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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. |
Satellite communication is a mode of communication of signal between transmitter and receiver through satellite. The satellite communication is like the line of sight microwave communication. A communication satellite isa space craft placed in an orbit around the earth provided with microwave transmitting and receiving equipment called Radio transponder. The range of microwave frequencies used in satellite communication for uplink is 5.925 GHz "to" 6.425 GHz and for downlink is 3.7 GHz "to" 4.2 GHz. (i) How is the line of sight microwave communication possible through satellite? (ii) Why can't a single satellitecover the whole earth for microwave communication? (iii) What is the practical utility of satellite communication? |
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Answer» Solution :(i) The line of sight communication stands for the transmitter and receiver in sight of each other. The line of sight microwave communication is POSSIBLE if the communication satellite is always at a fixed locations with respect to EARTH, i.e., the satellite which is acting as a radiotransponder must be at rest with radiotranspondermust be at rest with respect to earth. it is so for a statellite KNOWN as geo-stationary satellite. (ii) A single satellite cannot cover the whole earth because, the LARGE part of the earth is out of sight due to curvature of the earth. The satellite communication is like the line of sight communications. (iii) In order to have microwave communication link over the ENTIRE globe of earth, at least three geostationary satellites are required, which are 120° apart from one another. The satellite communication is also used effectively in mobile communication especially in remote hilly areas. It is highly economical compared to other communication systems. |
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| 2. |
a. Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. b. How does this magnetic energy compare with the electrostatic energy stored in a capacitor? |
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Answer» SOLUTION :a. Work done = Energy spent in ESTABLISHING current 1 in a solenoid is `U_(B)=(1)/(2)LI^(2)` `=(1)/(2)L((B)/(mu_(0)n))` (since `B=mu_(0)nI`, for a solenoid) `=(1)/(2)(mu_(0)n^(2)Al)((B)/(mu_(0)n))^(2)` (from EQ. `6.9.2(3))=(1)/(2mu_(0))B^(2)Al` b. The MAGNETIC energy per UNIT volume V is, `u_(B)=(U_(B))/(V)=(U_(B))/(Al)=(B^(2))/(2mu_(0))` We have already obtained the relation for the square of the field strength. |
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| 3. |
Which of the objects below best serves as a symbol of an Indian woman's suhag'? |
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Answer» bindi |
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| 4. |
Describe the coil and bar magnet experiment to demonstrate the phenomenon of electromagnetic induction. |
Answer» Solution :(a) Magnet and coil experiment: If one END of a magnet is suddenly introduced into a coil connected to a sensitive GALVANOMETER it shows a MOMENTARY deflection This shows that a momentary current and hence an INSTANTANEOUS emf is produced in the circuit and the direction of the current depends on the pole towards the coil. The current produced is called induced current. If the magnet is withdrawn suddenly, a sudden deflection is again observed but in the opposite direction. The deflection is greater if the magnet is introduced or withdrawn quickly. No induced emf is obtained when there is no relative motion between the coil and the magnet.  (b) Coil and coil experiment: Consider two coils Pand Placed close to each other Coil P connected to a cell through a key Sw is called the primary coil. The other coil S connected in series with a sensitive galvanometer G is called the secondary coil. On pressing Sw a momentary deflection is noticed in the galvanometer, When the key is released the galvanometer deflects in the opposite direction. However, no induced emf is obtained when a steady current flows through the primary, as LONG as the Secondary coil is of rest. 
