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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 fission process the ratio is mass of the products/mass of the reactants is |
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Answer» EQUAL to 1 |
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| 2. |
If a current i ampere flows in a long straight thin walled tube, then magnetic induction at any point inside the tube is |
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Answer» infinite |
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| 3. |
An ac source of rms voltage V is put across a series combination of an inductor L, capacitor C and a resistor R. If V_(L), V_(C ) and V_(R) are the rms voltage across L, C and R respectively then why is V ne V_(L) + V_(C ) + V_(R) ? Write correct relation among V_(L), V_( C) and V_(R). |
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Answer» Solution :`V_(L), V_( C) and V_(R )` are not in the same phase `V_( L) + V_(C ) + V_(R ) GT V` |
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| 4. |
Calculate the magnetic field inside a solenoid when the number of turns is halved and length of the solenoid and the area remain the same. |
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Answer» Solution :MAGNETIC field inside a solenoid is `B=(mu_0 NI)/L` If the NUMBER of turns is HALVED `B.=(mu_0(N/2)I)/L rArr B.=1/2((mu_0NI)/L)` |
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| 5. |
Match the object natures in column I with the natures of their images formed by a plane mirror in column II |
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Answer» |
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| 6. |
Whichstatementaboutradioactiveradiationsis true |
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Answer» Speed of `alpha`-PARTICLES is a CHARACTERISTIC property |
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| 7. |
If an object is thrown straight upward with an initial speed of 8 m//s and takes 3 seconds to strike the ground, from what height was the object thrown ? |
Answer» Solution :The figure on the next page SHOWS that the displacement is down, so we call down the positive DIRECTION . Therefore, `a=+g and v_(0) = -8 m//s` (because up is the negative direction). We're given `a, v_(0)` and t, and we need to find `Delta s`. Since v is MISSING, we use Big Five #2. 
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| 8. |
When a large amount of current is passing through solenoid, it contract, explain why ? |
| Answer» Solution :Current in two consecutive turns being in same DIRECTION make them to FORM unlike poles TOGETHER hence, they ATTRACT each other. | |
| 9. |
An image of a bright square is obtained on a screen with the aid of a convergent lens. The distance between the square and the lens is 40 cm. The area of the image is nine times larger than that of the square. Select the correct statement(s), |
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Answer» IMAGE is FORMED at a distance of 120 cm from the lens |
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| 10. |
The structure of solids is investigated by using |
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Answer» COSMIC rays |
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| 11. |
In an oscillaing L.C circuit.the maximum charge on the capacitors is Q.The charge on capacitor when the energy is stored equally between the lectric and magnetic field is ……. |
| Answer» Answer :C | |
| 12. |
In young' s double slit experiment, for the light of wavelength lambda_(1) fringe y_(1) and for the light of wavelength lambda_(2) fringe width is y_(2) . If the whole arrangement is dipped into a liquid of refractive index muit is founded that for the wavelength lambda_(1)fringe width becomes y_(3) . Now the correct relation isFour light waves are represented by (i) y= a_(1) "sin" omega t""(ii) y= a_(2) "sin" ( omegat+e)(iii) y=a_(1) "sin" 2 (omegat) "" (iv) y= a_(2) "sin" 2 (omega t +e) |
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Answer» `y_(2)=y_(1)(lambda_(1))/(lambda_(2))` |
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| 13. |
A wheel with 10 metallic spokes each 0.5m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth's magnetic field H_(E) at a place. If H_(E)=0.4G at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1G=10^(-4)T. |
