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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 copper disc of radius 0.1 m is rotated about its centre, with 10 revolutions per second, in a uniform magnetic field of 0.1 tesla with its plane perpendicular to the field. The e.m.f. induced across the radius of the disc is |
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Answer» `pi xx 10^(-1)` volt `e=(B omega R^(2))/(2)=(0.1 xx(2pi xx 10)xx(0.1)^(2))/(2)` `rArr e=0.1 xx(0.1)^(2)xx pi xx10=pi xx 10^(-2)V` |
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| 2. |
The speed of a motor increases from 1200 rpm to 1800 rpm in 205. How many revolutions does it make in this period of time? |
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Answer» 400 |
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| 3. |
The electric field componentsare E_(x)=ax^(-1//2),E_(y)=E_(z)=0in which a =800 N/C m^(1//2)calculate(a) theflux throughthe cube and (b) the chargewithin the cube asume that a=0.1 m |
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Answer» Solution :For any closed surface, all area vectors are to be shown perpendicularly outward. Here, out of six faces of a cube, each having area `a^2`, electric flux gets associated with faces 1 and 2 only (because remaining four faces are parallel to given electric field along X-axis. Electric flux LINKED with FACE-1, `phi_(1) = A_(1)E_(1)cos theta_(1) =A_(1) (alphax_(1)^(1//2)) cos theta_(1)` `=(a^(2)) (ALPHA a^(1//2))cos (180^(@)) (therefore vecA_(1) || -vecE_(1))` `therefore phi_(1) = -alpha a^(5//2)`........(1) `(therefore x_(1) =a)` Electric flux linked with face -2 is: `phi_(2) = A_(2)E_(2) cos theta_(2)` `therefore theta_(2) = A_(2)(alphax_(2)^(1//2)) cos theta_(2)` `phi_(2) = sqrt(2)(alpha a^(5//2))`......... (2) Net electric flux passing through given cube is, `phi = phi_(1) + phi_(2)` `=-alphaa^(5//2) + sqrt(2) alpha a^(5//2)` `= (800)(0.1)^(5//2) (1.414-1)` `=((800)(0.414))/(3.162)^(5)` `phi = 1.048 (Nm^(2))/C` (Here, `phi`comes out to be positive which indicates that this flux comes out of given cube) Now, if net charge enclosed by given cube is q then according to Gauss.s theorem, `phi = q/epsilon_(0)` `therefore q =-phi epsilon_(0) = (1.048) (8.85 xx 10^(-12))` `therefore q = 9.2748 xx 10^(-12)` C |
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| 4. |
Give any two applications of ultraviolet radiations. |
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Answer» Solution :1. USED in LASIK eye surgery 2. UV lamps are used to KILL germs in water purifiers 3. Disinfection for virus and bacteria 4. To produce PHOTO electric current in burglar alarm. |
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| 5. |
A circular metallic ring of radius R has a small gap of width d. The coefficient of thermal expansion of the metal isalpha in appropriate units. If we increase the temperature of the ring by a amount DeltaT, then width of the gap –(A)Will increases by an amount dalphaDeltaT |
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Answer» Will INCREASES by an amount `d alphaDeltaT` |
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| 6. |
What the different gains in a transistor ? |
| Answer» SOLUTION :CURRENT GAIN , VOLTAGE gain , POWER gain | |
| 7. |
If the magnetic field is parallel to the positive Y-axis and the charged particle is moving along the positive X-axis as per figure, which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge). |
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| 8. |
If Bohr's qunatisation postulate (angular momentum=nh//2pi) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do never speak of quantisation of orbits of plants around the sun? |
| Answer» Solution :Bohr's quantisation POSTULATE is in TERMS of Planck's constant (h). But angular momenta associated with planetary motion are `~~10^(70)h` (for earth). In terms of Bohr's quantisation postulate, this will correspond to `n~~10^(70)`. for such large VALUES of n, the differences in successive energies and angular momenta of the quantised levels are so small, that the levels can be considered as continuous and not discreate. | |
| 9. |
