//THETA` |
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| 41. |
A uniform dielectric ball is placed in a unifrom electricfileldof strength E_(0). Underthese conditionsteh dielectricbecomespolarizeduniformly. Findthe electricfieldstrength E inside teh ball and thepolarization P of the dielectricwhose permittively equals epsilon. Make use of the reslultobtainedin Problem 3.96 |
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Answer» SOLUTION :By superposition the field `vec(E)` INSIDE the ball is given by `vec(E) = vec(E_(0)) - (vec(P))/(3epsilon_(0))` On the other HAND, if the SPACE is not too small, the macrospic equation `vec(P) = (epsilon - 1) epsilon_(0)vec(E)` 0r, `vec(E) = (3 vec(E_(0)))/(epsilon + 2)` Also, `vec(P) = 3epsilon_(0) (epsilon - 1)/(epsilon + 2) vec(E_(0))` |
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| 42. |
(A) : Voltmeter put across a part of the circuit, it reads slightly less than the original voltage. (R) : Voltmeter is always connected in parallel in circuit across which voltage is to be measured. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 43. |
In a hydrogen atom, the centripetal force is furnished by the coloumb attraction between electron and proton. If a_(0) is the radius of first state orbit and permittivity of vacuum is in_(0),then the speed of the electron is.... [m = mass of electron, e = charge on electron] |
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Answer» 0 `(mv^(2))/(a_(0))=(e^(2))/(4piin_(0)a_(0)^(2))` `:.v^(2)=(e^(2))/(4piin_(0)ma_(0))` `:.v=(e)/(sqrt(4piin_(0)ma_(0)))` |
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| 44. |
A blockof icestarts slidingdown from the topof an inclinedroofof a alonga lineof the greatestslope. Theinclination of the roof withthe horizontal is 30^(@). Theheights of thehighestand lowerst points of the roof are 8.1 m and 5.6 mrespectively.Atwhathorizontal distancefrom the lowest point will the blockhitthe ground ? Neglect air friction .[g=9.8m//s^(2)]. |
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Answer» Solution :Accelerationof the block ALONGTHE greatestslope is equal to `a = g sin30^(@)` . Distancetravelled by the block along the greatestslope is equal to `S = ((8.1 - 5.6))/(sin 30^(@)) = 5m` Ifu be the speed of the blockwhen it justaboutto leavethe roofthen `u^(2) = 0+ 2 g sin 30^(@)30^(@) xx5rArru = 7 m//s` If thebe the timetaken to hitthe groundthen `5.6 = u sin 30^(@) t + (1)/(2) gt^(2)= (7)/(2)t + (1)/(2) XX 9.8 t^(2)` `rArr 7t^(2) + 5t - 8 = 0` ` t = (-5+pm sqrt(25-(4)(7)xx(8)))/(2xx7)` `rArr t = (-5pm 15.78)/(14) ` , -vevalueis to be rejected , `i.e., t = (-5+15.78)/(14) = 0.77 sec` Horizontal distancetravelled is equal to . `x = u cos 30^(@) t = (7 sqrt(3))/(2) xx (10.78)/(14) m RARRX = 4.67 m ` 
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| 45. |
If the percentage of modulation is 50% then the useful power in the AM wave is |
| Answer» ANSWER :D | |
| 46. |
A radioactive element decays by beta- emission. A detector records " beta particles in 2 second it records (0.75 n beta particles (Given In2 = 0.6931, In3 = 1.0986). The mean life correct to nearest whole number is........... Second |
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Answer» 3 |
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| 47. |
What is the angle between electric di-pole moment and intensity of electric field due to the dipole at its equatorial point? |
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Answer» `0^@` |
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| 48. |
A parallel stream of electrons accelerated by a potential difference V= 25 V falls normally on a diaphragm with two narrow slits separated by a distance d=50 mu m. Calculate the distance betweeb neighbouring maxima of the diffraction patterns on a screen located at a distance =100 cm from the slits. |
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Answer» Solution :From the Young slit formula `DELTAX=(l LAMBDA)/(d)=(l)/(d).(2piħ)/(sqrt(2meV))` SUBSTITUTION GIVES `Deltax= 4.90 mu m`. |
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| 49. |
Two point charges are separated by some distance , repel each other with a force F. What will be the force if distance between them is halved ? |
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Answer» SOLUTION :`F=1/(4piepsilon_0)(q_1q_2)/r^2` When `r=r/2` `F.=1/(4piepsilon_0) (q_1q_2)/((r/2)^2)=41/(4piepsilon_0) (q_1q_2)/r^2=4F` NEW FORCE F.=4F |
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| 50. |
Assertion: Insulating materials may not provide insulation when subjected to a very high electric field condition. Reason: When electric field is applied on a dielectric or insulating material then it aligns polar molecules of dielectric in the direction of electric field and at the same time positive and negative ends of the molecule experience force along opposite directions and molecule is stretched. Abnormally high electric field can stretch the molecule beyond permissible limits and negative ends or electrons are separated from molecules and material start conducting electricity. This phenomenon is called breakdown. |
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Answer» If both ASSERTION and REASON are CORRECT and reason is a correct EXPLANATION of the assertion . |
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