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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. |
Out of the following which one is not a possible energy for photon to be emitted by hydrogen atom according to Bohr's atomic model? |
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Answer» `1.9eV`  So 11.1 EV ENERGY is not POSSIBLE. |
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| 2. |
A body is moving in a circular orbit. It is just about to slide to the outer side and mu mg = (mv^(2))/(l). In this expression, mu represents |
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Answer» Coefficient of STATIC friction |
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| 3. |
A wheel isrotating at 60 rotations per minute. If 480 J of energy is required to double its rotational speed, calculate the moment of inertia of the wheel. |
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Answer» Solution :Data :' ` f_(1) = 60 rpm = 1Hz, f_(2) =2f_(1)= 2 Hz, W = 480 J` ` E_("rot") = 1/2 Iomega^(2) "" therefore W= 1/2 I (omega_(2)^(2) - omega_(1)^(2))` ` W = 1/2 I (2 pi F _(2))^(2) -(2pif_(1))^(2) = 2 pi^(2) I (f_(2)^(2) -f_(1)^(2))` ` I = W/(2pi^(2) (f_(2)^(2) -f_(1)^(2))) = 480 /(2pi^(2) (4-1)) = 80/ pi^(2)= 80/((4.142)^(2))= 8.104 kg.m^(2)` This is the moment of INERTIA of the wheel. |
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| 4. |
When charge is at rest then it creates only electric field not the magnetic field. A moving charge generates electric as well as magnetic field around it. If charge is moving with constant velocity then electric and magnetic field do not change with time, hence, it cannot produce electromagnetic wave. When charge is accelerated then time-varying electric and magnetic fields are created and thus electromagnetic wave is produced by accelerated charge. We know that the current-carrying wire creates magnetic field around it. If alternating current is flowing through the wire then it creates variable magnetic field around it and thus starts radiating electromagnetic waves. When a charged capacitor is connected across an inductor then charge starts oscillating between the plates of the capacitor. Normally one plate of the capacitor is connected to Earth and the other plate is connected to an antenna. Antenna radiates electromagnetic wave. Hertz successfully produced electromagnetic waves using such LC oscillator. Speed of electromagnetic wave in vacuum is constant and equal to 3 xx 10^8 m/s. But inside a medium the speed changes according to electrical and magnetic properties of the medium. Whichof the following cannotcreateelectromagneticwave ? |
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Answer» Wirecarryingalternatingcurrent |
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| 5. |
A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. a. What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero? b. What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 ms^2? |
| Answer» SOLUTION :`BI L =MG , B 0.26 T ""b. 2 mg =1.176 N` | |
| 6. |
COLOR OF FERROUS SULPHATE IS |
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Answer» GREEN |
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| 7. |
Explain how transistor can be used as a switch ? |
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Answer» Solution :To understand the operation of transistor as a SWITCH. (i) As long as `V_(i) ` is low and unable to forwardbias the transistor , `v_(0)` is high ( at `V _(C C)`. (ii) If `V_(i)` is high enough to drivethe transistor into saturation then `V_(0)` is low, very nearto ZERO.  (iii) When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on. (ii) We can say that a low INPUT to the transistorgives a high output and high input gives a low output. (v) When the transistor is used in CUTOFF ( or ) saturation state it acts as a switch.  |
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| 8. |
(A) : A car can run on roadbecauseof forceapplied by roadon the car . (R ) : Friction provides the necessaryforcefor translatorymotion for a car startingform rest . |
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Answer» Both (A) and (R ) are TRUE and (R ) is THECORRECT EXPLANATION of (A) |
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| 9. |
The figure shows a system of two concentric spheres of radii |
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Answer» `(r_(1)r_(2))/((r_(2)-r_(1)))` The radial rate of flow of heat through this shell in steady state will be `H=(dQ)/(dt)=-KA(dT)/(dr)=-K(4pir^(2))(dT)/(dr)` `rArrint_(r_(1))^(r_(2))(dr)/(r^(2))=-(4piK)/(H)int_(T_(1))^(T_(2))dT` This on integration and simplification gives, `H=(dQ)/(dt)=(4piKr_(1)r_(2)(T_(1)-T_(2)))/(r_(2)-r_(1))` 
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| 10. |
The power delivered by the source of a LCR circuit is maximam when' |