A transformer having efficiency 90% is working on 100 V and at 2.0 kW power. If the current in the secondary coil is 5A, calculate (i) the current in the primary coil and (ii) voltage across the secondary coil. |
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Answer» Solution :Here `eta= 90% = (9)/(10), I_(s) = 5A` `E_(p) = 100 V`, `E_(p)I_(p) = 2kW = 2000 W` (i) `I_(p) = (2000)/(E_(p))` or `I_(p) = (2000)/(100) = 20 A` (II) `eta = ("Output power")/("Input power") = (E_(s)I_(s))/(E_(p)I_(p))` or `E_(s)I_(s) = eta xx E_(p)I_(p)` `= (9)/(10) xx 2000 = 1800 W` `THEREFORE E_(s) = (1800)/(I_(s)) = (1800)/(5) = 360` volt |
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| 10. |
ABC is a right angled glass prism of refractive index, 1.5 /_A, /_B and /_C are 60^(@), 30^(@) and 90^(@) respectively. A think layer of liquid is on the AB. For a ray of light is incident normally on AC to be totally reflected at AB, the refractive index of the liquid on AB should be |
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Answer» 1.5 |
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| 11. |
A long solenoid is made by closely winding turns of wire of diameter d (together with insulation) (Fig. 27.5). The wire carries a current i. Find the magnetic field induction in the centre O and on the end A of the solenoid. out the calculation for d =0.1 mm, i =5 A. |
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Answer» |
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| 12. |
The process of vector addtion of individual displacements of light waves at a point when two or more than two waves arrive at two points is |
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Answer» principleof superposition of waves |
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| 13. |
The P.D in volts between the points A and B is |
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Answer» 6V |
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| 14. |
A stationary shell of mass M explodes in to two parts and their masses are in the ratio 2:3, then the ratio of their deBroglie wavelengths is |
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Answer» `2:3` |
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| 15. |
What is a valence band ? |
| Answer» Solution :The ENERGY band FORMED by a series of energy LEVELS containing the valence electrons is known as valence band. | |
| 16. |
Figure 25-43 shows a 24.0 V battery and four uncharged capacitors of capacitances C_1=1.00 mu F, C_2=2.00 mu F C_3=3.00 mu F and C_4=4.00 mu F IF only switch S_1 is closed, what is the charge, on (a) capacitor 1 (b) capacitor 2 (c ) capacitor 3 and (d) capacitor 4? IF both switches are enclosed , what is the charge on (e) capacitors1.(f) capacitor 2, (g) capacitor 3, and (h) capacitor 4? |
| Answer» Solution :`a)18.0 MU C b) 32.0 mu C c) 18.0 mu C d)32.0 mu C E) 16.8 mu C f) 33.6 mu C g) 21.6 mu C H) 28.8 mu C` | |
| 17. |
State the factors on which the refractive index of a material medium for a given wavelength depends. |
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Answer» Solution :Factors are : (i) MAGNETIC PERMEABILITY of the MEDIUM (II) Electricity permittivity of the medium |
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| 18. |
A particle starts from rest and has acceleration of 2 ms-2 for 10 s. After that it travels for 30 s. with constant speed and then undergoes a constant retardation of 4 ms and comes back to rest. The total distance covered by the particle is: |
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Answer» 650 m `v=u+at` `v_1=0+2xx10 =20 ms^(-1)` Also `s_1=(V_(1)^(2)-u_(1)^(2))/(2A)=(20xx20-0)/(2xx2)=100m` In SECOND case ,`v_(2)`be the final vel.Which remains constant `s_(2)=v_(2)xxt(2)` `=20xx30=600 m` In third case `v_(3)=0,u_(3)=20 ms^(-1)` `a=-4 ms^(-2)` Now `t=(v_(3)-u_(3))/(a)=(-20)/(-4)` TOTAL DISTANCE =`S_1+S_2+S_3`=100+600+50=750 m |
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| 19. |
Explain angle of dip. |
