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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. |
A parallel plate capacitor of plate area A and plate seperation is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then |
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Answer» `Q=(epsilon_0AV)/d` |
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| 2. |
An iron ball of mass 0.2 kg is heated to 100^@C when it is put in the ice block at 0^@C 25 gm of ice meled. The specific heat of iron is |
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Answer» 3.1 |
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| 3. |
A steel ruler is calibrated to read true at 20.0^(@)c. A draftsman uses the ruler at 40.0^(@)C to draw a line on a 40.0^(@)C copper plate. As indicated on the warm ruler, the length of the line is 0.50m. To what temperature should the plate be cooled , such that the length of the line truly becomes 0.50 m? |
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Answer» |
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| 4. |
What is the condition for total internal reflection to take place ? |
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Answer» Solution :The LIGHT RAY must TRAVEL from an optically denser medium to rarer medium. The ANGLE of INCIDENCE in the denser medium should be greater than the critical angle. |
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| 5. |
Three batteries are connected as shown in figure. Reading of ideal ammeters A_(1), A_(2) & A_(3) are : |
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Answer» `(1)/(3)A, 0,(1)/(3)A` 
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| 6. |
Explain with the help of a diagram, how a depletion layer and barrier potential are formed in a junction diode. |
Answer» Solution : Due to the diffusion of ELECTRONS and the holes, from their MAJORITY zone to minority zone, a layer of positive and negative SPACE charge region on either side on the junction is formed. This is called the depletion region. The loss of electrons, from n-region and gain of electrons by the p-region, causes a difference of potential across the junction. This tends to prevent the movement of charge CARRIER across the junction and is, therefore, TERMED as barrier potential. |
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| 7. |
A convexo-convexhas a focal length of f_1=10 cm. One of the lens surfaces having a radius of curvature of R=10 cm is coated with silver. Construct the image of the object produced by the given optical system and determine the position of the image if the object is at a distance of a=15 cm from the lens. Refractive index of lens =1.5. |
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Answer» `1/10=(1.5-1)(1/(R_1)-1/(-10))` `:. R_1=+10 cm` Now, using, `1/v+1/u=(2(n_2//n_1))/(R_2)-(2(n_2//n_1))/(R_1)` SUBSTITUTING the VALUES, `1/v-1/(-15)=(2(1.5))/(-10)-(2(1.5-1))/(+10)` `:. v=-2.14 cm` 
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| 8. |
A wheel of moment of inertia 5xx10^(-3)kgm^(2) is making 20 rps. It is stopped in 20 s. The angular retardation is : |
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Answer» `pi` rad `s^(-2)` `ALPHA=(omega-omega_(0))/(t)=(0-40pi)/(20)=-2pi" rad/"s^(2)` `therefore` angular retardation= `2pi" rad/"s^(2)` |
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| 9. |
The equivalent resistance between the points A and B in Athe circuit shown in Fig. is |
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Answer» `3.12 OMEGA` ` R_125 = (10 XX 5)/(10 + 5) = 3.33 , R_1253 = 3.33 + 5 = 8.33 "and" R_(EQ) = (8.33 xx 5)/(8.33 + 5) = 3.12 Omega` |
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| 10. |
In electromagnetic induction the induced charge is independent of |
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Answer» CHANGE of flux |
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| 11. |
A rectangular block of mass m and area of cross section A floats in a liquid of density rho.If it is given a small vertical displacement from equilibrium it undergoes with a time period T, then |
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Answer» `T prop 1/sqrt(m)` When the blocks is floating. `therefor mg = Al rho g`  If the block is GIVEN vertical displacement y then the effective RESTORING force is, `F =-[A(l +y)rho g - mg)] =-[A(l + y)rhog -Alrhog] =-Alrho gy` here inertia factor = mass of block =m Spring factor `=A rho g` `therefore` Time period `=T = 2pi sqrt(m/(A rho g))` i.e. `T prop 1/sqrt(A)` |
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| 12. |
A car A is moving with speed "40 km h"^(-1) along a straight line 30^(@) north of east and another car B is moving with same speed along a straight line 30^(@) south of east. The relative velocity of car A as observed from the car B is |
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Answer» `"40 KM h"^(-1)" NORTH - east"` |
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| 13. |
The acceleration due to gravity at pole and equator can be related as :- |
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Answer» `g_p LT g_e` |
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| 14. |
The ionisation potential of hydrogen atom is |
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Answer» 12.97V |
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| 15. |
The word livres stands for: |
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Answer» UNIT of CURRENCY in France |
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| 16. |
Show that electrostatic potential ls constantthroughout the volume of the conductor and has the same value (as inside) on its surface. |
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Answer» SOLUTION :Sine `vecE=0` inside the conductor and has no tangential component on the surface no work is done in moving a small test charge within the conductor and on Its surface. That is there is no potential DIFFERENCE between any TWO point inside or on the surface of the conductor. If the conductor is charged, ELECTRIC field normal to the swface exises this means potential will be different for the surface and a point just outside the surface. In a SYSTEM of conductors of arbitrary size, shape and charge configuration each conductor is characterised by a constant value of potential but this constant may differ from one conductor to the other. 