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Answer» `L_omega=C_omega` |
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| 11. |
If the wave length of light in an experiment on photo electric effect is doubled keeping the intensity constant |
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Answer» The photo ELECTRONS may or may not be EMITTED |
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| 12. |
What is electromagnetic damping? |
| Answer» Solution :GRADUAL reduction in AMPLITUDE of vibration of a metallic body in a magnetic field on account of FORMATION of EDDY currents is called the electromagnetic DAMPING. | |
| 13. |
A parralel beam of monochromatic light of wavelength 5000 Å is incident normally in a narrow slit of width 0.25 mm. The diffraction pattern is observed on a screen placed at a focal lens of a convex lens placed closed to the slit between slit and screen. Find the angular seperatiion between the first secondary maxima on either side of the central maximum. |
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Answer» |
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| 14. |
What was the most important conclusion of Rutherford's model of atom ? |
| Answer» SOLUTION :Every atom CONSISTS of a TINY central core, called nucleus which contains all of ther atom.s positive charge and most of it.s mass `(99.9%)`. | |
| 15. |
At time t = Os a particle starts moving along the x-direction. If its kinetic energy increases uniformly with time Y', the net force acting on it must be proportional to : |
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Answer» constant `1/2mv^2 prop t` or `1/2mv^2=kt`, where .k. is a constant. If p = MV is the linear MOMENTUM of the body,then `(p^2)/(2m)=kt` or `p=sqrt(2mk)xxt^(1/2)`or `F=(dp)/(dt)=sqrt(2mk)xx1/2xxt^(1/2)` or `F prop (1)/(sqrtt)`. |
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| 16. |
The electric current flowing in a wire in the direction from B to A is decreasing . Find out the direction of the induced current in the metallic loop kept above the wire as shown. |
| Answer» SOLUTION :CLOCKWISE. | |
| 17. |
निम्नलिखित में से कौन-सा एक रासायनिक परिवर्तन है? |
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Answer» दूध का खट्टा होना |
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| 18. |
भूपर्पटी पर सर्वाधिक मात्रा में पाया जाए वाला तत्त्व |
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Answer» एल्युमीनियम |
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| 19. |
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 Twith its plane perpendicular to the field. The emf induced across the radius of the dics is |
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Answer» `PI/10 V` |
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| 20. |
A circular hole is made on a copper disc. The diameter of this hole is 10 cm at 27^(@) C . If disc is heated upto 227^(@) C than calculated the change in diamter of hole . (Given coefficient of linear expansion of copper is 1.7 xx 10^(-5) ""^(@) C^(-1)) |
| Answer» SOLUTION :`3.4 XX 10^(-2) CM` | |
| 22. |
In previous question, the time of flight of the body is |
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Answer» `(2U sin THETA)/G` |
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| 23. |
A transistor is operated in common emitter configuration at V_c = 2 V, such that a change in the base current from 100 µA to 200 µA produces a change in the collector current from 5 mA to 10 mA. The current gain is |
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Answer» 100 |
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| 24. |
An eye has a defect of myopia if it does not see |
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Answer» NEAR OBJECTS |
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| 25. |
The clocks in moving ships will appear to go "_____________" clocks on earth |
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Answer» FASTER than |
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| 27. |
(A) : Earth's magnetic field does not affect the working of a moving coil galvanometer. (R) : The earth's magnetic field is quite weak as compared to magnetic field produced in the moving coil galvanometer. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 28. |
What is dimensional formula of potential ? |
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Answer» |
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| 30. |
Two parallel conductors A and B separated by 15 cm carry electric current of 9 A and 4 A in the same direction. Find the point between A and B where the field is zero. |
| Answer» SOLUTION :10.38 CM from A | |
| 31. |
In the circuit the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance R_G = 60.00 Omega, (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance r_s= 0.02 Omega , (c ) is an ideal ammeter with zero resistance? |
