Saved Bookmarks
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 gas expands adiabatically at constant pressure such that its temperature Talpha1/sqrtV What is the value of gamma=((C_p)/(C_v)) for the gas ? |
|
Answer» `2.00` |
|
| 2. |
a.What is an electromagnetic spectrum ?b.How the electromagnetic waves are arranged ?c.Mention some of the properties of E.M waves. |
|
Answer» Solution :a. Electromagnetic waves cover a wide range of frequencies or wavelengths. An ordely arrangement of electromagnetic radiations from gamma rays of very short wavelength to radio waves of long wavelength is known as electromagnetic spectrum. b.In the order of increasing wavelength, they are arranged as `gamma` - rays, X - rays, U.V. rays, visible light, infrared radiations, microwaves and radio waves. c.i.Produced by ACCELERATED charges. ii.Do not require any material for propagation. iii.Velocity of electromagnetic waves in free space is equal to that of light in free space. iv.Obey the principle of superposition. v.They carry ENERGY as they PROPAGATE through space. |
|
| 3. |
A TV Tower has a height of 75 m. What is the maximum distance and are area upto which this TV transmission can be received ? Take radius of the earth as 6.4xx10^6m. |
|
Answer» Solution :`d=SQRT(2rh) = sqrt(2xx6.4xx10^6xx75`,BR>`3.1xx10^4m = 31 km.` Area convered =`pid^2=2pirh=3018km^2` |
|
| 4. |
A body is executing S.H.M. with period 12 s. The time it takes in traversing a distance equal to half its amplitude is : |
|
Answer» 6 s `IMPLIES omega t=pi//6implies(2pi)/(12).t=(pi)/(6)impliest=1" s"` So correct CHOICE is (d). |
|
| 5. |
State the Kirchhoff's rules used in electric networks. How are these rules justified ? |
|
Answer» Solution :STATEMENT of Kirchhoff's laws Justification. Junction Rule: At any junction, the sum of currents, entering the junction, is equal to the sum of currents leaving the junction. Loop Rule: The algebraic sum of changes, in potential, AROUND any closed loop INVOLVING resistors and cells, in the loop is zero. `Sigma(DeltaV)=0` Justification: The first law is in accord with the law of conservation of charge SEconds law is in accord with the law of conservation of energy. |
|
| 6. |
Two candles of equal height h are placed in between vertical walls on a line perpendicular to the walls at a distance a from each other and also from the nearer wall. With what speed will the shadows of the candles move along the walls if one burns out completely in time t_(1) and the other in time t_(2). |
|
Answer» |
|
| 7. |
Two satellites of mass m and 9m are orbiting a planet in orbits of radius R.Their periods of revolution will be in the ratio of |
|
Answer» `9:1` Hence correct CHOICE is (C ). |
|
| 8. |
A radio nuclide sample has N_(0) nuclei at t=0. Its number of undecayed nuclei get reduced to N_(0)/e at t=tau. What does the term ‘tau’ stand for ? Write in terms of 'tau', the time interval T in which half of the original number of nuclei of this radio nuclide would have got decayed. |
|
Answer» SOLUTION :The time ‘`tau`’ in which number of undecayed nuclei get reduced to `N_(0)/e` is called the mean LIFE or average life period. The time interval T (or more appropriately `T_(1/2)`) in which half of the ORIGINAL number of nuclei would have got DECAYED is `T(orT_(1/2)=0.693/(lambda)=0.693lambda)` |
|
| 9. |
The difference between the nickel K_(alpha)-line wavelength and the continuous spectrum limit of the X-ray spectrum is 10%. Find the voltage applied to the X-ray tube. |
|
Answer» |
|
| 10. |
Hydrogen bomb is based on the principle of : |
| Answer» Answer :A | |
| 11. |