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| 17. |
Describe the image in convex mirror, of the image of rod PQ in the plane mirror (radius of curvature of mirror is 60 cm). |
| Answer» SOLUTION :Virtual, erect and 3 CM LONG | |
| 18. |
S.I. unit of permittivity (e0) ...... |
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Answer» `C^(2)N^(-1)M^(-2)` `therefore E_(0) = (q_(1)q_(2))/(4piFr^(2))` `therefore` Unit of `E_(0) = ("Coulomb")^(2)/(NM^(2)) = C^(2)N^(-1)m^(-2)` |
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| 19. |
Give the expression for magnetic field due to a toroid. |
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Answer» Solution :A toroid is a hollow circular right of finite thickness on which a large NUMBER of turns of an insulated WIRE are closely WOUND. `B=mu_(0)nI` |
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| 20. |
Which of the following is not an application of eddy currents ? |
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Answer» INDUCTION Furnace. |
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| 21. |
When a light wave in a rarer medium is reflected from the surface of an optically denser medium, it suffers a phase change of (in radian) |
| Answer» Answer :C | |
| 22. |
In the figure shown below each battery has emf = 5 V. Then the magnetic field at P is |
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Answer» ZERO |
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| 23. |
The relation between frequency ‘n’ of the string, if n_(1), n_(2), n_(3)… are the frequenciees of segments of the stretched string is. |
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Answer» `n = n_(1) + n_(2) + n_(3) + …` |
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| 24. |
(a) Derive the expression for the force acting per unit length between two long straight parallel current carrying conductors. Hence, define one ampere. (b) Two long parallel straight conductors are placed 12 cm apart in air. They carry equal currents of 3 A each. Find the magnitude and direction of the magnetic field at a point midway between them (drawing a figure) when the currents in them flow in opposite directions. |
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Answer» Solution :Consider two straight, parallel, long, current carrying conductors AB and CD carrying currents `I_(1)` and `I_(2)` respectively in same direction and let these be separated by a distance d. Now magnetic field B1 DEVELOPED at a point Q on 2nd conductor DUE to current `I_(1)` flowing in 1st conductor is `B_(1)=(mu_(0)I_(1))/(2pid)` As per right hand rule `B_(1)` is acting normal to the plane of the paper pointing inward. Thus, conductor CD carrying current `I_(2)` is in a magnetic field which is perpendicular to its length. Therefore, force experienced by 2nd conductor CD due to `B_(1)`. `F_(21)=B_(1)I_(2)l`, where l = length of the 2nd conductor or `F_(21)=(mu_(0)I_(1))/(2pid)I_(2)l=(mu_(0)I_(1)I_(2)l)/(2pid)` and force per unit length `(F_(21))/l=(mu_(0)I_(1)I_(2))/(2pid)=(mu_(0))/(4pi).(2I_(1)I_(2))/d` The force `F_(21)` in accordance with Fleming.s left hand rule is directed towards the conductor AB. In the same way, it is found that force experienced per unit length of wire AB is `(F_(12))/l=(mu_(0))/(4pi).(2I_(1)I_(2))/d` and is directed towards CD. IF `I_(1)=I_(2)=1A` and d=1 m, then `(F_(12))/p=(mu_(0))/(4pi)xx(2xx1xx1)/1=(mu_(0))//(2pi)=2xx10^(-7)Nm^(-1)`. Hence, SI base unit of current i.e., ampere is defined as the value of that steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross-section and placed 1 m apart in vacuum would produce on each of these conductors a force equal to `2xx10^(-7)Nm^(-1)`.  (b) Here `I_(1)=I_(2)=I=3A`, distance between two parallel conductors d = 12 cm = 0.12 m. So the normal distance of point P midway between them from either conductor r`=d/2=0.06m` Then `B_(1)=B_(2)=(mu_(0)l)/(2pir)=(4pixx10^(-7)xx3)/(2pixx0.06)=1XX10^(-5)T` 
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| 25. |
what is the direction of his total displacemetn vector with respect to due east ? |
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Answer» `43 ^(@)` SOUTH of east |
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| 26. |
When shunt connected to galvanometer has the value equal to resistance of galvanometer, range of current meter would become _______ |
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Answer» twice Here S = G `thereforen-1=G/S=1""(becauseS=G)` `thereforen=2` Now, `I=nI_(G)=2I_(G)` |