Answer» Solution :If a magnetic needle is perfectly balanced about a horizontal axis so that it can SWING in a plane of the magnetic meridian. The needle would make an angle with the horizontal. This is known as the angle of dip (also known as inclination).  Thus, dip is the angle that the total magnetic field `OVERSET(to) (B_E)` of the Earth makes with the SURFACE of the Earth.  Figure shows the magnetic meridian plane at a point P on the surface of the Earth. The plane is a section through the Earth. The magnetic field of Earth at point P is `overset(to) (B_E)` . There are two components of `overset(to) (B_E)` (1) Horizontal component `overset(to) (H_E)` (2) VERTICAL component `overset(to) ((Z_E))` The angle that `overset(to) (B_E)` makes with `overset(to) (H_E)` is the angle of dip I. In most of the northern hemisphere, the north pole of the dip needle tilts downwards. Likewise in most of southern hemisphere, the south pole of the dip needle tilts downwards. |
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| 20. |
Find the approximate length of the Golden Gate bridge if it is known that the steel in the roadbed expands by 0.53 m when the temperature changes from +2 to + 32^@C. |
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Answer» |
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| 21. |
A message signal is used to modulated a carrier signal of frequency 5 MHz and peak side-bands are produced seperated by 40 kHz. If the modulation index is 0.75 then the peak voltage and frequency of the messages singal, respectively are |
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Answer» 60 V, 10kHz and peak voltage , ` V_(c) = 40 V ` MODULATION index , ` mu = 0.75 ` ` thereforemu =(V_(m))/(V_(c)) "" [ becauseV_(m)` is peak voltage of message signal] ` 0.75 = (V_(m))/(40) ` ` rArr V_(m) = 40 xx 0.75 ` ` V_(m) = 30 V ` SINCE, difference of frequencies of TWO side bands is equal to the band width (2 fm). i.e., Band width = 40 kHz ` 2f_(m) = 40 kHz rArr f_(m) = 20 kHz ` Hence , the peak voltage and frequencey will be 30 V and 20 kHz . |
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| 22. |
Why are telescopes fitted with large aperture objective? |
| Answer» Solution :Greater the APERTURE, greater is the NUMBER of rays entering it. Thus, the image is brighter if the lens is of WISE aperture. | |
| 24. |
हेनरी का नियम मान्य है |
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Answer» सभी गैसों पर |
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| 25. |
The kinetic energy of alpha - particles emitted in the decay of ""_(88)Ra^(226) into ""_(86)Rn^(222) is measured to be 4.78 MeV. What is the total disintegration energy or the 'Q - value of this process ? |
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Answer» Solution :The STANDARD relation between the KINETIC ENERGY of the `alpha`-PARTICLE `(KE_alpha)` and the Q-value(or total disintegration energy ) is `KE_a=((A-4)/A).Q` `Q=(A/(A-4)).KE_a`. `=(226/(226-4))xx4.78` MeV= `226/222xx4.78` MeV Q=4.865 MeV `approx` 4.87 MeV |
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| 26. |
Which of the following pairs of space and time varying vecE and vecB fields would generate a plane electromagnetic wave travelling along the Z-direction ? |
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Answer» `E_(X), B_(x)` |
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| 27. |
Name the phenomenon which is responsible for bending of light around sharp corners of an obstacle. Under what conditions does this phenomenon take place ? Give one application of this phenomenon in everyday life. |
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Answer» Solution :DIFFRACTION, Condition : SIZE of the OBSTACLE sharpness should be comparable to the wavelength of the LIGHT falling. Application : The finite resolution of our eye. |
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| 28. |
Sound with intensity larger than 120 dB. appears painful to a person. A small speaker delivers 4W of audio output. How close can the person get to the speaker without hurting his ears? |
| Answer» SOLUTION :`1/sqrtpi m` | |
| 29. |
A motorcycle ( mass of cycle plus rider = 2.50 xx 10^(2) kg ) is travelling at a steady speed of 20.0 m/s. The force of air resistance acting on the cycle and rider is 2.00 xx10^(2) N. Find the power necessary to sustain this speed if the road is sloped upward at 3.70^(@) with respect to the horizontal . |