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Answer» Solution :The emf induced, `epsilon=((1)/(2))omegaBR^(2)=((1)/(2))xx4pixx0.4xx10^(-4)xx(0.5)^(2)=6.28xx10^(-5)V` [Note that the NUMBER of SPOKES is immaterial because the EMFS across the spokes are in parallel.] |
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| 14. |
Two capacitors of capacitance of 6 muF and 12 muF are connected in series with a battery. The voltage across the 6 muF capacitor is 2 V.Compute the total battery voltage. |
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Answer» Solution :In series ARRANGEMENT of CAPACITORS `V_1/V_2 = C_2/C_1` `rArr V_2 = (C_1 V_1)/(C_2) = (6muF xx 2V)/(12muF) = 1V` `:.` TOTAL voltage across the capacitors COMBINATION ` = V_1 + V_2 = 2 + 1 =3V` `:.` Battery voltage = 3 V 
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| 15. |
A magnetic needle of magnetic moment 0.9 Am^(2) and pole strength 5Am is pivoted so that it is free to turn in a horizontal plane at a position where the earth's B-field is 3.6 xx 10^(-5) tesla. It is in equilibrium at an angle 30^(@) from the magnetic meridian when it is pulley by a string attached to its north pole in the easterly direction. what is the tension of the string ? |
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Answer» |
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| 16. |
The coil of a galvanometer is 0.02 × 0.08 m2. It consists of 200 turns of fine wire and is in a magnetic field of 0.2 tesla. The restoring torque constant of the suspension fibre is 10^(–6) Nm per degree. Assuming the magnetic field to be radial. (i) What is the maximum current that can be measured by the galvanometer, if the scale can accommodate 30^@ deflection? (ii) What is the smallest, current that can be detected if the minimum observable deflection is 0.1^@? |
| Answer» SOLUTION :(i) `4.69xx10^(-4)` A, (II) `1.56 xx 10^(-6)` A | |
| 17. |
A projectile of mass m, charge Z, initial speed v and impact parameter b is scattered by a heavy nucleus of charge Z. Use angular momentum and energy conservation to obtain a formula connecting the minimum distance (s) of the projectile from the nucleus to these parameters. Show that for b = 0 , s reduces to the closest distance of approach r_(0). |
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Answer» SOLUTION :Fig. shows the conditions. Of the problem. Charge on the nucleus = Ze, Charge on the projectile = Z.e At infinity , the angular momentum of projectile about the nucleus = mvb When the projectile is at the minimum distance s from the nucleus, its velocity vector v. is normal to the radius vector (drawn from the centre of the nuclesu ). `therefore` Angular momentum of the projectile about the nucleus = mv.s. From the law of conservation of angular momentum , we have, mvb = mv.s or v. `= (vb)/(s) `(i) When the projectile is infinitely away from the nucleus, it has only kinetic energy (`because` P.E = 0 )given by, `E_(K) = (1)/(2) mv^(2)` when the projectile is at minimum distance s from the nucleus, it has both kinetic and potential energies given by, `E_(k)= (1)/(2) mv^(2) and E_(p) = (1)/(4 pi epsilon_(0)) . ("ZZ".e^(2))/(s)` From the law of conservation of energy, we have, `E_(k) + E_(p) = E_(k)` or `(1)/(2) mv^(2) + (1)/(4 pi epsilon_(0)) .("ZZ".e^(2))/(s) = (1)/(2) mv^(2)` putting v. = vb/s from eq. (i), we get, `(1)/(2) m (v^(2) b^(2))/(s^(2)) + (1)/( 4pi epsilon_(0)) .("ZZ".e^(2))/(s) = (1)/(2) mv^(2)` Dividing both sides by `mv^(2) `/2, we get, `(b^(2))/(s^(2)) + (1)/(4 pi epsilon_(0)) .("ZZ".e^(2))/(s(mv^(2)/2)) = " or " s^(2) = b^(2) + (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2)` This is the required formula. For a head- on COLLISION , b = 0. `therefore s^(2) = (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2) " or " therefore s = (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2)` Which is the distance of CLOSEST approach |
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| 18. |
In a reference frame Ka photon of frequency omega falls normally on a mirror approaching it with relativistic velocity V. Find the momentum inparted to the mirror during the reflection of the photon (a) in the reference frame fixed to the mirroe, (b) in the frame K. |
