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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. |
Same as problem 4 except the coil A is made to rotate about a vertical axis figure. No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counter clockwise and the coil A is as shown at this instant, t = 0, is |
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Answer» constant current clockwise This is possible only when prevailing current in loop A must be in clockwise and it should be constant because when loop A is stationary no current is induced in loop B. (Consider loop A as magnet and THINK about how magnetic field lines of loop A link with loop B). |
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| 2. |
In Fig. 16-43, an aluminum wire, of length L_1= 60.0 cm, cross-sectional area 1.25 xx 10^(-2)cm^(2) and density 2.60 "g/cm"^(3), is joined to a steel wire, of density 7.80 g/cm""^(3) and the same cross-sectional area. The compound wire, loaded with a block of mass m =10.0 kg, is arranged so that the distance L_2 from the joint to the supporting pulley is 86.6 cm. Transverse waves are set up on the wire by an external source of variable frequency: a node is located at the pulley, (a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes. (b) How many nodes are observed at this frequency? |
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| 3. |
Two small identical discs, each of mass m, lie on a smooth horizontal plane. The discs are interconnected by a light non-deformed spring of length l_0 and stiffness ϰ. At a certain moment one of the disc is set in motion in a horizontal direction perpendicular to the spring with velocity v_0. Find the maximum elongation of the spring in the process of motion, if it is known to be considerably less than unity. |
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Answer» Solution :In the C.M. frame of the system (both the discs+spring), the LINEAR momentum of the discs are related by the relation, `overset~vecp_1=-overset~vecp_2`, at all the moments of TIME. where, `overset~p_1=overset~p_2=overset~p=muv_(rel)` And the total kinetic energy of the system, `T=1/2muv^2_(rel)` Bearing in mind that at the moment of maximum deformation of the spring, the projection of `vecv_(rel)` ALONG the length of the spring becomes zero, i.e `v_(rel(x))=0`. The conservation of mechanical energy of the considered system in the C.M. frame gives. `1/2(m/2)v_0^2=1/2kx^2+1/2(m/2)v_(rel(y))^2` (1) Now from the conservation of angular momentum of the system about the C.M., `1/2(l_0/2)(m/2v_0)=2((l_0+x)/(2))m/2v_(rel(y))` or, `v_(rel(y))=(v_0l_0)/((l_0+x))=v_0(1+x/l_0)^-1~~v_0(1-x/l_0)`, as `xlt ltl_0` (2) Using (2) in (1), `1/2mv_0^2[1-(1-x/l_0)^2]=kx^2` or, `1/2mv_0^2[1-(1-(2X)/(l_0)+(x^2)/(l_0^2))^2]=kx^2` or, `(mv_0^2x)/(l_0)~~kx^2`, [neglecting `x^2//l_0^2`] As `x!=0`, thus `x=(mv_0^2)/(kl_0)` |
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| 4. |
Two slits in Young's interference experiment have widths in the ratio 1:16 . Deduce the ratio of intensity at maxima and minima . |
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| 5. |
Atomic lithium of concentration n=3.6.10^(16)cm^(-3) is at a temperature T=1500K. In this case the power emmitted at the resonat line's wavelength lambda=671nm(2Prarr 2S) per unit volume of gas is equal to P=0.30W//cm^(3). Find the mean lifetime of Li atoms in the resonance excitation state. |
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Answer» Solution :The number of excited atoms PER unit volume of the gase in `2P` state is `N=n(g_(p))/(g_(s))e^(-2pi hc//lambda kT)` Here `g_(p)=` degeneracy of the 2p state`=6,g_(s)=` degeneacy of the `2s` state`=2` and `lambda=` WAVELENGTH of the resonant line `2p rarr 2s`. The rate of decay of these atoms is `(N)/(TAU)` per sec. per unit volume, since each such atom EMITS light of wavelength `lambda`, we MUST have `(1)/(tau)(2pi ħc)/(lambda)n(g_(p))/(g_(s))e^(-2pi h c//lambda kT)=P` Thus `tau=(1)/(P)` Thus `tau=(1)/(P)(2pi ħc)/(lambda)n(g_(p))/(g_(s))e^(-2pi h c//lambdakT)=65.4xx10^(-9)s=65.4ns` |
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| 6. |
Ifmu= coefficient of friction and theta is the angle of friction, then |
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Answer» `MU=sinphi` |
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| 7. |
What is a 'magnet proof 'watch ? |
| Answer» SOLUTION :A ring of soft iron is fitted round a watch to make it free from external magnetic influence . That ring ACTS as a magnetic SCREEN . This KIND of watch is called a 'magnet PROOF ' watch . | |
| 8. |
A particle moves along a horizontal circle with constant speed. If 'a' is its acceleration and 'E' is its kinetic energy (A) a is constant(B) E is constant(C ) a is variable(D) E is variable |
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Answer» A and B are CORRECT |
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| 9. |
The velocity of combustion products ejected from the nozzle of a rocket is 2.0 km/s and the temperature is 600^@C. Find the temperature in the combustion chamber and the maximum efficiency. Assume the fuel combustion to be complete and carbon dioxide to be ejected from the nozzle. |
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| 10. |