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Answer» `5.75xx10^(3)`W |
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| 30. |
Which one is not an e.m. wave. |
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Answer» X-RAYS |
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| 31. |
(A): The direction of velocity vector remains unchanged though the coordinate system is changed (R): The direction of real vector is indepen dent of coordinate system |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 32. |
A plane electromagnetic wave travels in free space along 2-axis. At a particular point in space, the electric field along y-axis is 9.3 V m^(-1). The magnetic induction (B) along z-axis is |
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Answer» `3.1 xx 10^(-8) T` `C=(E)/(B) IMPLIES B=(E)/(C)=(9.3)/(3 xx 10^(8))=3.1 xx 10^(-8) T` |
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| 33. |
What is the maximum and minimum possible resistance which can be determined using the PO Box shown in above . |
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Answer» <P>`1111 kOmega, 0.1Omega` `(X)_(min)=((Q)_(min))/((P)_(max))( R)_(min)=(10)/(1000)(10)/(10000)(1)=0.01Omega` . |
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| 34. |
The number density of free electrons in a copper conductor estimatedis 8.5 xx 10^(28) m^(-3). How long does an electron take to drift from one end of a wire 3.0 m long to its other end ? The area of cross-section of the wire is 2.0 xx 10^(-6) m^(2) and it is carrying current of 3.0 A. |
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Answer» Solution :If free electrons drift through length l of a given wire in time t then their drift SPEED is, `v_(d) = (l)/(l)` Now, CURRENT passing through given wire is, `I = Av_(d) ` ne `therefore I = A ((l)/(l)) ` ne `therefore t = ("Alen")/(I)` `therefore t = (2 xx 10^(-6) xx 3 xx 8.5 xx 10^(28) xx 1.6 xx 10^(-19))/(3)` `therefore t = 27200 s ` Note : If above time is asked in h, min and s then : `therefore t = (27200)/(3600) ` h `therefore t = 7.556 `h ` therefore t = 7 h + (0.556 xx 60) ` min ` = 7 h 33.36 ` min 7h 33 min + `(0.36 xx 60)` s `therefore t = 7 h 33 ` min 22 s |
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| 35. |
(A) Electrons in the atom are held due to coulomb forces (R) : The atom is stable only because the centripetal force due to coulomb's law is balanced by the centrifugal force |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 36. |
The Electric flux through the surface: |
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Answer» (A) in figure (iv) is the LARGEST. `therefore phi = (sumq)/(epsilon_(0))` `therefore phi = sumq` Thus, electric flux through a surface doesn.t depend on the shape, SIZE or area of a surface but it depends on the amount of charge enclosed by the surface. In given figures, the charge enclosed are same that means the electric flux through all the surfaces should be the same. Hence, option (D) is correct. |
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| 37. |
In a carburator of an engine, aire is drawn from atmosphere through section A with a velocity of 10m//s. The narrow section B is connected to a petrol tank as shown. The minimum ratio of area of cross section A to area of cross section B, so that the petrol can just enter the carburator tube is sqrt(66/x). Then the value of x is : (density of petrol =1000kg//m^(3), density of air =2kg//m^(3), g=10m//s^(2) atmospheric pressure =10^(5) Pa and assume density of air remain constant) |
| Answer» SOLUTION :NA | |
| 38. |
What is the de-Broglie wavelength of a neutron at absolute temperature T K ? |
| Answer» SOLUTION :`lambda=h/(SQRT(2m_nE_k))=h/sqrt(2m_n3/2k_BT)=h/sqrt(3m_nk_BT),K_B to`Boltzmann.s CONSTANT | |
| 39. |
Match the items in column I to the items in column II given below {:(,"Column-I",,"Column-II"),((A),"Planet moving around the sun in the elliptical orbit",(p),"Linear momentum is conserved/constant"),((B),"Planet moving around the sun in the fixed circular orbit",(q),"Angular momentum is conserved about one point in space."),((C ),"A particle of mass " m_(1) " is projected towards a stationary mass " m_(2)." For " m_(1) "and " m_(2) " as system",(r ),"Kinetic energy is conserved"),((D),"A particle of mass m is projected by making an angle " theta " from the surface of the earth with speed less than escape velocity",(s),"Total energy is conserved"):} |
