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. |
15 kW power is supplied through a line of 0.5Omega resistance under 250 V potential difference .Fimd the efficiency of the supply in percentage . |
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Answer» Solution :power , p = 15 kW = 15000 W So , current N the supply line , `I=(P)/(V)=(15000)/(250)=60A` power loss due to inactive resistance in the line `I^(2)R=(60)^(2)xx0.5=1800W`, Therefore , total power in line= 15000 + 1800 = 16800 W `therefore"Efficiency"=("effective power")/("total power")xx100%` `= (15000)/(16800)xx100%=89%` |
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| 2. |
Heavy and stable nuclei have more neutrons than protons . This is because of the fact that |
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Answer» neutrons are heavier than PROTONS |
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| 3. |
A beam of light consisting of red, green and blue and is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will : |
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Answer» SEPARATE PART of the RED colors from the GREEN and blue colors. |
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| 4. |
Two identical cells send the same current in 2Omega resistance whether connected in series or parallel. The internal resistance of the cell should be: |
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Answer» `1OMEGA` |
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| 5. |
Which specific problem of Ruthorford atom model was attempted to be solved by Bohr atom model?Mention how the drawback is rectified. |
| Answer» Solution :Rutherford atom model COULD not EXPLAIN the stability of an atom and the line SPECTRA EMITTED by most of the atoms.Bohr rectified those by introducing the concept of non radiating orbits. | |
| 6. |
Light of wavelength 5000 A^(@) is diffracted by a slit. In diffraction pattern fifth minimum is at a distance of 5 mm from central maximum. If the distance between the screen and the slit is 1m. The slit width is |
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Answer» 0.5 mm |
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| 7. |
Train is runing in a metre gauge with a speed of 18okmh^-1 The vertical component of the earth's magnetic field is0.2xx10^-4Wb//m^2.The emf developed between its wheels is |
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Answer» 1mV |
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| 8. |
How does Ampere-Maxwell law explain the flow the of current through a capacitor when it is being charged by a battery? Write the expression for the displacement current in terms of the rate of change of electric flux. |
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Answer» Solution :Explanation of flow of current through CAPACITOR EXPRESSIONFOR displacement current. During charging, electric flux between the plates of capacitor KEEPS onchanging this results in the production of a displacement current between the plates. `I_(d)=epsilon(d varphi_(E))/(dt)(I_(d)=epsilon_(0)A (DE)/(dt))` |
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| 9. |
A radioactive sample has an activity of4 xx10^(7) Ci , Expross its activity in .becqueral. and .rutherford.. |
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Answer» Solution : Since 1 CI=`3.7 xx10^(10)` decays per second, activity `= 4 xx10^(7) Ci= 4 xx 10^(7) xx3.7 xx 10^(10) Bq` `= 1.48 xx 10^(18) Bq` Since , `1xx10^(6)`DECAY per second = 1 RD Activity = `1.48 xx10^(18) Bq= (1.48 xx10^(18))/(1xx10^(6))Rd = 1.48 xx10^(12)Rd`. |
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| 10. |
SI Unit of plane angle is- |
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Answer» Degree |
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| 11. |
Sn,C and Ge are all group XIV element.Yet,Sn is a conductor,C is an insulator while Si and Ge are semiconductors Why? |
| Answer» SOLUTION :Depending upon the ENERGY GAP value, an element can be distinguished and it has to be arranged in group XIV in ACCORDANCE with their value of Eg.Therefore SN is a conductor,C is an insulator while Si and Ge are semiconductor. | |
| 12. |
A particle is projected vertically upwards from point A with initial speed of 29.4 m/s. During what interval of time is the particle at more than 39.2m from A. |
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Answer» Solution :`(u^2 sin^2 theta)//(2G)` = (u^2sin ^2THETA)//(2g)` gives `tantheta=4 or `theta= tan^-1(4)` |
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| 13. |
A five wire potentiometer is connected to an accumulator of e.m.f 2.2 V and 1Omega internal resistance. The potentiometer wire has resistance 1Omega per metre. What is the maximum voltage that you can measure with this particular arrangement of potentiometer? What length of this potentiometer will balance the e.m.f of a daniell cell (e.fm.f =1.18V)? what resistance to balance this cell exactly at the centre of the last wire? |