The electric potential in volt due to an electric dipole of dipole moment 2 xx 10^(-8) coulomb-metre at a distance of 3m on a line making an angle of 60^(@) with the axis of the dipole is |
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Answer» ZERO |
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| 11. |
Pharaohs were rulers were ancient |
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Answer» Russia |
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| 12. |
What is interference og light ? |
| Answer» Solution :The phenomenon of redistribution of ENERGY in a MEDIUM by SUPERPOSITION of waves from TWO COHERENT sources. | |
| 13. |
State two important properties of photon which are used to write Einstein's photoelectric equation. Define (a) stopping potential and (b) threshold frequency, using Einstein's equation and drawing necessary plot between relevant quantities. |
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Answer» Solution :1st part: The PROPERTIES of photon used to write Einstein's PHOTOELECTRIC equation are (i) the rest MASS of the photon is zero, (II) the energy of the photon is E = HV. |
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| 14. |
Where does the energy of a capacitor reside ? |
| Answer» SOLUTION :In the dielectric material PRESENT between the plates of the CAPACITOR. | |
| 15. |
In the given reaction N_2 + 3H_2 → 2NH_3 |
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Answer» 1 MOLE of `H_2` gives 3 mole `NH_3` |
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| 16. |
AssertionIf the path of a charged particle in a region of uniform electric and magnetic field is not a circle, then its kinetic energy may remain constant. ReasonIn a combined electric and magnetic field regiona moving charge experiences a net force F=qE+q(VxxB), where symbols have their usual meanings. |
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| 17. |
The maximum illumination on a screen at a distance of 2 m from a lamp is 25 lux . The value of total luminous flux emitted by the lamp is |
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Answer» 1256 LUMEN |
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| 18. |
A thin sheet of a transparent material (mu= 1.60) is placed in the path of one of the interfering beams in a YDSE using sodium light, lambda= 5890 dotA. The central fringe shifts to a position originally occupied by the 12th bright fringe. Calculate the thickness of the sheet. |
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Answer» Solution :`triangle x= (mu-1)t= n lambda` It is GIVEN that `mu= 1.60,n= 12, lambda= 5890 dotA` `THEREFORE t= (12XX 5890 xx 10^(-10))/(1.60-1)= 1.18xx 10^(-5)m = 12 mu m`. |
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| 19. |
Discuss practical applications of eddy currents. |
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Answer» Solution :Eddy currents are used to advantage in certain applications like: (1) Magnetic braking in trains Strong electromagnets are situated above the rails in some electrically powered trains. When the electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train. As there are no mechanical linkages, the braking EFFECT is smooth. (2) Electromagnetic damping: Certain galvanometers have a fixed core made of non-magnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly (3) Induction furnace: Induction furnace can be used to produce HIGH temperatures and can be utilised to prepare alloys by melting the constituent metals. A high frequency alternating current is passed through a coil which SURROUNDS the metals to be melted. The eddy currents generated in the metals produce high temperatures sufficient to melt it. (4) Electric power meters : The shiny metal disc in the electric power meter (analogue type) ROTATES DUE to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil. You can observe the rotating shiny disc in the power meter of your house. |
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| 20. |
Two identical thin bar magnets each of length (L0 and pole strength (m) are placed at right angles to each other. Magnetic moment of system is : |
| Answer» ANSWER :C | |
| 21. |
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A ? |
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Answer» Solution :`R=15 CM, N = 3500,mu_r=800 ""B = ? , I = 1.2 A` `B=(mu_0mu_rNI)/(2piR)=(4PIXX10^(-60)xx800xx3500xx1.2)/(2pixx15xx10^(-12))=4.48T` |
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| 22. |
Given below is a list of electromagnetic spectrum and it's mode of production. Which one does not match: |
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Answer» Gamma RAYS (Radioactive DECAY of NUCLEUS) |
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| 23. |
If mu_(0) is permeability of free space and epsilon_(0) is permittivity of free space , the speed of light in vacuum is given by |
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Answer» `sqrt((1)/(mu_(0)epsilon_(0)))` |
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| 24. |
The width of incident wavefront is found to be doubled in a denser medium. If it makes and angle of 70^@ with the interface, the refractive index of the denser medium is: |
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Answer» 1.288 |
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| 25. |