A long solenoid carring a current, produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is doubled. The new value of the magnetic field is |
|
Answer» 2B `B.=mu_(0)(2n)(2I)` `thereforeB.=4(mu_(0)nI)` `thereforeB.=4B`. |
|
| 12. |
A srot soleniod of radius a, number of turns per unit length n_(2), and length Lis kept coaxially inside a very long solenoid of radius b, number of turns per unit length n_(2). What is the mutual inductance of the system? |
|
Answer» `mu_(0)pib^(2)n_(1)n_(2)L` `M=mu_(0)pia^(2)n_(1)n_(2)L` |
|
| 13. |
If 2.2 kW of power having 22000 V is transmitted on a transmission line with 10Omega resistance then power dissipated in the form of thermal energy will be …… |
|
Answer» 0.1 W P=VI `therefore I=P/V=2200/22000`=0.1 A Power CONSUMED in FORM of THERMAL energy `=I^2R=0.01 xx 10` =0.1 W |
|
| 14. |
A blue ray of light enters an optically denser medium from air. What happens to its frequency in denser medium? |
| Answer» SOLUTION :The FREQUENCY of LIGHT REMAINS same. | |
| 15. |
The direction of the induced emf is determined by |
|
Answer» MAXWELL's law |
|
| 16. |
Verify that the cyclotron frequency omega=(eB)/m has the correct dimensions of [T]^(-1). |
|
Answer» Solution :When charge particle enter perpendicular magnetic field it PROVIDES the centripetal force for revolution. So, `(MV^(2))/R=qvB` `therefore(mv)/R=qB` but `v/R=omegaandq=e` `omega=(eB)/m=v/R` `therefore[omega]=[(eB)/m]=([v])/([R])` = `(M^(0)L^(1)T^(-1))/L^(1)` = `M^(0)L^(0)T^(-1)=[T^(-1)]` |
|
| 17. |
"Newton"/"coulomb" is equivalent to : |
|
Answer» 1C/V |
|
| 18. |
The fraction of the initial mumber of radioactive nuclei which remain undecayed after half of a half-life of the radioactive sample is : |
|
Answer» `1/4` `N=N_(0)/2^(n)=N_(0)/2^(4//2)=N_(0)/sqrt2` |
|
| 19. |
The potential difference between points A and B in the circuit shown in figure will be |
|
Answer» Solution :`i=(10-5)/(2.5+2.5+40)=1/9A` (CLOCKWISE) `V_B-15i-25i=V_A` `:. V_A-V_BH=40 i = - 40/9V` |
|
| 20. |
Choose the following statement a True of False. a. The real image formed by a lens is always inverted. b. The image formedby a lens is always inverted. c. The image formed by a concave lens is always erect and diminished. d. An air bubble insidewater acts likea concave lens. e. The distance between a real object and its real image formed by a single lens cannot be more than 4f. f. I an object is moved at a constant speed toward a convex lens from infinity to focus, then its image moves slower in the begninning and faster later on, away from the lens. g. The focal length of a glass (mu=1.5) lens is 10 cm in air. When it is completely immersed in water its focal length will become 40cm. |
|
Answer» Solution : a. False. Real images of real object are always INVERTED. But for virtual object the image may be erect. b. False. Real images are always inverted. c. True. From equation, we have `m=(f_(0))/(x+f_(0))` The VALUE of mis always positive and less that 1. d. True . The air BUBBLE in water diverges a parallel beam of light.  E. True. The lease distance between a real object and its real image is 4f. f. True. `(1)/(y)+(1)/(x)=(1)/(f),(dx)/(dt)=-u, (dy)/(dt)=v` `-(1)/(y^(2))(dy)/(dt)-(1)/(x^(2))(dx)/(dt)=0` `v=(y^(2))/(x^(2))u` Initially, x is large and y is small. So, initially v is small. But as the object moves from `oo` to focus, x decreases and y increases. So, v also increases.  g. True. Using equation, we have `f=f_(0)[(mu(mu_(0)-1))/(mu_(0)-mu)]` Hence, `mu_(0)=(3)/(2), mu=(4)/(3)` `rArr f=4f_(0)` The effective FOCAL length comes out to be positive. |
|
| 21. |
An electric lamp which runs at 40 V consumes 10 A current is connected to AC mains at 100 V,50 cycles per second. Calculate the inductance of the choke. |
