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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 a diode AM detector with the output circuit consists of R=1M Omega and C=1 pF would be more suitable for detecting a carrier signal of |
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Answer» 1MHz for DEMODULATION, `(1)/(v_(c)ltltRC` `v _(c)=(1)/(RC)` or `v_(c)gtgt(1)/(10^(6))=10^(6)Hz`. `THEREFORE v_(c)gtgt10^(6)Hz` |
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| 2. |
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon? |
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Answer» SOLUTION :(a) `3.14xx10^(-19)J, 1.05xx10^(-27)kgm//s` (B) `3xx10^(16)"photons/s"` (C) `0.63m//s` |
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| 3. |
Define RMS value of AC. Derive a relation between RMS value of AC voltage & maximum voltatge. |
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Answer» Solution :i. RMS value is ALSO defined as that value of the steady current which when flowing through a given circuit for a given time produces the same amount of heat as produced by the alternating current when flowing through the same circuit for the same time. The effective value of an alternating voltage is represented by `V_(eff).` ii. The alternating current `i=I_(m)sinomegat""ori=I_(m)sintheta,` is represented GRAPHICALLY in Figure. The corresponding squared current wave is also shown by the dotted lines. iii. The SUM of the squares of all currents over one cycle is given by the area of one cycle of squared wave. Therefore, `I_(RMS)=sqrt(("Area of one cycle of squared wave")/("Base length of one cycle"))` IV. An elementary area of thickness `d""theta` is considered in the first half-cycle of the squared current wave as shown in Figure. Let `i^(2)` be the element `=i^(2)d""theta` Area of one cycle of squared `" wave "=int_(0)^(2pi)i^(2)d""theta`  `=int_(0)^(2pi)I_(m)^(2)SIN^(2)thetad""theta=I_(m)^(2)int_(0)^(2pi)sin^(2)thetad""theta` `=I_(m)^(2)int_(0)^(2pi)[(1-cos2theta)/(2)]d""theta` `"since"sin^(2)theta=(1-cos2theta)/(2)` `=(I_(m)^(2))/(2)[int_(0)^(2pi)d""theta-int_(0)^(2pi)cos2thetad""theta]` `=(I_(m)^(2))/(2)[theta-(sin2theta)/(2)]_(0)^(2pi)` `=(I_(m)^(2))/(2)[(2pi-(sin2xx2pi)/(2))-((0-sin0)/(2))]` `=(I_(m)^(2))/(2)xx2pi=I_(m)^(2)pi[becausesin0=sin4pi=0]` Substituting this in equation (1), we get `I_(RMS)=sqrt((I_(m)^(2)pi)/(2pi))=(I_(m))/(sqrt(2))[" Base length of one cycle is "2pi]` `I_(RMS)=0.707I_(m).` v. Thus we find that for a symmetrical sinusoidal current rms value of current is `70.7%` of its peak value. Similarly for alternating voltage, it can be shown that `V_(rms)=0.707I_(m).` |
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| 4. |
The kinetic energy of the body is increased by 300%. The percentage increases in momentum of the body will be |
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Answer» 300 |
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| 5. |
Refer to the arrangement of charged in and a Gaussian surface of radius R with Q at the centre. Then |
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Answer» total FLUX through the surface of the SPHERE is `(+Q)/(epsilon_0)` |
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| 6. |
500 J of work is done in sliding a 4 kg block up along a rough inclined plane to a vertical height of 10 m slowly. The work done against friction is (g = 10m/s^2) |
| Answer» Answer :a | |
| 7. |
A parallel plate capacitor is filled with a dielectric of dielectric constant (relative permittivity) 5 between its plates and is charged to acquire an energy E. Then it is isolated (disconnected from the battery) and the dielectric is replaced by another dielectric of dielectric constant (relative permittivity) 2. The new energy stored in the capacitor |
| Answer» ANSWER :B | |
| 8. |
(I) Wavefront division used to produce two coherent sources. (II) Destructive interference occurs at the centre of the shadow. (III) Newton used a prism to produce dispersion. (IV) Atmospheric particles changes the direction of the sunlight. |
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Answer» I and II only |
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| 9. |
(A): A domestic electrical appliances, working on a three pin, will continue working even if the top pin is removed. (R): The third pin is used only as a safety device. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A' |
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| 10. |
A resistance coil of 60 ohm is immersed in 42000 gm of water. A current of 7A is passed through it. Calculate the rise in temperature per minute. The sp. Heat of water = 4200(J)/(kg- ""^(@)K) |
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Answer» `5^(@)C` |
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| 11. |
If you made a map of magnetic field lines atMelbourne in Australia, then the magnetic fieldlines seem to be |
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Answer» GO into the ground |
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| 12. |
How infrared rays are detected ? |
| Answer» SOLUTION :THERMOPILES and BALOMETER. | |
| 13. |
Define the terms : Work function. |