| Answer» SOLUTION :put `R_(1)=1018` (one thousand times the e.m.f of Cd-cell), adjust `R_(2)` till no deflection in the GALVANOMETER `R_(2)=972Omega,5.1m)` | |
| 14. |
An electromagnetic wave emitted by an elementary dipole propagates in vacuum so that in the far fold zone the mean value of the enrgy flow density is equal to S_(0) at the point removes from the dipole by a distance r along the perpendicualr draws to the dipole's axis. Find the mean radiation power of the dipole. |
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Answer» Solution :We know that `S_(0)(r ) prop (1)/(r^(2))` AT other angles `S(r, THETA) prop sin^(2) theta` Thus `S(r, theta) = S_(0) (r )sin^(2) theta = S_(0)sin^(2) theta` Average power radiated `= S_(0) xx 4PI r^(2) xx (2)/(3) = (8pi)/(3) S_(0)r^(2)` (Average of `sin^(2) theta` over whole spher is `(2)/(3))`
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| 15. |
A toroid has a mean radius R equal to 20//pi cm, and a total of 400 turns of wire carrying a current of 2.0A. An aluminium ring at temperature 280 K inside the toroid provides the cure. (a) If the magnetization I is 4.8 xx 10^(-2) A/m, find the susceptibility of aluminium at 280 K. (b) If the temperature of the aluminium ring is raised to 320 K, what will be the magnetization ? |
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Answer» Solution :( a) The number of turns per unit length of the toroid is `n=(400)/(2piR)` The magnetic intensity H in the core is `H=ni=(400xx2.0A)/(2pixx20/pixx10^(-2)m)=2000A/m` The susceptibility is `CHI=I//H=(4.8xx10^(-2)A//m)/(2000A//m)=2.4xx10^(-5)` (B) The susceptibility X of a paramagnetic SUBSTANCE VARIES with absolute temperature as `chi = c //T`. Thus, `chi_(2)//chi_(1)=T_(1)//T_(2)` The susceptibility of aluminium at temperature 320 K is, THEREFORE, `chi=280/320xx2.4xx10^(-5)=2.1xx10^(-5)` Thus, the magnetization at 320 K is `I=chiH=2.1xx10^(-5)xx2000A//m=4.2xx10^(-2)A//m` |
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| 16. |
What do you mean by self-induction? |
| Answer» Solution :If the MAGNETIC FLUX is CHANGED by changing the current, an EMF is induced in that same coil. This phenomenon is known as self-induction. | |
| 17. |
A solenoid has radius r and length l carries a current I. If the number of its turns is 200, the energy stored in the solenoid is ….. |
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Answer» `(N^2A^2I)/(2pir)` `THEREFORE U=1/2 (mu_0N^2A I^2)/l` |
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| 18. |
If earth suddenly stops revolving and whole of its K.E. is used up for raising its temperature and if S = specific heat, R = radius, omega = angular velocity of earth, the rise of temperature of earth is : (J = joules constant) |
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Answer» `(R^(2)omega^(2))/(5S)` `therefore theta=(R^(2)omega^(2))/(5JS)` |
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| 19. |
Einstein's photoelectric effect is based on __ of energy |
| Answer» SOLUTION :CONSERVATION | |
| 20. |
What is the name of the master percussionist who praised Evelyn? |
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Answer» JAMES Brown |
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| 21. |
An unstable particle moves in the reference frame K^' along its y^' axis with a velocity v^'. In its turn, the frame K^' moves relative to the frame K in the positive direction of its x axis with a velocity V. The x^' and x axes of the two reference frames coincide, the y^' and y axes are parallel. Find the distance which the particle traverses in the frame K, if its proper lifetime is equal to Deltat_0. |
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Answer» SOLUTION :The COMPONENTS of the velocity of the unstable particle in the FRAME K are `(V, v^'SQRT(1-V^2/c^2),0)` so the velocity relative to K is `sqrt(V^2+v^('^2)-(v^('^2V^2)/(c^2))` The life time in this frame dilates to `Deltat_0//sqrt(1-V^2/c^2-v^('^2)/c^2+(v^('^2)V^2)/(c^4))` and the distance traversed is `Deltat_0(sqrt(V^2+v^('^2)-(v^('^2)V^2)//c^2))/(sqrt(1-V^2//c^2)sqrt(1-v^('^2)//c^2))`
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| 22. |
A spring of force constant K is first stretched by distance a from its natural length and then further by distance b. The work done in stretching the part b is |
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Answer» `(1)/(2)KA(a-b)` |
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| 23. |
Write a note on characteristic x-ray spectra : |
| Answer» Solution :X-rays are produced when fast moving electrons strike the metal target. The INTENSITY of the X-rays when plotted against its wavelength gives a curve CALLED X-ray SPECTRUM. | |
| 24. |
A plane wave that falls on a barrier containing small circular opening of dimension d. What can be said about the behaviour of the wave if (i) lamda lt lt d, (ii) lamda = d ,(iii) lamda gt gt d |