|
Answer» Solution : Data supplied, I = 10 A, V= 40 V, E= 100 V, f=50 Hz, `OMEGA = 2pi f= 100 pi ` `I = (E )/(SQRT(R^2 + omega^2 t^2) )` ` 10 = (100)/(sqrt(16+ (100 pi L)^2)` solving we get ` sqrt(16 + (10 pi L)^2)= 10` `16 + (100 pi L)^2 = 100` `L = 0.02916 H` |
|
| 22. |
What is the magnitude of the induced current in the circular loop KLMN of radius 'r' if the straight wire PQ carries a steady current ofmagnitude 'i' ampere? |
Answer» SOLUTION :ZERO 
|
|
| 23. |
Give the characteristics of image formed by a plane mirror . |
|
Answer» Solution :(I)The image formed by a plane mirror is virtual ,erect and laterally inverted. (ii)The size of the image is equal to the size of the object (III)The iamge DIATANCE FAR behind the mirror is equal to the object distance in front of it. (IV)If an object is placed between two plane mirror inclined angle `theta` ,then the number of image n formed is as `n((360^(@))/(theta)-1)` |
|
| 24. |
मानव के पाचन में सबसे बड़ी ग्रंथि है - |
|
Answer» यकृत |
|
| 25. |
In Fig 15.5, the block has a kinetic energy of 3J and the spring has an elastic potential energy of 2J when the block is at x= +2.0cm. (a) What is the kinetic energy when the block is at x= 0? What is the elastic potential energy when the block is at (b) x= - 2.0cm and (c ) x= -x_(m)? |
| Answer» SOLUTION :(a) 5J, (B) 2J, (C ) 5J | |
| 26. |
The earth's magnetic field at a given point is 0.5 xx 10^(-5) Wb m^(2) . This field is to be annulled by magnetic induction at the center of circular conducing loop of radius 5 . The current required to be flown in the loop is nearly |
|
Answer» 0.2A |
|
| 27. |
A photon of frequency omega_(0) is emitted from the surface of a star whose mass is M and radius R. Find the gravirational shift of frequency Delta omega//omega_(0) of the photon at a very great distance from the star. |
|
Answer» Solution :We shall only consider stars which are not too compact so that the graviational field at their surface is weak: `(gammaM)/(c^(2)R)lt lt 1` We shall also clarify the PROBLEM by making clear the MEANING of the (slightly changed) NOTATION. SUPPOSE the photon is emitted by some atom whose total relativistic energies(including the rest mass) are `E_(1)&E_(2)` with `E_(1) lt E_(2)`. These energies are defined in the absence of gravitational field and we have `omega_(0) = (E_(2)-E_(1))/(cancelh)` as the frequency at infinity of the photon that is emitted in `2rarr1` transition. On the surface of the star, the energies have the values `E'_(2) = E_(2)-(E_(2))/(c^(2)).(gammaM)/(R ) = E_(2) (1-(gammaM)/(c^(2)R))` `E'_(1) = E_(1)(1-(gammaM)/(c^(2)R))` Thus, from `cancelh omega = E'_(2) - E'_(2)` we get `omega = omega_(0)(1-(gammaM)/(c^(2)R))` Here `omega` is the frequency of the photon emitted in the transition `2rarr1` when the atom is on the surface of the star. In shown that the frequency of spectral lines emitted by atoms on the surface of some star is less than the frequency of lines emitted by atoms here on earth (where the gravitational effect is quite small). Finally `(Delta omega)/(omega_(0)) =- (gammaM)/(c^(2)R)`. The answer given in the book is incorrect is general through it agrees with the above result for `(gammaM)/(c^(2)R) lt lt1`. |
|
| 28. |
A homogenous disc of mass 2 kg and radius 10 cm is rotating about its axis with an angular velocity of 4 rad s^(-1). The angular momentum of the disc is : |
|
Answer» 1 KG `ms^(-1)` Now ANGULAR momentum `L=Iomega=0.01xx4=0.04kgm^(2)s^(-1)` |
|
| 29. |
If the speed of the electron in a hydrogen atom orbit of principal quantum number n be V, then the curve showing variation of V with n is: |