A rough L- shaped rod is located in a horizontal plane and a sleeve of mass m is inserted in the rod. The rod is rotated with a constant angular velocity omega in the horizontal plane. The length l_(1) andl_(2) are shown in figure. The normal reaction and frictional force acting on the sleeve when it starts slipping are (mu: coefficient of friction between rod and sleeve) |
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Answer» `N=MOMEGA^(2)l_(1)` `N_(2)=mg` `momega^(2)xxsinphi=` `f=momega^(2)l_(2)` `N_(1)=momega^(2)l_(1)` 
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| 26. |
A capacitor of capacitance 'C' is being charged by connecting it across a d.c. source along with an ammeter. Will the ammeter show a momentary deflection during the process of charging ? If so, how would you explain this momentary deflectin and the resulting continuity of current in the circuit ? Write the expression for the current inside the capacitor. |
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Answer» Solution :Yes, ammeter will show a momentary deflection. The momentary deflection is DUE to the FLOW of electrons in the circuit during the charging process. During the charging process the electric FIELD between the capacitor plates is changing and hence a displacement current flows in the gap. Hence we can say that there is a continuity of current in the circuit. EXPRESSION `I_d = epsi_0 (d THETA)/(dt)` |
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| 27. |
A rod immersed horizontally in water does not appear small when is viewed normally but appears small if it is kept vertically. Explain. |
| Answer» Solution :When the ROD is kept horizontally under water, every point of the rod appears to be upward equally due to refraction. So, length of the rod appears to be same. But if the rod is immersed vertically, different points of the rod appear to be upward with different DISTANCES and from image. Since apparent displacement of a point depends on its depth, apparent displacement of the lower point of the rod is larger than the higher point. Thus, apparent displacement of the lowest and higher points of the rod are maximum and minimum respectively. So the rod appears to be SMALL. | |
| 28. |
When glass rod is rubbed against silk, silk becomes negatively charged of 320 nC, then how much electrons lost by the glass rod ? |
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Answer» (A) `2 xx 10^(10)` `e = -1.6 xx 10^(-19)` C Q = ne `-320 xx 10^(-9) = N xx (-1.6 xx 10^(-19))` `THEREFORE n = (320 xx 10^(-9))/(1.6 xx 10^(-19)) = 2 xx 10^(12)` |
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| 29. |
The electric field components in figure are E_(x) =ax, E_(y) = E_(z) =0in which a - 500 N/C m. Calculate (a) the flux through the cube and (b) the charge within the cube, length of side is aa=0.1m |
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| 30. |
The discharging current in the atmosphere due to the small conductivity of air is known tobe 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral ? In other words, what keeps theatmosphere charged ? |
| Answer» SOLUTION :The atmosphere is continually being charged by thunderstorms and LIGHTNING all over the globe and discharged through regions of ordinary weather. The two opposing CURRENTS are, on an AVERAGE, in equilibrium. | |
| 31. |
A vehicle is driven along a straight horizontal track by a motor which exerts a constant driving force. The vehicle starts from rest and the effects of friction and air resistance are negligible. Which of these graphs represents the vehicles kinetic energy with time 'r'? |
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Answer»
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| 32. |
What is the potential difference between the surface and interior point of charged conductor? |
| Answer» Solution :The POTENTIAL at EVERY POINT in a charged conductor is same and is EQUAL to the potential on the SURFACE. Therefore potential difference is zero. | |
| 33. |
Two identical calorimeters A and B contain an equal quantity of water at 20^(@)C. A 5 g piece of metal X of specific heat 0.2" cal g"^(-1).^(@)C^(-1) is dropped into A and 5 g piece of metal Y is dropped into B. The equilibrium temperature in A is 22^(@)C and that in B is 23^(@)C. The intial temperature of both the metals was 40^(@)C. The specific heat of metal Y ("in cal g"^(-1).^(@)C^(-1)) is |
| Answer» Answer :A | |
| 34. |
The wave that are bent down by the ionosphere are |
| Answer» Answer :D | |
| 35. |
Ratio of thermal conductivities of two different materials is 4:3 To have same thermal resistance of the two rods of having equal thickness, what should be the ratio of their lengths ? |
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Answer» `3:4` `:.R_(1)=R_(2).(l_(1))/(k_(1)A_(1))=(l_(2))/(k_(2)A_(2))` Here `A_(1)=A_(2)` as thickness is same. `:.(l_(1))/(l_(2))=(k_(1))/(k_(2))=4/3`. THUS correct choice is (b). |
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| 36. |
Consider a spherical planet of radius R . Its density varies with the distance of its centre t as rho=A-Br, where A and B are positive constants. Now answer the following questions. For B=0 , pressure due to gravitation force at a distance r( ltR) from the centre of the planet is |
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Answer» `(2)/(3)piGA^(2)(R^(2)-r^(2))` |