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| 5. |
The meter bridge can be used a) to determine an unknown resistance b) to compare two resistances c) to determine the specific resistance of the material of a wire. |
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Answer» Only (a), (C) are CORRECT |
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| 6. |
The specific characteristic of a common emitter amplifier is |
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Answer» HIGH INPUT resistance |
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| 7. |
A solid of constant heat capacity 1J"/"^(@)C is being heated by keeping it in contact with reservoirs in two ways : (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount ofheat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat. In both the cases body is brought from initial temperature 100^(@)C to final temperature 200^(@)C. Entropy change of the body in the two cases respectively is: |
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Answer» IN2, In2 Case (ii) `int dS =C [overset(385.5)underset(373)int (dT)/(T)+overset(398)underset(385.5)int (dT)/(T)+overset(410.5)underset(398)int (dT)/(T) +overset(423)underset(410.5)int (dT)/(T)` `overset(435.5)underset(423)int (dT)/(T)+overset(448)underset(435.5)int (dT)/(T)+overset(460.5)underset(448) int (dT)/(T)+overset(473)underset(460.5)int (dT)/(T)]=" In (473/373)"` None of the choice is correct. If GIVEN TEMPERATURE are in Kelvin then answer will be (a). |
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| 8. |
A cylinder is filled with water of density p upto a height h. If the beaker is at rest, the average pressure at the walls is: |
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Answer» 0 Thus CORRECT choice is (C). |
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| 9. |
Discuss Earth's magnetic field in detail. |
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Answer» Solution :(i) Gover suggested that the Earth's magnetic field is due to HOT rays coming out from the Sun. (II) These rays will heat up the air near EQUATORIAL region. (iii) Once air becomes hotter , it rises above and will movetowards northern and southern hemispheres and get electrified. (iv) This may be responsible to magnetize the ferromagnetic materials near the Earth's surface. (v) Till date, so many theories have been proposed. (vi) But none of the THEORY completely explains the cause for the Earth's magnetism. The north pole of magneticcompass needle is attracted towards the magnetic south pole of the Earth which is near the geographic north pole. (vii) Similarly , the south pole of magnetic compass needle is attracted towards the geographic north pole of the Earth which is near magnetic north - pole. (viii) The branch of physics which deals with the Earth's magnetic field is called Geomagnetism or Terrestrial magnetism. (ix) There are three QUANTITIES required to specify the magnetic field of the Earth on its surface, which are often called as the elements of the Earth's magnetic field They are (a) magnetic declination (D) (b) magnetic dip or inclination (I) (c ) the horizontal component of the Earth's magnetic field `(B_(H))` The angle subtended by the Earth's total magnetic field `vecB`with the horizontal direction in the magnetic meridian is called dip or magnetic inclination (I) . The component of Earth's magnetic field along the horizontal direction in the magnetic meridian is called horizontal component of Earth's magnetic field , denoted by `B_(H)`. |
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| 10. |
In RLC circuit the emf may _____ or may be in with current depending upon the value of circuit elements. |
| Answer» SOLUTION :Lead/lag, PHASE | |
| 11. |
We have two capacitors C_1 and C_2of capacitance 12 muF and 6 muF respectively. As shown in the given circuit arrangement the two capacitors are joined to a power supply of 6 volts. Initially the switch S_1 is closed but switch S_2 is open. After some time, the switch S_1 is opened and simultaneously switch S_2 is closed. Now answer the following questions : The final value of charge on capacitor C_2 is |
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Answer» `24 muC` |
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| 12. |
A spring of mass m lies on a smooth table. One, end of it is clamped to the vertical wall and other end is attached to a mass m ( as shown in fig) and is placed on a horizontalfrictionlesssurface. If the block of mass M is given a velocity v_(0) towards right find the K.E. of the system . ( AsSigmae velocity in spring is varying linearly). |
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Answer» |
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| 13. |