|
Answer» a straight line `(mv^(2))/(r_(n))=(1)/(4pi epsi_(0)) ((ZE).e)/(r_(n)^(2))` `mv^(2) r_(n)=(Ze^(2))/(4pi epsi_(0))` `mv^(2)r_(n)=(Ze^(2))/(4pi epsi_(0))..........(1)` Bohr.s quantum condition gives `mvr_(n)=(nh)/(2pi) ..........(2)` Dividing eq. (1) by (2), `(mv^(2)r_(n))/(mv r_(n))=((Ze^(2))/(4pi epsi_(0)))/((nh)/(2pi))` `V=(Ze^(2))/(2n epsi_(0)H)` `v xx n =(Ze^(2))/(2epsi_(0) h)="constant"` i.e. `v xx n ="constant" .......(3)` Equation (3) represent a reactangular hyperbola. |
|
| 30. |
An image of a linear object due to a convex mirror is 1/4 th of the length of the object . If focal length of the mirror is 10 cm , find the distance between the object and the image. The linear object is kept perpedicular to the axis of the mirror. |
|
Answer» Solution :Mgnification of image `m = - (v)/(U) = (H.)/(h) = 1/4` `THEREFORE v = - u/4` Here, f = 10 cm ,`m = 1/4` v = ? , u +v = ? Using Gauss LAW `1/f = 1/u + 1/v` `therefore 1/10 = 1/u -(4)/u` `therefore1/10 = - (3)/u` `therefore u = - 30 cm ` Againusing Gauss law , `(1)/(v) = 1/f - (1)/(u) = (1)/(10) + (1)/(30)` `thereforev = 30/4 = 7.5 cm` `therefore ` Distance between object and image ` h = u + v = | - 30 | + 7.5` `thereforeu + v = 37.5 cm ` |
|
| 31. |
The phenomena of spontaneous emission of radiation from radioactive substance is known as : |
| Answer» Answer :D | |
| 32. |
A copper rod length 1 am and area of cross - section 0.1" m"^(2) has its end maintained at 100^(@)C by means of 0.4 KW electric heater. What is temperature of other end in steady state ? (K=400W//mK) : |
|
Answer» `90^(@)C` i.e `(Q)/(t)=0.4xx10^(3)` If T be the TEMPERATURE of other end of ROD then `("k A"(100-T))/(d)=0.4xx10^(3)` `(400xx0.1xx(100-T))/(1)=0.4xx10^(3)` `rArrT=90^(@)C` Hence correct CHOICE is (a). |
|
| 33. |
A ball of radius R has a uniformly distributed charge Q. The surrounding space of the ball is filled with a volume charge density rho=(b)/(r), where b is a constant and r is the distance from the centre of the ball. It was found that the magnitude of electric field outside the ball is independent of distance r. (a) Find the value of Q. (b) Find the magnitude of electric field outside the ball. |
|
Answer» (B). `E=(b)/(2epsilon_(0))` |
|
| 34. |
A sourse of gamma quanta is placed at a height h= 20m above an absorber. With what velocity should the source be displaced upward to counterbalance completely the gravitational variation of gamma quanta energy due to the Earth's gravity at the point where the absorber is located? |
|
Answer» Solution :Because of the gravitational shift the FREQUENCY of the gamma ray at the location of the absorber in increased by `(delta omega)/(omega)=(gh)/(c^(2))` For this to be compensated by the Doppler shift (assuming that resonant absorption is possible in the ABSENCE of gravitational field) we MUST have `(gh)/(c^(2))=(V)/(c ) or v(gh)/(c )= 0.65 mu m//s` |
|
| 35. |
x-y plane separates two media z ge 0 contains a medium of refractive index 1 and z le 0 contains a medium of refractive index 2. A ray of light is incident from first medium alogn a vector hati+hatj-hatk, the unit vector along refracted ray is |
|
Answer» `1/(2sqrt(3))hati+1/(2sqrt(3))hatj-sqrt(5/6)HATK` |
|
| 36. |
जीवाणुओ को पूर्णतः समाप्त करने की क्रिया कहलाती है - |
|
Answer» निर्जलीकरण (DEHYDRATION ) |
|
| 37. |
If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of: |
|
Answer» 20 m `v=144 KMH^(-1)=40MS^(-1),u=0` `:. A=(v)/(t)=(40)/(20)=2 ms^(-2)` Now `s=0+(1)/(2)xx2xx(20)^(2)=400m` |
|
| 38. |
Two bodies of masses 4 kg and 2 kg are moving with velocities2ms^(-1) and 10ms^(-1) towards each other due to gravitationalattraction, what is the velocity of their centre of mass? |