| Answer» Solution :(i) When `lamda lt lt d `i.e., wavelength of the wave is very very small when compared to the circular BARRIER or the diameter of the circular opening is lager than the wavelength of the waves COMING out of the opening continue to move in straight LINES, i.e., the ray approximation CONTINUES to be valid and we need not consider diffraction effects or wave nature. (ii) When `lamda = d`i.e., the diameter of the opening is of the order of the wavelength, the waves spread out from the opening in all directions the wave nature is predominant. (iii) `lamda gt gt d ` i.e., if the circular opening is very small relative to the wavelength, the opening behaves as a point source of waves, and emits spherical waves. Diffraction is more pronounced. The above ARGUEMENT is valid for all types of waves. | |
| 25. |
Define electric flux, Is it a scalar or a vector quantity?A point charg q is at a distanceof(d)/(2)directlyabove the centre of a square of side 'd' Use Gauss' lawto obtain the expression for the electric flux through the square. (b) If the point charge is now moved to a distance 'd' fromthe centre of the square and the side of the square is doubled , explain how the electric flux will be affected. |
| Answer» Solution :If the point charges q is now moved to DISTANCE .d. from the centre of the SQUARE and the side of the square is DOUBLED from .d. to .2d.,the charge q still remains at the square will remain UNCHARGED at ` (q)/( 6 in _0) ` | |
| 26. |
Name the physical quantity whose SI unit is F m^(-1) (farad/metre). |
| Answer» SOLUTION :ELECTRIC PERMITTIVITY `epsi_0 or EPSI`. | |
| 27. |
A ball is dropped from a height of 20m above the surface of water in a lake. The refractive index of water is 4//3. A fish inside the lake , in the line fall of the ball, is looking at the ball. At an instant when the balll is 12.8m above the water surface, the fish sees the speed of ball as . |
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Answer» `9ms^(-1)` `X_("image of ball")=(4)/(3)X_("ball")` `V_("image of ball")=(4)/(3)V_("ball")=(4)/(3)xx12=16ms^(-1)` |
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| 28. |
For which combination of working temperatures, the efficiency of 'Carnot's engine' is the least? |
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Answer» 60 K, 40 K `eta = (1-(T_(L))/(T_(H)))XX100` According to trial and error method `eta=(1-(80)/(100))xx100` `eta = ((100-80)/(100))xx100` `eta = (20)/(100)xx100=20%` |
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| 29. |
A glass slab when placed in air has a critical angle of 40^(@). Calculate the critical angle of the same glass slab if the medium around it is replaced by a liquid of refractive index 1.28. |
| Answer» SOLUTION :`55.69^(@)` | |
| 30. |
A large number of bullets are fired in all directions with same speed. What is the maximum area of their spread ? |
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Answer» `(PIV^(4))/g^(2)` |
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| 31. |
A short bar magnet has a magnetic moment of 0.48 JT^(-1). Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet. |
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Answer» Solution :Here m = 0.48 J `T^(-1)` and r = 10 cm and 0.1 m (a) Magnetic FIELD at a point on its AXIS `B=(mu_0)/(4pi) * (2m)/(r^3) = (10^(-7) XX 2 xx 0.48)/((0.1)^3)` `=9.6 xx 10^(-5)` T along S-N or 0.96 G along S-N direction. (b) Magnetic field at a point at a point on the equatorial line `B =(mu_0)/(4pi)* (m)/(r_3) =(10^(-7) xx 0.48)/((0.1)^3)` `=4.8 xx 10^(-5) ` T along N-S or 0.48 G along N-S direction. |
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| 32. |
Mass of bob of a pendulum is m and it is suspended from the ceiling inside a uniform external electric field of magnitude E directed in horizontal direction. Initially the bob has no charge and it takes time T to complete one oscillation. Now it is given a positive charge q. How much time will the bob take to complete one oscillation now? |
| Answer» SOLUTION :`T/([1+("QE"/"MG")^2]^(1//4))` | |
| 33. |
The objectives with larger apertures are used in telescope for : |
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Answer» BRIGHT image |
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| 34. |
Direct current is passed through a copper sulphate solution using platinum electrodes. The elements liberated at the electrones are |
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Answer» COPPER at ANODE and SULPHUR at cathode |
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| 35. |
Ampere's circuital law can be derived from |
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Answer» OHM's LAW |
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| 36. |
Small identical balls with equal charges are fixed at the vertices of a regular polygon of N sides, each of length d. At a certain instant, one of the ball is released. After long time interval, the adjacent ball to the previous one is released. The difference in kinetic energies of the two released balls is K at a sufficiently long distance from the polygon. |
