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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 : AC can be transmitted over long distances at high voltage without much power loss. R : The average value of AC is defined over any half cycle. |
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Answer» If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then MARK (1) |
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| 2. |
A uniform circular disc of mass 400 g and radius 4.0 cm is rotated about one of its diameter at an anglar speed of 10 rot/s. The kinetic energy of the disc is : |
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Answer» `3.2xx10^(-5)J` `=(1)/(2)((1)/(4)MR^(2)OMEGA^(2))` `=(1)/(8)xx400xx(4)^(2)XX(20pi)^(2)` `=800xx400pi^(2)ergs`. `=320000pi^(2)xx10^(-7)J` `=32xx(3.142)^(2)xx10^(-3)` `=3.2xx10^(-1)J` |
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| 3. |
A greenhouse has an electronic system (block diagram is given below ) which automatically switches ON a heater if the air temperature in the greenhouse drops too low . A manual switch is included so that the automatic system can be switched off . (Hint , The temperature sensor gives a logic -I output when the air temperature is normal and logic 0 when it is too cold ) a. What is meant by 1 and 0 in digital circuit ? b. Name the logic gate X . Why is it used ? c. Name the logic gate Y. d. Construct a truth table of this electronic system by taking A and B as inputs and D as output |
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Answer» Solution :a - 1 means maximum voltage 0-means minimum voltage B. NOT GATE To give 1 voltage at Y C .AND gate d 
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| 4. |
A cubical region of side .a. has its centre at the origin. It encloses three fixed point charges of charge -q at (0, -a/4,0), + 3q at (0,0,0) and -q at (0, + a/4, 0). Choose the correct options (s) |
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Answer» The NET electric flux crossing the plane x = + a/2 is EQUAL to |
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| 5. |
Current density in a copper wire is 2.5 xx 10^(8) Am^(-2). If 8A current if flowing through it diameter of the wire is ...... |
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Answer» 0.2 mm J = `(I)/(A) = (I)/(pi R^(2))` `thereforer =SQRT((I)/(pi J))` `= sqrt( (8)/(3.14 xx 2.5 xx 10^(8)) )` ` = sqrt( 1.009 xx 10^(-8) ) ` ` therefore r = 1 1.009 xx 10^(-4) ` m `therefore r = 1 xx 10^(-4)`m `therefore "Diameter " D = 2 xx 10^(-4 ) m = 2 xx 10^(-2)` CM = 0.02 cm = 0.2 mm |
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| 6. |
White light of spectral range 4000 - 7000 Å is incident normally on a soap film of uniform thickness 0.0004 mm and mu = 1.3. The wvaelengths of the light which are strongly reflected are : |
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Answer» 4000 and 6000 `Å` Thickness of film, `t = 0.0004 MM` `= 4 xx 10^(-5) cm` and `mu = 1.3` For normal incidence` i = r = 0^(@)` Using condition for brightness `2mu t cos r = (2n + 1)(LAMBDA)/(2)` `2 xx 1.3 xx 4 xx 10^(-5) = (2n + 1).(lambda)/(2)` `therefore lambda = (20,800)/(2n + 1) Å` When `n = 0, lambda = 20,800 Å` `n = 1, lambda = (20,800)/(3) Å = 6933 Å` `n = 2 , lambda = (20,800)/(5) Å = 4160 Å` `n = 3 , lambda = (20,800)/(7) Å = 2972 Å` So required wavelengths are `6933 Å` and `4160 Å`. |
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| 7. |
Fraunhoffer diffraction experiment at a single slit using light of wavelength 400 nm , the first minimum is formed at an angle of 30^@ . Find the direction of the secondary maximum. |
| Answer» SOLUTION :`SIN^(-1)3/4` | |
| 8. |
A wave represented by the equation y = a cos (kx- omega t)is superimposed with another wave to form a In stationary wave such that point x=0 is a node. The equation for the other wave is |
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Answer» `a SIN (kx + omega t)` |
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| 9. |
Two wires 'A' and 'B' of the same material have their lengths in the ratio 1: 2 and radii in the ratio 2:1. The two wires are connected in parallel across a battery. The ratio of the heat produced in 'A' to the heat produced in 'B' for the same time is |