We have two capacitors C_1 and C_2of capacitance 12 muF and 6 muF respectively. As shown in the given circuit arrangement the two capacitors are joined to a power supply of 6 volts. Initially the switch S_1 is closed but switch S_2 is open. After some time, the switch S_1 is opened and simultaneously switch S_2 is closed. Now answer the following questions : The final value of potential of capacitor C_1 is |
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Answer» 6 V |
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| 14. |
In an L–C circuit, current is oscillating with frequency 4 xx 10^6 Hz. What is the frequency with which magnetic energy is oscillating? |
| Answer» SOLUTION :`upsilon_(m) = 2 UPSILON=8 XX 10^6` HZ. | |
| 15. |
A screen is placed 80 cm from an object. The image of the object on the screen is formed by a convex lens placed between them at two different locations separated by a distance 20 cm. Determine the focal length of the lens. |
Answer» Solution :We can obtain image of a given point OBJECT O at another fixed point I for two different positions of a convex LENS provided that values of u and v are interchanged in two positions as SHOWN here.  Thus in PRESENT question `OI = |mu | + v|=80 cm =x +(x+20)=2x+20` `implies x =30 cm ` and so `mu_(1)` =- 30 cm and `v_(1)=+50` cm `:. (1)/(f) = (1)/(v) -(1)/(u) = (1)/(+50)-(1)/(-30)=(1)/(50)+(1)/(30) = (80)/(1500) implies f= (1500)/(80) = 18.75 cm =19` cm |
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| 16. |
A ray is incident on a sphere of refractive index sqrt(2) sown in figure. Angle of refraction of the ray inside sphere is 30^(@). If total deviation suffered by the ray is 30x^(@), then find x |
Answer»  TOTAL deviation `=(45.^(@)-30.^(@))+180.^(@)-2(30.^@)+(45^(@)-30^(@))` `(30^(@)+120^(@)` `=150^(@)=30x` `x=5`. |
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| 17. |
We have two capacitors C_1 and C_2of capacitance 12 muF and 6 muF respectively. As shown in the given circuit arrangement the two capacitors are joined to a power supply of 6 volts. Initially the switch S_1 is closed but switch S_2 is open. After some time, the switch S_1 is opened and simultaneously switch S_2 is closed. Now answer the following questions : The fractional loss in electrostatic energy of the arrangement after closing the switch S_2 is |
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Answer» `1/4` `u_i= 1/2 C_1 V_1^2 = 1/2 XX (12 mu F) xx (6 V)^2 = 126muJ` and final value of electrostatic energy of the arrangement `u_f = 1/2 (C_1 + C_2) V^2 = 1/2 (12 muF+ 6 mu F) xx (4V)^2 = 144 MUJ` `:.` Loss in electrostatic energy `Delta u_i = u_i = (216 - 144) = 72 mu J` `rArr (Delta u) /u_1 = 72/216 = 1/3` |
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| 18. |
A driver in a stationary car horns which produces sound waves having frequency 2000 Hz. Thewaves are directed normally towards a reflecting wall. If the wall approaches the car with a velocity u = 3.3 m/s. The change in the frequency of sound after reflection from the wall is x%. Then x |
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Answer» |
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| 19. |
Define positive electric flux. |
| Answer» SOLUTION :Positive electric flux : When the LINES of force are EMERGING out of the SURFACE , the flux is positive . In this case the lines of force DIVERGE. | |
| 20. |
Internal resistance of a cell is _____ proportional to surface area of electrodes immersedinto the electrolyte and ____ proportional to the separation between the electrodes. |
| Answer» SOLUTION :INVERSELY , DIRECTLY | |
| 21. |
Members of the Jacobin Club were known as |
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Answer» Conservatives |
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| 22. |
Today the asset value of the trees stands at more than |
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Answer» 1.5 MILLION rupees |
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| 23. |
A conductor of linear mass dendity 0.2 g m^(-1) suspended by two flexiblewire as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1T whose direction is into the page. Compute the current inside the current and also the direction for the current . Assume g = 10" ms"^(-2) |
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Answer» Solution :Downward force, F = mg Linear mass density, `m/l = 0।2 GM^(-1)` `m/l=0।2 X 10^(-2) KG m^(-1)` ` m=(0।2 x 10^(-3) x l) kg m^(-1)` `f=(0।2 x 10^(-3)x l x 10) N` Upward magnetic force acting on the wire F= BIl `0।2 x 10^(-3)xxlxx10=1xxIxxl` `I=2xx10^(-3)` `I-2mA` |
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| 24. |
An ammeter is connected to measure the current intensity in a circuit with a resistance R. What relative error will be made if connection of the ammeter does not change the current intensity in the circuit? The voltate across the ends of the circuit is kept constant. |
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Answer» `(R )/(R_(0))` |
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| 25. |