|
Answer» `5MS^(-1)` Correct choice is (d). |
|
| 39. |
In a Zener regulated power supply a Zener diode with V_(z)= 6.0 Vis used forregulation. The load current is to be 4.0 mA and the unregulated input is 10.0 V. What should be the value of series resistorR_(s)? |
| Answer» SOLUTION :The value of`R_(s)` should be such that the current through the Zener diode is much larger thanthe load current. This is tohave goodload regulation. Choose Zener current as five times the load current. This is to have good load regulation. Choose Zener current as five TIMESTHE load current, i.e., `I_(Z) = 20 mA`. The total current through `R_(S)` is therefore 24mA.The voltage drop across`R_(S)`is `10.0- 6.0 =4.0 V`.This gives `R_(S) = 4.0 V ( 24 xx 10^(-3) )A167 Omega `.The nearest value of CARBON resistoris `150 Omega `. So, a seriesof `150 Omega ` is appropriate . NOTE that slight variationin the value of the resistordoes not matter, what is important is that the current `I_(Z)` should be sufficiently larger than `I_(L)`. | |
| 40. |
If the energy of photons corresponding to the wavelength of 6000 A^@ is 3.2 xx 10^(-19) J, the photon energy for the wavelength of 4000 A^@ will be |
|
Answer» `1.11 XX 10^(-19) J` |
|
| 41. |
Diameter of lens of telescope is 0.16 m. Wavelength of light is 7500Å, so resolving power of telescope will be .. |
|
Answer» `1.73xx10^(5)` `=(D)/(1.22 LAMBDA)` `=(0.61)/(1.22xx75xx10^(-8))` `=0.00666xx10^(8)` `=6.67xx10^(5)` |
|
| 42. |
In a LCR series circuit capacitance is changed from C to 2C. To maintain the resonant frequency unchanged the inductance L be changed to…………….. |
|
Answer» |
|
| 43. |
A fish rising vertically to the surface of water in a lake uniformly at the rate of 3 m/s observes a bird diving vertically towards the water at a rate of 9 m/s vertically above it. If the refractive index of water is (4/3), find the actual velocity of the dive of the bird. |
|
Answer» Solution :If at any instant, the fish is at a depth x below the water SURFACE while the bird is at a height y above the surface, then the apparent height of the bird above the surface as seen - by fish will be `y_(ap) = mu_y` So the total apparent distance of the bird as seen by fish in water at depth x will be `h= x + MU y `(or) `(DH)/( dt) =(DX)/(dt) + mu ( dy )/(dt),i.e.,9= 3 + mu(dy )/( dt)` `i.e.,(dy)/(dt) = (6)/( (4//3)) = 4.5m//s` 
|
|
| 44. |
A gas has been subjected to an isochoric-isobaric cycle 1-2-3-4-1 (Fig.). Plot the graph of this cycle in the p-rho, V-Tand p-T coordinates. |
Answer» SOLUTION :The CYCLE are DEPICTED in FIGS.
|
|
| 45. |
An airplane flies from a town A to a town B when there is no wind and takes a total time T_(0) for a return trip. When there is a wind blowing in a direction from town A to town B, the plane's time for a similar return trip, T_(w), would satisfy |
|
Answer» `T_(0)LT T_(W)` `T_(w)= (s)/(v+ v_(w)) + (s)/(v + v_(w)) = s [(2v)/(v^(2) -v_(w)^(2))] = (2s)/(v) [(v^(2))/(v^(2)-v_(w)^(2))]` `T_(w) = T_(0) [(1)/(1-(v_(w)//v)^(2))]` |
|
| 46. |
Figure shows four points A, B, C and D in uniform electric field vec(E ) . The line AB and CD are perpendicular and BC and AD are parallel to the field lines. If V_(A), V_(B) , V_(C ) and V_(D) are electric potential at A, B,C,D choose the correct option. |
|
Answer» `V_(A) = V_(B) = V_(C ) = V_(D)` |
|
| 47. |
In Amplitude modulation |
|
Answer» The AMPLITUDE of the carrier wave varies in ACCORDANCE with the amplitude of the modulating SIGNAL |
|
| 48. |
Which of the following is true: |
| Answer» Answer :D | |
| 49. |
In stationary wave intensity of sound heard is maximum at |
|
Answer» antinodes |
|