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Answer» `1:2` |
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| 10. |
A particle having charge 'q' and mass 'm' is projected with velocity (4hati-6hatj+3hatk) m//sec from the origin in a region occupied by electric field 'E' and magnetic field 'B' such that vecB=B_(0)hati and vecE=E_(0)hatj (take (qE_(0))/m=2). Find the time (in sec) when the magnitude of velocity of the charge particle becomes 5sqrt(5) m//sec. (neglect the gravity) |
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Answer» `0=6-(qE_(0))/m t_(1)` `t_(1)=3` sec `v_(X)^(2)+v_(y)^(2)+v_(z)^(2)=(5sqrt(5))^(2)` `4^(2)+v_(y)^(2)+3^(2)=125` `v_(y)=10 m//sec` `v_(y)=u_(y)+a_(y)t` `10=0+(qE_(0))/m t_(2)` `impliest_(2)=5` sec So, `t=t_(1)+t_(2)=8` sec |
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| 11. |
A point traversed half a circle of radius R=160cm during time interval tau=10.0s. Calculate the following quantities averaged over that time: (a) the mean velocity ltlt v gtgt, (b) the modulus of the mean velocity vector |ltlt v gtgt|, (c) the modulus of the mean vector of the total acceleration |ltlt w gtgt| if the point moved with constant tangent acceleration. |
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Answer» `LT v gt = ("Total distance COVERED")/("Time elapsed")` `=s/t=(piR)/(tau)=50cm//s` (1) (b) Modulus of mean velocity VECTOR `|lt vecv gt|=(|Deltavecr|)/(Deltat)=(2R)/(tau)=32cm//s` (2) (c) Let the point moves from i to f along the half circle (figure) and `v_0` and `v` be the spe at the points respectively. We have `(dv)/(dt)=w_t` or, `v=v_0+w_t t` (as `w_t` is constant, according to the problem) So, `lt v gt =(underset0oversettint(v_0+w_t t)dt)/(underset0oversettintdt)=(v_0+(v_0+w_t t))/(2)=(v_0+v)/(2)` (3) So, from (1) and (3) `(v_0+v)/(2)=(piR)/(tau)` (4) Now the modulus of the mean vector of total acceleration `|lt vecw gt|=(|Deltavecv|)/(Deltat)=(|vecv-vecv_0|)/(tau)=(v_0+v)/(tau)` (see Figure) (5) Using (4) and (5), we get: `|lt vecw gt|=(2piR)/(tau^2)` 
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| 12. |
Two cells of same emf E but internal resistance r_(1) and r_(2) are connected in series to an external resistor R(figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero? |
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Answer» Solution :Applying OHM's law Effective resistance `=R+r_(1)+r_(2)` and effective emf of two CELLS `=E+E=2E`, so the electric current is given by `I=(E+E)/(E+r_(1)+r_(2))` The potential difference across the TERMINALS of the first cell and putting it equal to zero. `V_(1)=E-Ir_(1)=E-(2E)/(r_(1)+r_(2)+R)r_(1)=0` or `E=(2Er_(1))/(r_(1)+r_(2)+R)implies1=(2r_(1))/(r_(1)+r_(2)+R)` `r_(1)+r_(2)+R=2r_(1) impliesR=r_(1)-r_(2)` this is the required relation. |
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| 13. |
______ is known as current sensitivity of agalvanometer. |
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Answer» Current PER UNIT ANGULAR DEFLECTION |
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| 14. |
Statement-1 A lens has two principal focal lengths which may differ because Statement -2 : Light can fall on either surface of the lens. The two principal focal lengths differ when medium on the two sides has different refractive indices. |
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Answer» Statement -1 is True, statement -2 is True, Statement -2 is a CORRECT EXPLANATION for Statement -1. |
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| 15. |
A and B are the two radioactive elements. The mixture of these elements show a total activity of 1200 disintegrations/minute. The half-life of A is 1 day and that of B is 2 days. What will be the total activity after 4 days ? Given, the initial number of atom in A and B are equal : |
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Answer» 200 dis/min Initial number of atoms is A and B are same `A_(0) alpha 1/T_(1//2)` `rArr A_(0)/A_(0) ((A))/((B)) =(48 hr)/(24 hr)=2` ALSO `A_(0) (A)+A_(0) (B)=1200` `rArr 3A_(0) (B)=1200` `A_(0) (B)=400` and `A_(0) (A)=800` So, `A(A)=(A_(0)(A))/(2^(4))=(800)/(16)=50` `and A(B)=(A_(0)(B))/((2)^(2))=(400)/(4)=100` Hence total activity after 4 days is `=A_(0) (A)+A_(0)(B)=50+100=150"dis/min"` |
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| 16. |