A doubleslit isillumated by lightof wave length 6000 A^(@) .The sitesare 0.1cmapartand the screenis placed1 m away . Then the angularposition of 10^(th) maxima is |
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Answer» `6xx10^(3)` RAD |
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| 26. |
If the plates of a parallel plate capacitor are not equal in area, then |
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Answer» quantity of charge on the plates will be the same, but nature of charge will DIFFER. |
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| 27. |
The series corresponding to minimum wavelength transition in H atom ...... |
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Answer» Balmer `(1)/(lambda)=Rz[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` For H, Z = 1 and R is the Rydberg.s CONSTANT `R= 1.097 xx 10^(7) m^(-1)`. In this series the shortest wavelength or the limit of this series for `n_(1)=1 and n_(2), =OO lambda=911 Å` |
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| 28. |
Explain under what conditions a real image of a virtual object and a virtual image of a virtual object are formed in a convex mirror. |
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Answer» SOLUTION :According to the equation of a SPHERICAL mirror, `(1)/(v) + (1)/(u) = (1)/(f) or, (1)/(v) = (1)/(f) - (1)/(u) or, v= (uf)/(u-f)` Now is case of a convex mirror, `f gt 0` and for virtual object, `u gt 0`. Now if `f gt u " then" " " v LT 0` , i.e., if the virtual object is SITUATED between the POLE and the focus, real image will be formed. And if `f lt u " then" " " v gt 0` i.e., if the virtual object is situated between the focus and infinity, virtual image will be formed. |
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| 29. |
Consider two slabs of current shown in the figure. Both slabs have thickness b in y direction and extend up to infinity in x and z directions. The common face of the two slabs is y = 0 plane. The slab in the region 0 lt y lt d has a constant current density = J_(0)hat(k)and the other slab in the region – d lt y lt 0 has a constant current density = J_(0) (–hat(k)). (a)Find magnetic field at y = 0 (b)Plot the variation of magnetic field (B) along the y axis. |
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Answer» `(##IJA_PHY_V02_C10_E01_035_A01##)` |
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| 30. |
Explain the refraction of plane wavefront of light through a convex lens. |
Answer» Solution : Let AB represent a plane wavefront. The SECONDARY WAVELETS from A and B reach the points A' and B' respectively in same the time as the secondary wavelets reach from `P_1 and P_2` . Thus the points `A', P_2 and B'` are in the same phase. This explains why a parallel BEAM of LIGHTS on refraction through a convex lens gets converged and gives a spherical wavefront, WHEREAS the plane wavefront diverges in the case of concave lens. |
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| 31. |
A capacitor, made of two parallel plates each of plate are A and separation d, is being charged by an external a.c. source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor. |
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Answer» Solution :Let a parallel plate capacitor of capacitance `C=(in_(0)A)/(d)` is being charged by an EXTERNAL source whose voltage changes sinusoidally as `V=V_(m)sin omegat`. Then instantaneous charge on capacitor plate `q=CV=CV_(m)sin omegat` `therefore" Conduction current"` `I_(C )=(dq)/(dt)=(d)/(dt)(CV_(m)sin omegat)=CB_(m)OMEGA cos omega t"...(i)"` In the free space between the PLATES of capacitor a displacement current `I_(D)` is set up, whose magnitude is given as : `I_(D)=in_(0)(dphi_(E))/(dt)=in_(0)(d)/(dt)(EA)=in_(0)(d)/(dt)((V)/(d)A)=(in_(0)A)/(d)(dV)/(dt)=C(dV)/(dt)=C(dV)/(dt)=C(d)/(dt)=C(d)/(dt)(V_(m)sin omegat)` `=CV_(m)cos omega t"...(ii)"` Comparing (i) and (ii), we find that `I_(C)=I_(D)` i.e., the displacement current inside the capacitor is the Comparing (i) and (ii), we find that `I_(C )=I_(D)` i.e., the displacement current inside the capacitor is the same as the current charging the capacitor. |
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| 32. |
A circular loop of area 0.01 m^(2) and carrying a current of 10A is placed parallel to a magnetic field of intensity 0.1 T. The torque acting on the loop, in Nm is |
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Answer» 1.1 |
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| 33. |
What does PET stand for? Whereis it used? |
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Answer» Solution :PET stands for Postitron EMISSION Tomography. PET scans are used in examining the brain. A patient is given a does of FDG (fluoro-deoxy-glucose)which has a flurino-18 atom attached to a moleculeof gucose. The flurine atom decays with emission of positron and neutrino with a half life of 110 minutes.`._(9)F^(18) to ._(8)O^(18)+._(-1)e^(0)+V_(e)` The positron encounters an electron and the two annhilate to produce two powerful `gamma`-ray photons, i.e., `e^(+)+e^(-)to 2gamma` The two photons so produced move in opposite directions and are DETECTEDBY specialized detectors, which determine their point of origin inthe pateint's body. Thus the area in the patients body, where glucose metabolism is most intese is DETECTED. Thus PET ISA diagonistic technique for the brain, which is BASED on radioactive decay. |