| Answer» Solution :The MINIMUM NEGATIVE potential of the anode for which all PHOTO electrons are prevented from reaching the anode is called the STOPPING potential.Stopping potential is directly proportional to the frequency above the threshold frequency. | |
| 14. |
A black hole, absorbs |
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Answer» all radiation |
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| 15. |
Check that the ratio (k e^(2))/( Gm_em_p)is dimensionless. Look upa Table of Physical Constant and determine the value of this ratio. What does the ratio signify? |
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Answer» Solution :From the RELATION `F= ( kq_1q_2)/(r^(2)) ` we find units of k as `Nm^(2) C^(2) `. Again units of F are `Nm^(2) kg^(-2) ` ` THEREFORE`Units of ` (ke^(2))/(Gm_rm_p)= ( Nm^(2) C^(-2) (C^(2)))/( (Nm^(2) kg^(-2) )(kg)(kg)) = `units less and hence dimensionless. We know that ` k= 9 xx 10 ^(9) Nm ^(2) C^(2) .,e= 1.6 xx 10 ^(-19)C , G = 6.67 xx 10 ^(-11)Nm ^(2)kg ^(-2), m _e= 9.1 xx 10 ^(-31) kgand m_p = 1.67 xx 10 ^(-27)kg ` ` therefore` value of `(k e^(2))/( Gm_e,m_p) =( 9xx 10^(9) xx (1.6 xx 10 ^(-19))^(2) )/(6.67 xx 10 ^(-11)xx 9.1 xx10 ^(-31)xx 1. 67 xx 10 ^(-27))= 2.4 xx 10 ^(39 ) ` The ratio signifies the ratio of electricforce and the GRAVITATIONAL FORCE ACTING between an electron and a proton separated by a finite distance. |
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| 16. |
500g of ice at 0°C is mixed with 1g steam at 100°C. The final temperature of the mixture is |
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Answer» `0°C` |
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| 17. |
The refractive index of water is 1.33. What will be the speed of light in water? |
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Answer» `3xx10 ^8 MS^(-1)` |
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| 18. |
A sphere of mass 50 gm is suspended by a string in an electric field of intensity 5NC^(-1) acting vertically upward. If the tension in the string is 520 millinewton, the charge on the sphere is (g = 10ms^(-2)) |
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Answer» `-4 XX 10^(-3)C` |
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| 19. |
Statement I : In propagation of light waves, the angle between plane of vibration and plane of polarisation is (pi)/(2) radians. Statement II : Plane polarised light is incident on an analyser.The intensity becomes 3/4th.The angle of axis of analyser with beam is then 30^(@). |
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Answer» Statement I is true, statement II is false. |
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| 20. |
When two waves interfering with each other have phase difference of 2pi and have amplitude a_1 and a2 the resultant amplitude is : |
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Answer» `a_1+a_2` |
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| 21. |
A satellite is taken to a height equal to radius of the earth and is projected horizontally with a speed of 7km//s. What is the nature of its orbit (G=6.67xxNm^2//kg^2,R=6400km,M=5.98xx10^24kg) |
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Answer» a)`6326m//s` |
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| 22. |
What do you mean by leakage current in a diode ? |
| Answer» Solution :The leakage current in a diode is the current that the diode will LEAK when a REVERSE VOLTAGE is applied to it. Under the reverse bias, a very small current in LA flows across the junction This is due to the FLOW of the minority charge carriers called the leakage current or REVERS saturation current | |
| 23. |
A single slit is illuminated by light of wavelengths lambda_(a) and lambda_(b) chosen so that the first diffraction minimum of the lambda_(a) component coincides with the second minimum of the lambda_(b) component. (a) If lambda_(b)= 420 nm, what is lambda_(a)? For what order number m_(b) (if any) does a minimum of the lambda_(b) component coincide with the minimum of the lambda_(a) component in the order number (b) m_(a) = 2 and (c) m_(a) = 3? |
| Answer» SOLUTION :(a) 840 NM, (b) 4, (c) 6 | |
| 24. |
The gravitational potential energy of a body of mass m at a distance r from the centre of the earth is U. What is the weight of the body at this distance? |
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Answer» `-U` `U=-gmr RARR mg=(-U)/(r )` So the correct choice is (c ). |
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| 25. |
A beam of monochromatic blue light of wavelength 4200Å in air travels in water (mu=4//3). Its wavelength in water will be |
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Answer» `2800 Å` |
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| 26. |
The angular speed that should the earth have to rotate about its axis so that apparent weight of a body on the equator will be zero. What would be length of the day at that time (Radius of earth= 6400 km, g=9-8 m/s_2) |
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Answer» 4077 sec |
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| 27. |
In a sound wave, a displacement mode is a pressure antitode. Explain. |