Consider the diagram shown below. A voltmeter of resistance 150 Omega is connected across A and B. The potential drop across B and C measured by voltmeter is |
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Answer» 29 V |
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| 17. |
The sky would appear red instead of blue if |
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Answer» ATMOSPHERIC particles scatter blue light more than red light |
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| 18. |
A long solenoid with 40 turns per cm carries a current of 1A. The magnetic energy stored per unit volume is __________J//m^(3) |
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Answer» `3.2pi` Energy per UNIT volume = `(1)/(2mu_(0))B=(1)/(2mu_(0))(munI)=(n^(2)I^(2))/(2)` `=(4pixx10^(7)XX(4xx10^(3))xx(1))/(2)` `=32pixx10^(-1)=3.2piJm^(-3)` |
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| 19. |
A galvanometer coil has a resistance of 12 Omega and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V ? |
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Answer» Solution :Here `R_G = 12 OMEGA , I_g = 3 mA = 3 xx 10^(-3)A` and VOLTMETER.s rangeV = 18 V To convert galvanometer into a voltmeter of desired range, we should join a resistance R in SERIES with the galvonometer, where `R = V/(I_g) - R_G = 18/(3 xx 10^(-3)) - 12 = 6000 - 12 = 5988 Omega`. |
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| 20. |
In a plane electromagnetic wave, the electric field oscillates sin usoidally at a frequency of 2.0xx10^(10) Hz and amplitude 48 V m^(-1).a.What is the wavelength of the wave ?b.What is the amplitude of the oscillating magnetic field ?c.Show that the average energy density of the E field equals the average energy density of the B field. [c=3xx10^(8)ms^(-1)]. |
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Answer» SOLUTION :a.`lambda = (c )/(upsilon)=(3xx10^(8))/(2xx10^(10))=1.5xx10^(-2)m` B.`c=(E_(0))/(B_(0)), B_(0)=(E_(0))/(c )=(48)/(3xx10^(8))=16xx10^(-8)T` c.`U_(E )=(1)/(2)epsilon_(0)E^(2), U_(B)=(1)/(2mu_(0))B^(2)` But `E=cB, c=(1)/(sqrt(mu_(0)epsilon_(0))) therefore U_(E )=U_(B)` |
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| 21. |
Three prop -particles (m_prop = 4.0026) amu are combined to form a C^12 nucleus. The energy released will be |
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Answer» 4.02 Mev |
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| 22. |
Two particles A and B of masses m and 2m have charges q and 2q respectively. Both these particle moving with velocities v_1 and v_2 respectively in the same directionenter the same magnetic field B acting normally to their direction of motion. If the two force F_A and F_B acting on them are in the ratio 1:2 , find the ratio of their velocities. |
| Answer» SOLUTION :As FORCE `F = qvB`, hence `(F_A)/(F_B) = 1/2 = (q_A v_A B)/(q_B v_B B) = (qv_A)/(2qv_B) = (v_A)/(2v_B) implies v_A = v_B` | |
| 23. |
The interaction of electric charges q_1 and q_2 separated by a distance r is F=1/(4piepsilon_0)(q_1q_2)/r^2 If p_1 and p_2 are the properties of two magnetic poles , separated by a distance r , the force is F=(mu_0)/(4pi)(p_1p_2)/r^2 a. Write two similar properties of these two forces. b. Name the quantities p_1,p_2 and mu_0 . c. Show that 'p' has the ampere metre^2. |
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Answer» SOLUTION :a. i. Both are inverse square law forces II. Both are long range , can be attractive or repulsive. B. `p_1 and p_2` are called pole strengths `mu_0` - is the property of vacuum (or air) and is called permeability . c. Force between conductors `F=(mu_0I^2l)/(2A)` To find unit we take `[(mu_0I^2l)/(2a)]=[(mu_0)/(4PI)(p^2)/r^2]` `([I^2][L])/([L])=([P^2])/([L^2])"":.[P^2]=[I^2][L^2]` `:.` P has unit, A m or Dimensions given , `[F]=[q^2/(4piepsilon_0r^2)]=[(mu_0p^2)/(4pir^2)]` `([q^2])/([epsilon_0])=[mu_0].[P^2]"":.[P^2]=([q^2])/([epsilon_0][mu_0])=([q^2])/([epsilon_0mu_0])=[q^2c^2]=[I^2T^2xxL^2T^(-2)]` Since `1/(epsilon_0mu_0)=c^2"":.[P]=[I.L]` |
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| 24. |
David is observing his image in a plane mirror. The distance between the mirror and his image is 4 m. If he moves 1 m towards the mirror the path distance between David and his image will be |