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Answer» Solution :(a) In the reference frame fixed to the MIRROR, the frequency of the photons is, by the Doppler shift formula `overset_(OMEGA) = omega sqrt((1+beta)/(1-beta)) (=omegasqrt(1-beta^(2))/(1-beta))` In this frame momentum imparted to the mirror is `(2cancelh overset_(omega))/(c) = (2cancelh omega)/(c) sqrt((1+beta)/(1-beta))`, (b) In the `K` frame, the incident particle CARRIES a momentum of `cancelh omega//c`and returns with momentum `(cancelh omega)/(c)(1+beta)/(1-beta)` The momentum inparted to the mirror, then, has the magnitude `(cancelh omega)/(c)[(1+beta)/(1-beta)+1] = (2cancelh omega)/(c)(1)/(1-beta)` Hence `beta = (V)/(c)`. |
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| 19. |
In a metre bridge, the balancing length from the left end is 20 cm. The resistance in the right gap is 1 Omega. The value of the unknown resistance is |
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Answer» `0.8 OMEGA` |
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| 20. |
A resistor of 6kOmega with tolerance 10% and another of 4kOmega with tolerance 10% are connected in series. The tolerance of combination is about : |
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Answer» `5%` Similarly `(DeltaR_(2))/(R_(2))xx100=10` `impliesDeltaR_(2)=(4xx10)/(100)=0*4kOmega` In series, `R_(s)=R_(1)+R_(2)=6+4=10kOmega` `DeltaR_(s)=DeltaR_(1)+DeltaR_(2)=0*6+0*4` `=1kOmega` `:.(DeltaR_(s))/(R_(s))xx100=(1)/(10)xx100=10%` Hence CORRECT CHOICE is `(b)`. |
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| 21. |
An equiconvex lens of focal length20 cm is cut into two equal halves perpendicular to the principal axis and kept at a separation of 10 cm co-axially. What ils the focal length of the system? |
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Answer» 35 cm |
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| 22. |
The turns ratio of a transformer used in a half wave rectifier ils 12:1. The primary is connected to the power mains 220V, 50Hz. Assuming the diode resistance in forward bias to be zero, calculate the d.c. voltage across the load. |
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Answer» |
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| 24. |
A student is given 4 identical batteries having emf 1.5 Veach and internal resistance of 0.1Omega each. The student is asked to connect them in assisting manner. By mistake he connects one battery in reverse way. The resultant emf and resultant internal resistance offered by the combination is .... |
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Answer» 3V, 0.2 `Omega`  Equivalent emf `esillon = 1.5 + 1.5 + 1.5 = 3.0 V ` and equivalent internal RESISTANCE r = `r_(1) + r_(2) + r_(3) + r_(4) ` = 0.1 + 0.1 + 0.1 + 0.1 = 0.4 `Omega` |
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| 25. |
Value of work function depend on…… |
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Answer» TYPE of METAL |
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| 26. |
What is the force of attraction between the two plates of a parallel plate capacitor? Assume that, area of each plate of the capacitor is A and one plate is charged with +Q and the other with -Q. |
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Answer» Solution :Intensity of electric field at a point near a CHARGED plate having surface DENSITY of charge `sigma` is, `E = (sigma)/(2in_0)` Now, `sigma = (Q)/(A)` `:. E = (Q)/(2A in_0)` MAGNITUDE of the charge on the other plate of the CAPACITOR = Q. So, the force experienced by the plate is, `F = QE = Q*(Q)/(2A in_0)= (Q^2)/(aA in_0)`. |
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| 27. |
(A) : TV waves become attenuated with increasing distance. (R) : The power radiated by TV transmitter varies inversely as the distance from the transmitter. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT explanation of 'A'. |
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| 28. |
Two plane mirrors are placed at some angle. There are five images formed when an object is placed symmetrically between them . Find the angle between the mirrors. |
| Answer» ANSWER :A | |
| 29. |
B .E. of a satellite is always |
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Answer» infinity |
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| 30. |
Which of the following relations represent Biot-Savart's law? |
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Answer» `d VEC(B) = (mu_(0))/(4pi) I (d vec(l) xx vec( r))/(r )` |
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| 31. |