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| 37. |
In the problem No.133, the velocity of the wave is |
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Answer» 30 m/s |
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| 38. |
A convex lens is made up of three different materials as shown in figure. A point obiect, placed on its axis, the number of images formed are ........... |
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Answer» 1 So, a point object placed on its axis, the numbe of images formed are THREE. |
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| 39. |
A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic field. If z is the vertical direction, the z-component of magnetic field is B_Z = B_O (I + lambdaz). If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, lambdaand acceleration due to gravity g. |
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Answer» Solution :Magnetic flux linked with ring, `phi=AB costheta (because theta =0)` `theta =Bzpil^2` `phi=B_0(1+lambdaz)pil^2` Now according to Faraday.s law induced emf, `epsilon=(dphi)/(DT)=d/(dt)[B_0 (1+lambdaz)pil^2]` `epsilon=B_0 pil^2 lambda (dz)/(dt)` `epsilon=B_0 pil^2 lambdav` Induced current `I=S/R` `I=(B_0pil^2lambdav)/R` Heat produced PER unit time, `H=I^2R=((Bpil^2lambdav)/R)^2R` `H=(B_0^2pi^2l^4lambda^2v^2)/R` This heat per unit time is produced at the cost of rate of change of potential energy. Rate of change of P.E. = Heat produced per unit time (`because`Here that K.E. remains constant) `d/(dt)(mgz)=H` `mg(dz)/(dt)=(B_0^2pi^2l^4lambda^2v^2)/R` `THEREFORE mgv=(B_0^2pi^2l^4lambda^2v^2)/R` `therefore v=(mgR)/(B_0^2pi^2l^4lambda^2)` which is required equation of terminal velocity . |
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| 40. |
The radius of soap bubble is 'r' and the surface tension of soap solution is S, the energy required to double the radius is |
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Answer» `6pir^2//S` |
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| 41. |
SI unit of electromotive force of a cell is NC^(-1) |
| Answer» Solution :False - SI unit of electromotive FORCE (emf) of a CELL is VOLT. | |
| 42. |
The magnification produced by the objective lens of a compound microscope is 25. The focal length of eye piece is 5 cm and it forms find image at least distance of distinct vision. The magnifying power of the compound microscope is |
| Answer» ANSWER :3 | |
| 43. |
Which one of the following is a volt ? |
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Answer» ERG PER cm |
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| 44. |
candela is the unit of |
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Answer» MAGNETIC flux |
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| 45. |
Number of properties given below have higher value for protium than deuterium ? Relative abundance, Atomic mass, Melting point, Boiling point, Density, Enthalpy of fusion, Bonddissociation energy, Bond length. |
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| 46. |
Two point objects carry positive charges of magnitude 5e and 20e . Separation between the charges is 1 cm. Find the location of points on the line joining the two charges where the electric field intensity is zero. |
Answer» Solution :As the TWO charges are of same polarity thus, electric field will be zero at a point between the two charges . Let a point M be LOCATED between the two charges where electric field is zero. Given : `Q_1=5e=5xx1.6xx10^(-19)` C `Q_2=20e=20xx1.6xx10^(-19)`C At M, `E_1=E_2` `rArr 1/(4piepsilon_0)Q_1/x^2 =1/(4piepsilon_0)Q_2/(0.001-x)^2` `rArr Q_1/x^2=Q_2/(0.01-x)^2` `rArr (5xx1.6xx10^(-19))/x^2 =(20xx1.6xx10^(-19))/(0.01-x)^2` `(0.01-x)^2/x^2=20//5`=4 `(0.01-x)/x=pm2` `rArr x=+1/300m =+1/3 cm` and `x=-1/100` m =-1 cm x=-1 cm lies on the LEFT side of charge `Q_1`, but on the left side of `Q_1`, the electric fields due to `Q_1` and `Q_2` will be in the same direction, so the net electric field cannot be zero. The electric field is zero only at `x=+1/3` cm from charge `Q_1` between the two charges. |
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| 47. |
The work done in rotating a magent of the magnetic moment 2Am^(2) in a magnetic field of induction 5 xx 10^(-3)T from the direction along the magnetic field to the direction opposite to the field , is |
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Answer» ZERO |
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| 48. |
The fundamental frequency of a closed pipe is equal to the frequency of the second harmonic of an open pipe. The ratio of their lengths is |
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Answer» `1:2` `f_(1) =v/(4l)` For OPEN pipe, frequency of second harmonic `f_(2) =v/L` ACCORDING to question, `f_(1) =f_(2)` `v/(4l) =v/L` or `l/L =1/4` |
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| 49. |
Two wires of the same material have length 6cm and 10cm and radii 0.5 mm and 1.5 mm respectively. They are connected in seriesacross a battery of 16 V. The p.d. across the shorter wire is |
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Answer» 5V |
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| 50. |
An ac genrator |
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Answer» CREATES ELECTRICAL ENERGY |
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