| Answer» Solution :When LONGITUDINAL stationary waves are set up in a medium, (e.g., starionary waves in air columns), there are some points at which the particles of the medium do not VIBRATE at all. Such points of zero displacement are called the nodes or the displacement nodes. However at these points, the CHANGE in pressure RELATIVE to the normal pressure is maximum. Therefore, these points are also called the pressure antinodes. Thus, the displacement nodes are the pressure antinodes. | |
| 28. |
In above question, if cavity is not concentric and centred at point P then repeat all the steps. |
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Answer» Solution :Again assume `rho` and `-rho` in the cavity, (similar to the previous EXAMPLE) : (i) `vec(E_(A))=vec(E_(rho))+vec(E_(-rho))=(rho [vec(OA)])/(3 epsi_(0))+((-rho)vec(PA))/(3 epsi_(0))` `rho/(3 epsi_(0)) [vec(OA)-vec(PA)]=rho/(3 epsi_(0)) [vec(OP)]` Note : Here, we can can see that the electric field INTENSITY at point A is independent of position of point a inside the cavity. Also the electric field is ALONG the line joining the centres of the sphere and the spherical cavity. (ii) `vec(E_(B))=vec(E_(rho))+vec(E_(-rho))=(rho(vec(OB)))/(3 epsi_(0))+(K[4/3pir^(3)(-rho)])/([PB]^(3)) vec(PB)` (iii) `vec(E_(C))=vec(E_(rho))+vec(E_(-rho))=(K[4/3pi R^(3) rho])/([OC]^(3)) vec(OC)+ (K[4/3 pir^(3) (-rho)])/([PC]^(3))vec(PC)` (iv) `vec(E_(O))=vec(E_(rho))+vec(E_(-rho))=0+ (K[4/3 pi r^(3) (-rho)])/([PO]^(3)) vec(PO)` 
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| 29. |
Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil? |
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Answer» Solution :Since the coil is tightly wound, we may take each circular ELEMENT to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is `B= (mu_0 NI)/(2R) = (4PI xx 10^(-7) xx 10^2 xx 1)/(2 xx 10^(-1) ) = 2pi xx 10^(-4) = 6.28 xx 10^(-4)T` |
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| 30. |
In the previous question the angle phi is equal to |
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Answer» `theta` |
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| 31. |
Potential gradient is defined as |
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Answer» fall of potential PER unit LENGTH of the WIRE |
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| 32. |
Figure-5.77. shows a fixed coil of N turns and radius a carrying a current I. At a distance x from its centre another small coaxial coil of radius b(b lt lt a) and resistance R is moving toward the first coil at a uniform speed v. Find the induced current in smaller coil. |
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Answer» |
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| 33. |
A point object is placed at a distance of 12 cm on the axis of a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that the image formed by the combination coincides with the object itself. What is the focal length of convex mirror? |
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Answer» Solution :For convex lens, `(1)/(V)- (1)/(-12) = (1)/(10)` `i.e.,v=60 CM` i.e., in the absence of convex mirror, convex lens will form the image I, at a distance of 60 cm behind the lens. Since, the mirror is at a distance of 10 cm from the lens, I will be at a distance of 60 - 10 = 50 cm from the mirror, i.e., `MI_1 = 50 cm`.  Now as the final image `I_2` is FORMED at the OBJECT O itself, the rays after reflection from the mirror RETRACES its path, i.e., rays on the mirror are incident normally, i.e., I, is the centre of the mirror, so that `R=MI_1= 50 `cm and hence `F=(R//2) = (50//2) = 25 `cm |
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| 34. |
A copper rod of length l is rotated about the end perpendicular to a uniform magnetic field B with: |
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Answer» `B omega L^(2)` |
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| 35. |
EQUIPOTENTIAL SURFACES DUE TO ISOLATED POSITIVE POINT CHARGE ARE |
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Answer» spherical |
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| 36. |
On what factors does magnifying power of a compound microscope depend? |
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Answer» <P> Solution :MAGNIFYING power P = `-L/f_0(1+D/f_0)`.Therefore, it depends on focal lengths of objective and eye piece |
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| 37. |
In Young's double slit experiment if the width of 4^(th) bright fringe is 2 xx 10^(-2)cm, then the width of 6th bright fringe will be ...... |
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Answer» `3XX10^(-2)` |
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| 38. |
An alternating voltage given by V = 300 sqrt(2) sin (50t) ("in volts") is connected across a 1 mu F capacitor through an AC ammeter. The reading of the ammeter will be |
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Answer» 10 mA 
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| 39. |
Obtain the expression for the energy stored in a parallel plate capacitor. |