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| 34. |
Truth table for the given circuit(See figure) is |
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Answer»
 For above logic CIRCUIT, TRUTH table is OBTAINED as follows:  Above output E, is same as output E given in option (C ) and so option (C ) is currect answer. |
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| 36. |
Newton's law of cooling is also applicable to |
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Answer» FORCED CONVECTION losses |
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| 37. |
Potential diffference across the terminals of the batery shown in Figyre is (r = internal resistance of battey) |
| Answer» ANSWER :D | |
| 38. |
What happens to the stability of nucleus, when BE increases ? |
| Answer» SOLUTION :NUCLEUS BECOMES more STABLE. | |
| 39. |
(A) : In a measurement, two readings obtai- ned are 10.001 and 10.0001. The second measurement is more precise. (R) : Measurement having more decimal places is more precise. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 40. |
A body of mass 50 kg is suspended using aspring balance inside a lift at rest. If the lift starts falling freely, the reading of the spring balance is |
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Answer» `= 50 kg` MASS of the BODY, m = 50 kg When LIFT is MOVING dowbn, Scale reading `= (m(g-a))/(g)` If lift starts falling freely, then a = g Now, scale reading `= (m(g-g))/(g)=0` |
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| 41. |
What is the minimum distance, in terms of focal length of lens, between an object and its real image formed by a convex lens ? |
| Answer» Solution :Four TIMES the FOCAL length of the LENS. | |
| 42. |
When two nuclei (A le 10) fuse together to form a heavier nucleus, the |
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Answer» binding energy PER UNCLEON increases. |
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| 43. |
The figures shows twopoints source whichemitlight of wavelength lambdain phasewill each other and are at a distance d = 5.5 lambdaapart along a line which is perpendicularto a largescreen at a distanceLfrom the centre of the source . Assume that dis muchless thanL Whichof the following statementis (are) correct ? |
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Answer» Onlyfive BRIGHT fringes appear on the screen |
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| 44. |
The equivalent capacitance between A and B, where two concentric spherical shells having radius a and b are connected as shown in the figure. |
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Answer» `(4PI epsilon_(0)ab)/((b-a))+4pi epsilon_(0)b` |
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| 45. |
Calculate the binding energy per nucleon of ""_(79^(197)Au. Mass of Au nucleus is 197. |
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Answer» |
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| 46. |
A mixture of light, consisting of wavelength 590 nmand an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the bright fringe of known light coincides with the 4th bright fringe of the unknown light. From the data, the wavelength of the unknonw light is |
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Answer» 885.0 nm |
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| 47. |
WHO WAS AUNG SAN SUU KYI? |
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Answer» PRO-DEMOCRACY LEADER |
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| 48. |
A cubical box is filled with sand and weighs 520 N. We wish to "roll" the box by pushing horizontally on one of the upper edges. (a) What minimum force is required? (b) What minimum coefficient of static friction between box and floor is required? (c) If there is a more efficient way to roll the box, find the smallest possible force that would have to be applied directly to the box to roll it. (Hint: At the onset of tipping, where is the normal force located?) |
| Answer» SOLUTION :(a) 260 N, (B) 0.50, (C) 184 N | |
| 49. |
One conducing U tube can be slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed v, then the emf induced in the circuit in terms of B, l and v where l is the width of each tube, will be ........ |
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Answer» zero  INDUCED emf DUE to bothh TUBES , `epsilon=Bvl + Bvl` `THEREFORE epsilon` =2Bvl (both batteries tubes considered as in series ) 
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