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Answer» Solution :(a) Total resistance in the CIRCUIT is `R_G + 3 = 63 Omega`. HENCE `I = 3/63 = 0.048 A` (B) Resistance of the galvanometer converted to an ammeter is `(R_G r_s)/(R_G + r_s) = (60 Omega xx 0.02 Omega)/((60+ 0.02) Omega) ~~ 0.02 Omega` Total resistance in the circuit is `0.02 Omega + 3 Omega = 3.02 Omega. ` Hence I = 3/3.02 = 0.99 A (c ) For the ideal ammeter with ZERO resistance . I = 3/3 = 1.00 A |
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| 32. |
The frictional force between two. surfaces is independent of |
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Answer» NATURE of SURFACE |
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| 33. |
Radius of gyration of the disc about a transverse axis passing through its centre is 14.14 cm. Then its radius of gyration about its diameter is |
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Answer» 10 cm |
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| 34. |
(A) : A moving charge is a source of magnetic field. (R) : A current element is a source of magnetic field . |
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Answer» Both 'A' and 'R' are true and 'R' is the correct EXPLANATION of 'A'. |
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| 35. |
Explain surface wave propagation of radio waves. Why does this method fail in case of short waves ? |
| Answer» Solution :GROUND WAVES are the waves which propagate through earth from one point to another. These waves are actually GUIDED by the earth and follows its curved SURFACE from transmitter to receiver. | |
| 36. |
An object is placed 21 cm in fron of a concave mirror of radis of curvature 20 cm.A glass slab of thicknes 3 cm and refractive index 1.5 is palced close to the mirror in the space between the object and the mirror. Find the position of the final image fromed. The distance of the nearer surface of the slab from the mirror is 10 cm. |
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Answer» <P> Solution : Shift by GLASS slab `=3(1-(1)/(3//2))=1cm` The MIRROR will be shifted towards slab by `1cm`. Distance of object O from VIRTUAL mirror `A^'P^'B^'=20cm`. Image of O is formed on O itself by `A^'P^'B^'` (since O is at centre of curvature of mirror `A^'P^'B^'`. Distance of image from original mirror APB is `21cm`. In fact, object and image COINCIDE. |
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| 37. |
Explain polarisation of polar molecule in uniform electric field. |
Answer» Solution :The permanent dipole moment of polai molecules is zero because the differen permanent dipoles are oriented randomly due to thermal agitation, so the total dipole moment is zero. When an external field is applied the individual dipole moments tend to ALIGN with the field. When summed over all the molecules there is then a net dipole moment in the DIRECTION of the external field means the dielectric is polarized which is shown in above FIGURE. The EXTENT of polarization depends on the relative strength of two mutually opposite FACTORS. The dipole potential energy in the external field tending to align the dipoles with the field and thermal energy tending to disturb the alignment. Generally, the alignment effect ls more important for polar molecules. |
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| 38. |
A coil of copper wire is connected in series with a bulb, a battery and a switch. When the circuit is completed thebulb lights up immediately . The circuit is switched off and a rod of soft iron is placed inside the coil . On completing the circuit again . It is observed that :- |
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Answer» Bulb is not so BRIGHT |
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| 39. |
The electron emitted in beta radiation originates from |
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Answer» INNER ORBITS of atom |
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| 40. |
A circuit, containing an 80 mH inductor and a 250 uF capacitor in series, is connected to a 240 V, 100 rad s^(-1) supply. The resistance of the circuit is negligible. (i) Obtain rms value of current. (ii) What is the total average power consumed by the circuit ? |
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Answer» Solution :Here L = 80 mH = 0.08 H, C = 250 `muF = 250 xx 10^(-6) F, R =0, V_(rms) = 240 V` and `omega = 100 rad s^(-1)` (i) `therefore` Impedance of CIRCUIT `Z = sqrt(R^(2) + (X_(L)-X_(C))^(2)) = sqrt(R^(2) + (L omega - 1/(C omega)^(2))` `=sqrt(0 + (0.08 xx 100 -1/(250 xx 10^(-6) xx 100)^(2))) = (8-40) = 32 Omega` `therefore` rms value of current `I_(rms) = V_(rms)/Z = 240/32 A = 7.5 A` (ii) The power factor `cos PHI = R/Z = 0/(32 Omega) =0` Since power factor of circuit is zero, hence the total average power consumed by the circuit is zero. |
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| 41. |