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Answer» FINAL kinetic energy of the first ball is greater than that of the second ball |
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| 37. |
What are the ideal materials for fabrication of solar cell? and give the criteria for selection. |
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Answer» Solution :Semiconductor with band gap close to 1.5 eV are ideal materials for SOLAR cell fabrication. These materials are as follows: `Si (E_(g)=1.1eV)`, GaAs`(E_(g)=1.43eV)`, CdTE `(E_(g)=1.45eV)`, CuIn`Se_(2)(E_(g)=1.04eV)` etc. semiconductors. The following are the important criteria for selecting the right materials for fabrication solar cell. (1) Band gap `E_(g)` about 1.0eV to 1.8eV. (2) High optical absorption about `10^(-4)CM^(-1)` (3) The electrical conductivity should be high. (4) The raw material should be easily AVAILABLE and (5) The cost should be lower. |
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| 38. |
Height of the image formed on the retina is directly proportional to? |
| Answer» SOLUTION :VISUAL ANGLE | |
| 39. |
A particle with charge +Q and mass m enters a magnetic field to magnitude B existing only of the right of the boundary YZ The direction of the motion of the particle is perpendicular to the direction of B Let T =2pi(m)/(QB) The time spent by the particle in the field will be . |
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Answer» `(2 PI m)/(QB)` |
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| 40. |
A transistor is used in common emitter configuration. Given alpha=0.9. Calculate the change in collector current when the base current changes by 2muA. |
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Answer» `1muA` |
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| 41. |
A linearly polarized transverse wave is propagating in z direction through a fixed -point P in space . At time t_(0) the x component E_(x) and the y component E_(y) of the displacement at P are 3 and 4 units respectively . At a later time t_(1) if E_(x) at P is 2 units the value of E_(y) will be |
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Answer» 5 UNITS |
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| 42. |
A flat coil of area A and n turns is placed at the centre of a ring fo radius r(r^(2)gtgtA)and resistance R. The coil and the ring are coplaner. When the current in the coil increases from zero to i , the total charge circulating in the ring is |
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Answer» `(mu_(0)nAi)/(2rR)` |
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| 43. |
One straight wire carries 5 A. Angles made by line segments joining point P at 10 cm on perpendicular bisector and the two ends of wire, with the wire are 60^(@) each. Then magnetic field at this point will be ______ T. |
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Answer» `3mu_(0)` = `(mu_(0)xx5)/(4pixx0.1)[cos60^(@)+cos60^(@)]` = `(50mu_(0))/(4xx3.14)[2COS60^(@)]=3.98089xx2xx1/2mu_(0)` `thereforeB~~3.98mu_(0)T` |
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| 44. |
Centre of mass is defined as the point at which whole mass of the body is supposed to be _____. |
| Answer» SOLUTION :[CONCENTRATED] | |
| 45. |
The permeability of a magnetic material is 0.9983. |
| Answer» SOLUTION :DIAMAGNETIC | |
| 46. |
Using the Rydberg formula, calculate the wavelengths of the gamma-line in the Lyman series of the hydrogen spectrum. |
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Answer» `1218Å` For answer see PHYSICS DARPAN, Part-2, Section-B, ILLUSTRATION of TEXTBOOK EXAMPLE no. 12.6 |
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| 47. |
For television transmission, the frequency employed is normally in the range : |
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Answer» `30-300MHz` |
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| 48. |
Shown a metallic square frame of edga a in a vertical plane. A uniform magnetic field B exists in the space in a directon perpendicular to the plane of the figure. Two boys pull the opposite corners of the square ot deform it intoarhombus. They start pulling the corners t t=0and sisplace the corners at a uniform speed u. (a) find the induced emf in the frame at the instant when the angles at these corners reduce to 60^@. (b) find the induced current in the frame at this instatnt if the totalresistance of the frame is R. (c) Find the total charge which flows through a side of the frame by the time the square is deformed intoa straight line. |
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Answer» Solution :(a) The PERPENDICULAR component i.e a sin theta is to be taken which is perpendicular to velocity` `So, l=a sin 30^(@) =(a/2)` `Net 'a' CHARGE =4XX(a/2)=2a` `So, induced emf= Bul=2auB` `(b) Current=(e/R)=(2auB)/(R )` |
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| 49. |
How long ago were the old schools around? |
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Answer» CENTURIES ago |
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