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| 27. |
In the arrangement shown above, all surfaces are frictionless. The rod R is constrained to move vertically. The vertically acceleration of R is a_(1) and the horizontal accelration of the wedge W is a_(2). The ratio a_(1)//a_(2) is equal to |
| Answer» Answer :A | |
| 28. |
If an air conditioner is put in the middle of a room and started working |
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Answer» the room can be cooled slightly |
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| 29. |
What is the work done in increasing the charge from Q_(1)" to "Q_(1) +delta_(1)+deltaQ_(1), if the potential difference between the plates is V? |
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Answer» |
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| 30. |
When electron jumps from n = 4 to n = 1 orbit, we get |
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Answer» SECOND line of Lyman series |
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| 31. |
Define angle of dip at a place . What will be the value of the angle at the poles and at the equator of the earth? At what place on the earth's surface will the horizontal component of the earth's magnetic field and its vertical component be equal ? |
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Answer» Angle of dip at the equator`=0^(@)` Now , `TANTHETA(V)/(H)[theta=` angle of dip ] For `V=H, tantheta=1tan45^(@) " or, " theta=45^(@)` |
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| 32. |
A dam provides water fall for a power house 300 m below. The catchment area of 10^(10)m^(2) has annual rainfall of 0.6 m. 25%+ 15% + 10% wateris lost due to different reasons. Then, |
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Answer» mass of water FALL is `3XX10^(12)`kg `W=3xx10^(12) xx 9.8xx300=9xx9.8=8.82xx10^(15)` `:.` (A) & (B) |
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| 33. |
Does the change in magnetic flux induce e.m.f. or current? |
| Answer» SOLUTION :The change in magnetic FLUX always induces e.m.f.. HOWEVER, the current is induced only when the circuit is CLOSED. | |
| 34. |
A particle of charge q is placed at rest in a uniform electric field E and then released . The kinetic energy attained by the particle after moving a distance y is |
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Answer» `q E^(2)`y KE = Workdone by the force = F.y = qEy |
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| 35. |
Calculate de-Broglie wavelength associated with a rifle bullet of mass 2 gram moving with a speed of 400 ms^(-1). |
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Answer» SOLUTION :`lambda=(h)/(MV)` `lambda=(6.625xx10^(-34))/(2XX10^(-3)xx400)` `lambda-8.281xx10^(-34)`m |
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| 36. |
A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. If A^(1)B^(1) is the image after refraction from the lens and the reflection from the mirror, find the distance of A^(1)B^(1) from the pole of the mirror and obtain its magnification. (b) How much part of image A^(1)B^(1) is above RS. |
| Answer» SOLUTION : (a) 15 CM, -3/2, (6) 0.3 cm | |
| 37. |
If the buoyant force of the water Is equal to the weight of the water displaced, then the piece of glass suspended in water weighs |
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Answer» Solution :`1666 cm^3 XX "1 gm"/(cm^3)`=1666 G of `H_2O` 1.7 kg `H_2O xx 9.8 m//sec^2`=16.66 N THEREFORE 100 N-17 N = 83 N |
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| 38. |
In a p-n junction diode, the current I can be expressed as I= I_(0) "exp" ((eV)/(k_(B)T)-1) where I_(0)is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, k_(B)is the Boltzmann constant (8.6 xx 10^(–5) eV//K) and T is the absolute temperature. If for a given diode I_(0) = 5 xx 10^(-12) A and T = 300 K, then (a) What will be the forward current at a forward voltage of 0.6 V? (b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V? (c) What is the dynamic resistance? (d) What will be the current if reverse bias voltage changes from 1 V to 2 V? |
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Answer» Solution :(a) 0.0629 A, (b) 2.97 A, ( C) 0.336 `Omega` (d) For both the voltages, the CURRENT I will be almost EQUAL to `I_(0)`, SHOWING almost infinite dynamic resistance in the REVERSE bias. |
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| 39. |