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Answer» |
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| 40. |
A physical quantity Q is related to measurable quantities L,M and N as Q = KL^(-2)M^(2)N^(3), where k is constant. If the % errors in the measurements of L, M and N are respectively 1%, 2% and 1%, then the % error in hc mcasurement of Q is |
| Answer» ANSWER :A | |
| 41. |
The tension in a wire clamped at both ends is halved with out appreciably changing the wire length between the clamps. What is the ratio of the new to the old wave speed for transverse waves traveling along this wire? |
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| 42. |
Two wires A and B of same material having their lengths in the ratio 6:1 are connected in series. The p.d. across the wires are 3V and 2V respectively. If r_Aand r_Bare radii of A and B respectively, ther_B //r_Ais |
| Answer» ANSWER :B | |
| 43. |
Statement- 1 : The resistance of super-conductor is zero Statement- 2 : The super-conductors are used for the transmission of electric power |
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Answer» Statement - 1: is true ,Statement - 2: is true ,Statement - 2: is CORRECT explanation of statement - 1, |
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| 44. |
During average life (tau) of a radioactive material, the number of nuclei left intact is .................... where N_(0)= number of nuclides at time t =0. |
| Answer» SOLUTION :`N_(0)/E` | |
| 45. |
Statement-I : Cause of production of continuous X-rays is the loss in kinetic energy of electrons during collisions with different nuclei of target Statement-II : A decelerating charged particle radiates electromagnetic waves. |
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Answer» If both Statement- I and Statement- II are TRUE, and Statement - II is the CORRECT EXPLANATION of Statement– I. |
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| 46. |
How do you classify p-type semiconductors ? |
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Answer» |
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| 47. |
A small telescope has an objective lens of focal length 150 cm and and eye piece of focal length 5 cm. If his telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece. |
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Answer» Solution :`tan alpha=100/3000=1/30` radian `tan alpha=h/f_0` `1/30=h/150` `h=5 CM` h height of image of tower `m_e=(1+ alpha/f_e)=(1+25/5)=6` `m_e=h'/h` `h'=5 TIMES 6=30 cm` h height of final image |
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| 48. |
Which of the following relationship between the force F on a particle's position 'x' implies simple harmonic oscillation? |
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Answer» `F = -5X` |
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| 49. |
Does ultrasonic waves show polarisastion. |
| Answer» SOLUTION :No, ultrasonic waves are LONGITUDINAL in nature. They cannot be POLARISED. | |
| 50. |
A ground receiving station is receiving a signal at 6MHz transmitted from a ground transmitter at a height of 500m located at a distance of 100km. If radius of earth is 6.4xx10^(6)m, maxim um number density of electron in ionosphere is 10^(12)m^(-3). the signal is coming via: |
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Answer» ground wave `d=sqrt(2Rh)=sqrt(2xx(6.4xx10^(6))xx300)=80xx10^(3)m=80km` Since the distance between transm itter and receiver is 100km, HENCE for the GIVEN frequency signal of6 MHz. the propagation is not possible via space propagation. for sky wave propagation the critical frequency. `v_(C)=9(N_("MAX"))^(1//2)=9xx(10^(12))^(1//2)=9xx10^(6)Hz=9MHz` Since 6 MHz `lt9 MHz`. so the propagation of signal of frequency 6 MHzis possible via sky wave. |
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