A cyclist speeding at 4.5 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the road and the tyres is 0.1. (i) Will he slip while taking the turn? (ii) Will he slip if his speed is 9 km/h? |
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Answer» SOLUTION :Frictional force PROVIDES the necessary CENTRIPETAL force. He will slip if the TURN is too sharp (i.e., too small a radius) or if his SPEED is too large. Maximum speed for not slipping is `v_("max")=sqrt(mu_(s)Rg)=sqrt(0.1xx3xx9.8)=1.71 m//s` (i) If `v=4.5 km//h=(5)/(4)m//s=1.25m//s,` he will not slip. (ii) If `v=9 km//h =(5)/(2) m//s=2.5m//s,` he will slip. |
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| 42. |
Answer the question regarding earth's magnetism: Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all? |
| Answer» Solution :(f) Besides MAGNETIC dipole FIELD, Earth ALSO has many local N-S poles due to DEPOSITS of magnetic mineral in its GROUND. | |
| 43. |
A small body of mass 0.1 kg isundergoing S.H.M. of amplitude 1 metre and period 0.2 s. The maximum force acting on it is nearly : |
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Answer» 99 N `:. F_("max")=ma_("max")=(4pi^(2)mr)/(T^(2))=(4pi^(2)xx0.1xx1)/((0.2)^(2))` = 99 N approx. THUS correct choice is (a). |
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| 44. |
A long straight wire carries a current I_(0). At distance a and b =3a from it there are two other wires, parallel to the former one, which are interconnected by a resistance R ( figure ). A connector slides without friction along the wire with a constant velocity v. Assuming the resistance of the wire, the connector, the sliding contact, and the self-inductance of the frame to be negligible , The point of application (distance from the long wire ) of magnetic force on sliding wire due to the long wire is (2a)/(lnx) from long wire. Then findout value of x. |
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Answer» |
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| 45. |
An object is subjected to an acceleration a = 4 + 3v. It is given that the displacement S = 0when v=0. The value of displacement when v = 3m//s is { 1- ( 4)/( 9) ln ( k ) }. The value of k is |
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Answer» |
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| 46. |
Two small drops of mercury, each of radius R coalesce to form a single large drop. The ratio of the total surface energies before and after the change is : |
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Answer» `1:2^((1)/(3))` |
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| 47. |
A proton, a deuteron and an a particle having same momentum enter a uniform magnetic field at right angles to the field. Then the ratio of their angular momenta during their motion in the magnetic field is |
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Answer» `2:2:1` |
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| 48. |
The range of voltmeter is 10 V and its internal resistance is 50 Omega. To convert it to voltmeter of range 15 V, how much resistance is to be added ? |
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Answer» Add `25 Omega` RESISTOR in parallel |
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| 49. |
Explain how current is formed. Give suitable example. |
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Answer» Solution :`rArr` CHARGE on particle is responsible for electric force between two particles. `rArr` Electric charge is of two type :(1) POSITIVE and (2) Negative `rArr ` In nature also in certain condition electric current is produced. For example lightening in the SKY. During this there is no steady flow of charge but changes unevenly. Sometimes lightening is dreadful. `rArr`In devices used by us in day to day LIFE there is steady flow of charge. Torch and cell driven clock are examples of this. |
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| 50. |
A point mass in and charge q is connected with a spring of negligible mass with natural length L. Initially spring is in its natural length. Now a horizontal uniform electric field E is switched on as shown. Find a) the maximum separation between the mass and the wall b) Find the separation of the point mass and wall at the equilibrium position of mass c) Find the energy stored in the spring at the equilibrium position of the point mass. |
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Answer» Solution :At maximum separation, velocity of point mass is zero. From work energy THEOREM, `W_("spring") + W_("fileld") = 0` `q Ex_0 - 1/2 kx_0^2 = 0 ` (`x_0` is maximum elongation) `implies x_0 = (2qE)/(K)"" :. ` separation = `L + (2qE)/(k)` b) At equilibrium position, `Eq = kx implies X = (qE)/(k) implies ` separation = `L + (qE)/(k)` C) `U = 1/2 kx^2 = 1/2k ((qE)/(k))^(2) = (q^2 E^2)/(2k)` . |
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