In the above problem the ratio of the initial rate of fall of the temperatures is nearly |
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Answer» 2.49 |
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| 40. |
A nuclearpower reactor generates electric power of 100 MW. How many number of fissions occur per second if nuclear fuel used in the reactor is uranium ? |
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Answer» <P> SOLUTION :Power of a reactor using URANIUM is`P=(n/t)xx200xx10^6xx1.6xx10^(-19)` J `100xx10^6 = (n/t)xx200xx1.6xx10^(-13)` (`because` P=100 MW) `n/t=(100xx10^6)/(200xx1.6xx10^(-13))= 3.125xx10^18`/s. |
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| 41. |
Define self-inductance. Write its SI unit. Derive an expression for self-inductance of a long, aircored solenoid of length l, cross-sectional area A (radius r), and having N number of turns. |
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Answer» Solution :Self-inductance of a coil is numerically equal to the amount of magnetic flux linked with it when unit current flows through the coil. Alternately, self-inductance of a coil is numerically equal to the emf induced in the coil when rate of change of current through the coil is unity. SI unit of self-inductance is henry (H). Expression for self-inductance: Let a current Iis passed through a solenoid coil of N turns, LENGTH I and A the area enclosed by each turn of coil. The magnetic field B at any point inside the solenoid is: `B = (mu_(0)NI)/l` `THEREFORE` Magnetic flux through each turn of the solenoid `phi_(B) = B.A = (mu_(0)NI)/l. A` `therefore` TOTAL magnetic flux linked with the coil `Nphi_(B)= ((mu_(0)NI)/l.A).N = (mu_(0)N^(2)IA)/l` By definition `Nphi_(B) = L I` ` therefore (mu_(0)N^(2)IA)/l = LI` `implies` Self-inductance of solenoid `L = (mu_(0)N^(2)A)/l` If r be the radius of solenoid coil, then `A = pir^(2)` and we can write `L = (mu_(0)N^(2))/l.pir^(2)` If instead of air the solenoid coil has been wound on a ferromagnetic material (say soft iron) of relative PERMEABILITY `mu_(r), then we have ` L = (mu_(0)mu_(r).N^(2)A)/l = (mu_(0)mu_(r).N^(2)pir^(2))/l` |
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| 42. |
B may be _____ than _____ tan H depending on _____ of the medium . |
| Answer» SOLUTION :GREATER , LESS , PERMEABILITY | |
| 43. |
If one sphere collides head - on with another sphere of the same mass at rest inelastically. The ratio of their speeds ((v_(2))/(v_(1))) after collision shall be |
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Answer» `((1-e))/((1+e))` |
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| 44. |
Find the magnitude of the magnetic induction B of a magnetic field generated by a system of thin conductors, alon which a curren i is flowing at the point A (0,R,0) which is the centre of the circular conductor of radius R. The ring is yz plane. |
| Answer» SOLUTION :`B = (mu_0i)/(4 PI R) sqrt(2(2 pi^2 - 2pi + 1))` | |
| 45. |
The incident ray ,normula and reflected ray in reflection lies : |
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Answer» PERPENDICULAR to each other |
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| 46. |
In certain place H=(1)/sqrt(3)v the angleof dip atthis place is |
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Answer» 0 |
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| 47. |
Dimensional representation for modulus of rigidity is given by |
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Answer» `ML^2T^-2` |
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| 48. |
The maximum value of resultant of two forces is 5 N and minimum is 1 N. The resultant of the two forces when they act at angle of 60^@ . |
| Answer» Answer :C | |
| 49. |
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 :Torque EXERTED on current carrying pivoted rectangular COIL when SUBJECTED to external magnetic field is, `tau=BINAsintheta` (Where `theta` = angle between `vecAandvecB`) `thereforetau=BIN(l^(2))sintheta` `thereforetau=(0.8)(12)(20)(0.1)^(2)sin30^(@)" "(becausetheta=30^(@))` `thereforetau=0.96Nm` |
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| 50. |
(A): The typres of aircrafts are slightly conducting. (R ): If a conductor is connected to ground, the extra charge induced on conductor will flow to ground. |
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Answer» Both .A. and .R. are TRUE and .R. is the correct EXPLANATION of .A. |
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