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Answer» Solution :Energystored in the capacitor : Capacitor not only stores the charge but also it stores energy.When a battery is connected to the capacitor electrons of total charge - Q are transferred from One plate to the other plate . To transfer the the charge work is done by the battery . This work done is stored as ELECTROSTATIC potential energy in the capacitor . To transfer an INFINITESIMAL charge DQ for a potential difference V the work done is given by dW = VdQ where `V=(Q)/(C ) ` The total work done to charge a capacitor is ` W = int_(0)^(Q) (Q)/(C) dQ = (Q^(2))/(2C)` This work done is stored as electrostatic potential energy `(U_(E))` in the capacitor. `U_(E)= (Q^(2))/(2C) = (1)/(2) CV^(2) ( :. Q = CV)` where Q = CV is used . This stored energy is thus directly proportional to the CAPACITANCE of the capacitor and the square of the voltage between the plates of the capacitor. But where is this energy stored in the capacitor ? To understand this question the equation (3) is rewritten as follows using the results `C = (epsilon_(0)A)/(d) `and V = Ed `U_(E) =(1)/(2) ((epsilon_(0)A)/(d) (Ed)^(2) = (1)/(2) epsilon_(0)(Ad) E^(2)` where Ad = volume of the space between the capacitor plates . The energystrored per unit volume of space is defined as energy density volume . From equation (4) we get `mu_(E) = (1)/(2) epsilon_(0)E^(2)` From equation (5) we infer that the energy is stored in the ELECTRIC field existing between between the plates of the capacitor . Once the capacitor is allowed to discharge the energy is retrieved . |
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| 40. |
A source of ac volatge V = V_(0) cos omega t delivers enegry yo a consummer by means of along straight coaxial cable with negligible active resistance. The current in the circuit varies as I = I_(0) cos omegat - varphi. Find the time-averaged enegry flux through the cross-section of the cable. The sheath is thin. |
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Answer» SOLUTION :As in the PREVIOUS problem `E_(r) = (V_(0)COS omegat)/(rIn(r_(2))/(r_(1))` and `H_(theta) = (I_(0)cos(omegat - varphi))/(2pir)` Hence TIME averged power flux (along the `z` axis) `= (1)/(2)V_(0)I_(0)cos varphi` On USING `lt cos omega t cos (omegat - varphi) gt = (1)/(2)cos varphi`. |
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| 41. |
What's importance of propagation of radio waves? |
| Answer» Solution :In all MODERN FORMS of communication such as RADIO, TV and microwaves ETC. | |
| 42. |
A certain oscillation results from the addition of coherent oscillations of the same direction y= a cos [omegat+(k-1)phi] where k is the numberof oscillation [ k=1,2………N] and phi is the phase difference between k^("th") and (K-1)^("th") oscillations. The amplitude of resultant oscillation will be |
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Answer» `(a SIN (N//2))/(sin (phi//2))` |
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| 43. |
f P is X-ray unit and Q is micron, then (P)/(Q) = |
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Answer» `10^(7)` |
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| 45. |
The effective capacitances of two condensers are 3muF and 16muF, when they are connected in series and parallel respectively. Compute the capacitance of each condenser. |
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Answer» SOLUTION :`C_s = (C_1C_2)/(C_1 + C_2)` ` 3 = (C_1 C_2)/(C_1 + C_2)`….(1) `C_P= C_1 + C_2` ` 4 = C_1 + C_2`…(2) Now on SOLVING eq (1) and (2) we will GET ` C_1 = 12muF` ` C_2 = 4muF` |
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| 46. |
Using the conservation laws, demonstrate that a free electron cannot absorb a photon completely. |
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Answer» Solution :We consider the collision in the rest frame of the initial ELECTRON. Then the reaction is `gamma+e (rest)rarr e("MOVING")` Energy momentum conservation GIVES `cancelh omega +m_(0)c^(2) = m_(0)c^(2)//SQRT(1-beta^(2))` `(cancelh omega)/(c ) = (m_(0)cbeta)/(sqrt(1-beta^(2)))` where `omega` is the ANGULAR frequency of the photon. Eliminating `h omega` we get `m_(0)c^(2) = m_(0)c^(2)(1-beta)/(sqrt(1-beta^(2))) =m_(0)c^(2) sqrt((1-beta)/(1+beta))` This gives `beta = 0` which implies `cancelh omega = 0`. But a zero photon means a photon. |
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| 47. |
(A) : The average magnitude of poynting vector vecS is the intensity of electromagnetic wave.(R) : Poynting vector is given by vecS=(vecExxvecB)/(2mu_(0)) |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A'. |
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| 48. |
Two objects move toward each other, collide, and separate. If there was not net external force acting on the object, but some kinetic energy was lost, then |
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Answer» the COLLISION was ELASTIC and total linear momentum was CONSERVED |
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| 49. |
If 10% of the radio active material decays in 5 days. The percentage of original material left after 20 days |
| Answer» ANSWER :C | |
| 50. |
Ray diverging from a point source forms a wavefront which is |
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Answer» Spherical |
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