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Answer» 3m |
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| 25. |
A particle moving along straight line has velocity v=mus^(2) wheres is the displacement.If S=S_(0) then which of the following graph best represent s versus t |
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Answer»
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| 26. |
A pivoted small magnetic needle of moment m is placed at a distance r along the axis of a short bar magnet of moment M. Find the force on the pivot of the needle. |
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Answer» |
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| 27. |
STATEMENT-1 : A transformer is used for impedance impedance matching so as to deliver maximum current to a circuit. and STATEMENT-2 : The resistance R ofexternal circuit connected across secondary, as seen by the generator in primary is R_(eq)=((N_(P))/(N_(S)))^(2)R . By adjusting . (N_(P))/(N_(S)), impedance of generator can be matched to external circuit. |
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Answer» Statement-1 is True, Statement-2 ISTRUE, Statement-2 is a correct EXPLANATION for Statement-1 |
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| 28. |
A U-tube is partially filled with water. Oil which does not mix with water is next poured into one side, until water rises by 25 cm on the other side. If the density of the oil is 0.8 g/cc, the oil level will stand higher than the water level by : |
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Answer» 6.25 cm Here pressure DUE to (h + 50) cm of oil = pressure due to 50 cm of water. `(h+50)xx0.8xxg=50xx1xxg` `rArrh=12.5` cm of Hg. Thus correct CHOICE is (c). |
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| 29. |
In a pipe closed at both ends the maximum amplitude of vibration is 5 mm and the amplitude of vibration at a distance 5 cm from one end is 4.33 mm. The length of the pipe is 120cm. To what mode of vibration does it correspond ? What is the frequency of the note emitted by the pipe ? Velocity of sound in the gas enclosed in the pipe 336 m/s. |
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Answer» |
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| 30. |
A capacitor is being charged through a resistance of 3 mega ohm. If it reaches 75% of its final potential in 0.5 sec, find its capacitance. |
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Answer» SOLUTION :At any moment during the charging process, the potential difference across the plates of a capacitor is givenby `V= V_0 (1- E^(-t//CR))` where `V_0` is the final potential Give that `V/V_0 = 75% = 0.75 , t = 0.5 ` sec and `R= 3xx10^6 Omega` Hence , `0.75 = 1-e^(-0.5//(3xx10^6)C)` or ` e^(-0.5//(3xx10^6)C) = 1 0 0.75 = 0.25` or, `e^(-0.5//(3xx10^6)C) =1/(0.25) = 4` or , `(0.5)/(3xx10^6 C)= log_e 4 = 2.3026xx log_10^4` or , `C= (0.5)/((3xx10^6) xx 2.3026 xx 0.6021)` `= 0.12xx10^(-6) F = 0.1 12 muF` |
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| 31. |
In an interference experiment , two parallel verticals slits S_(1) and S_(2) are used. A thin glass plate is introduced in the path of light from S_(1). Then |
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Answer» fringe pattern remains unaltered |
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| 32. |
Colour in discharge tube filled with gas at low pressure is due to … |
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Answer» ELECTRON in EXCITED state in atom |
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| 33. |
A hollow circular tube of radius R and negligible internal diameter is fixed on horizontal surface ball A of mas m is given velocity v_(o) in the shown direction. It collides with ball B of mass 2 m. Collision is perfectly elastic. If centre of loop is origin of co-ordinates system, then co-ordinate of next collision is |
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Answer» `((sqrt3)/(2)R, (- R)/(2))` `-v_(o) = v_(1) + 2v_(2)`….(i) & `e = (v_(2) - v_(1))/(-v_(o)) = 1` `:. V_(2) - v_(1) = -v_(o)`…. (ii) from (i) & (ii) `v_(1) = (v_(o))/(3), v_(2) = (-2v_(o))/(3)` `:. T = (2piR)/(v_(o))` Angle rotated by by `A = (2PI)/(3)`, Angle rotated by `B = (4pi)/(3)`rad. 