If a particle is thrown horizontally at a speed of3 xx 10^8 ms^(-1)deduce the vertical fall in traveling 1 km distance. Given : g = 10 ms^(-2). Does that result depend upon the mass of the particle? Comment on your result, considering that Newton thought light is made up of corpusclesthat at a very large speed by the source. |
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Answer» Solution : If y be the vertical fall in time t, then `y = 1/2 "gt"^2 = 1/2 xx 10 ( (1)/( 3xx 10^5))^2 m = 5.5 xx 10^11 m` This result is INDEPENDENT of the mass of the particle. As it is clear from the above result, the downward DISTANCE covered by the CORPUSCLE is only `5.5 xx 10^(-11)`m for 103 m of horizontal distance covered by the corpuscle. So, we can neglect the effect of earth's gravitation and safely assume that the corpuscles TRAVELS along a straight line. |
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| 32. |
In a p - n junction diode, the increase in forward voltage of 0.19 V increases the forward current by 37.6mA. The dynamic resistance of the junction is : |
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Answer» 2 ohm |
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| 33. |
Approximate value of wavelength of electron waves in Division experiment at maximum diffraction is : |
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Answer» `1.67 A^@` |
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| 34. |
A moving-coil galvanometer of resistance 200 ohms gives full scale deflection of 100 divisons for a current of 50 milliamperes. How will you convert it into an ammeter to read 2 amperes for 20 divisons? |
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Answer» Solution :Data: `G=200 Omega, I_(g) =50 mA = 50 xx 10^(-3)A` The total NUMBER of scale divisions is 100. The ammeter has to read 20 divisons for a CURRENT of 2A. HENCE, for 100 divisions, the current MUST be I=10A. To convert the galvanometer into an ammeter, a low resistance (shunt) must be connected in parallel with the galvanometer coil. The required resistance, `S=(I_(g)/(I-I_(g)))G` `=(50 xx 10^(-3))/(10-50 xx 10^(-3)) xx 200` `=(50 xx 10^(-3))/((10000 -50) xx 10^(-3)) xx 200` `=(200)/(200-1)= 200/199 = 1.005 Omega` |
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| 35. |
Draw a graph showing variation of photoelectric current (I) with anode potential ( V ) for different intensities of incident radiation. Name the characteristic of incident radiation that is kept constant in this experiment. |
| Answer» SOLUTION :The frequency of INCIDENT radiation was KEPT CONSTANT. | |
| 36. |
A sphere is suspended from the end A of a rod AB and the rod is balanced by placing weights on a scale pan suspended from the end B. AB = 22 cm. The flucrum of the balance is at a distance of 2 cm from the end A and 20 cm from the end B. Another sphere is placed directly below the sphere suspended from A. The distance between the centres of the two spheres is 2 cm. The two spheres are now charged with +100 esu. To make the rod AB balanced again, what should be the change of weight in the scale pan? |
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Answer» |
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| 37. |
The forbidden gap in silicon crystal is |
| Answer» Answer :D | |
| 38. |
Assertion : If two long wires, hanging freely are connected to a battery in series, they come closer to each other. Reason : Force of attraction acts between the two wires carrying current. |
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Answer» If both ASSERTION and REASON are TRUE and the reason is the CORRECTEXPLANATION of the assertion. |
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| 39. |
In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole? |
| Answer» Solution :A COMPASS FREE to move in a vertical plane, WOULD stand EXACTLY vertical on the geomagnetic north or south pole. It is so because angle of DIP at poles is `90^@` and earth.s magnetic field is exactly in vertical direction. | |
| 40. |
A wave is represented by the equation : y = A sin(10 pi x + 15 pi t + pi//3)where, x is in metre and t is in second. The expression represents. |
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Answer» a WAVE travelling in the positive x-direction with a VELOCITY 1.5m/s |
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| 41. |
A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle 2 theta . The earth's magnetic field component in the direction perpendicular to swing is B . Maximum potential difference induced across the pendulum is |
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Answer» `2BL "SIN"((THETA)/(2))(gL)^(1//2)` |
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| 42. |
(A): More energy is released in fusion than fission. (R): More number of nucleons take part in fission. |
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Answer» Both .A. and .R. are TRUE and .R. is the correct explanation of .A. |
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| 43. |
The output of a step down transformer is measured to be 24 V when connected to a 12 watt light bulb. The value of the peak current is |
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Answer» <P>`1/sqrt(2)` A `therefore` Value of PEAK current `I_(m) = sqrt(2) I = sqrt(2) xx 0.5 = 1/sqrt(2)` A. |
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| 44. |
A projector lamp can be used at a maximum voltage of 60V, its resistance is 20Omega, the series resistance (in ohm) required to operate the lamp from a 75 V supply is |
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Answer» 2 |
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| 45. |
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Omega , what is the maximum current that can be drawn from the battery ? |
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Answer» Solution :Consider a battery with EMF e and INTERNAL resistance r, connected with external resistance R. Applying KVL in above CLOSED loop, we get - IR- Ir = -` epsilon` `therefore IR + Ir = epsilon` `therefore I(R + r) = epsilon ` `therefore I = (epsilon)/(R + r) ""` .... (1) For a given battery, values of £ and r are constants. Here as R decreases I increases.Hence , when R = minimum = 0,we get I = maximum= `I_(MAX)`, thus , `I_(max)= (epsilon)/(r)"" `... (2) `therefore I_(max) = (12)/(0.4) =30` A |
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| 46. |
When ac source V = 220 sin (100 pi t) volt is connected to a parallel plate capacitor, displacement current is obtained maximum equal to 7600 mA. If diameter of plate of a capacitor is 36 cm then find perpendicular distance between the plates of this capacitor. |
| Answer» SOLUTION :`d=0.0082 m` | |
| 47. |
A particle of mass 120 g moving at speed of 750 cm/s is acted upon by a variable force opposite to its direction of motion. If the velocity of the particle becomes 250 cm/s along the direction of force, find the value of time 't' for which acted. |
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Answer» SOLUTION :Area of Ft GRAPH = IMPULSE = change in MOMENTUM `(1)/(2)(t+6)xx10^(-2)xx10` `=120xx10^(-3)(7.5+2.5)` `therefore t=18s` 
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| 48. |
Power radiated by a transmitting antenna is proportional to |
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Answer» `(lambda/l)^2` |
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| 49. |
A satellite is moving in a circular orbit at a certainheight above the earth's surface. It takes 5.26xx10^3 s to complete one revolution with a centripetal acceleration equal to 9.32 m s^(-2). The height of the satellite orbit above the earth's surface is (Take radius of earth =6.37xx10^6 m ) |
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Answer» Solution :As, `T=2pisqrt(((R+h)^3)/(GM))` `T^2/(4pi^2)=(R+h)^3/(GM)` …(i) Centripetal acceleration , `a=(GM)/(R+h)^2` `(R+h)^2/(GM)=1/a` `(R+h)=T^2/(4pi^2)XXA` [Using (i)] `=(5.26xx10^3 // 2pi)^2 xx 9.32=6.54xx10^6` m `therefore h=6.54xx10^6 - R = 6.54xx10^6 - 6.37xx10^6` `= 0.17xx10^6` m = 170 km |
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| 50. |
A population inversion for two energy levels is often described by assigning a negative Kelvin temperature to the system. What negative temperature would describe a system in which population of the upper energy level exceeds that of the lower by 10% and the energy difference between the two levels is 2.2 eV? |
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Answer» Solution :As `N_(X)=N_(0)e^(-(E_(x)-E_(0))//kT) :. (N_(x))/(N_(0))e^(-(E_(x)-E_(0))//kT)` `ln ((N_(x))/(N_(0)))=-(E_(x)-E_(0))/(kT)` or `-T=(E_(x)-E_(0))/(K ln((N_(x))/(N_(0))))` Now, `E_(x)-E_(0)=2.2eV and (N_(x))/(N_(0))=1.10, k=8.62xx10^(-5)eV//K` `:. -T=(2.2eV)/((8.62xx10^(-5)eV//K)ln(1.10))=2.67xx10^(5)K or T=-2.67xx10^(5)K` |
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