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| 34. |
A small signal voltage V(t) = V_(0)sin omegat is applied across an ideal capacitor C |
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Answer» CURRENT l(t) leads VOLTAGE V(t) by `180^(@)` |
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| 35. |
Which one of the following has the dimensions of ML^(–1)T^(–2) ? |
| Answer» Answer :D | |
| 36. |
The figure shows four pairs of polarizing sheets, seen face-on. Each pair is mounted in the path of initially unpolarised light. The polarizing direction of each sheet (indicated by the dashed line) is referenced to either a horizontal x-axis or a vertical y axis. Rank the pair according to the fraction of the initial intensity that they pass, greatest first |
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Answer» (a) (i)gt(ii)gt(iii)gt(IV) 
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| 37. |
Explain with a suitable diagram the concepts of valence band and conduction band. |
Answer» Solution :A crystal consists of a three dimensional periodically repeating arrangement of a large number, say N, of the atoms, ions or molecules of the solid. An isolated atom is characterized by discrete and well-defined allowed electron energy levels. As the atomns come close together in a solid, for each of the discrete energy level of an atom, there are N such levels in the crystal. The energies of these N levels DIFFER so slightly that theirdistribution is very nearly continuous and said to CONSTITUTE an energy bands energies are FORMED which the electrons may occupy. These bands are separated bybands of forbidden energies.  The energy levels of inner shell electrons are not affected much by thepresence of neighbouring atoms and hardly widen into bands. They play no part in determiningthe electrical properties of solids The TWO energy bands which are important for electrical conduction a crystal are the highest almost-filled band and the lowest almost-empyt band, respectively called the valence and conduction bands. At or see absolute zero of TEMPERATURE, the energy band whose all available states a occupied by valence electrons is called the valence band while the empyt electrical properties of solids. energy band above it is called the conduction band. |
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| 38. |
A toy cyclist completes one round of a square track of side 2m in 40 seconds. What will be the displacement at the end of3 minutes. |
| Answer» ANSWER :D | |
| 39. |
Answer the following questions: (a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2//3)e:(-1//3)e]. Why do they not show up in Millikan's oil -drop experiment? (b) What is so special about the combination e//m? Why do we not simply talk of e and m separately? (c) Why should gasses be insulators at ordinary pressure and starts conducting at very low pressures? (d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons? (e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: E=hv, p=h//lambda. But while the value of lambda is physically significant, the value of v ( and therefore, the value of phase speed v lambda) has no physical significance. Why ? |
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Answer» Solution :(a) The quarks having fractional charges are thought to be CONFINED within a proton and a neutron. These quarks are bounded by forces. These forces become stronger when the quarks are tried to be pulled apart That is why, the quarks always remain together. It is due to this reason that through fractional charges does exist in nature but the observable charges are always intergral multiple of charges of electron. (b) The motion of electron in the electric and magnetic field is related with the basic equations `eV=1/2mv^(2) and BEV=(MV^(2))/r` All these equation involve e and m together, i.e., there is no equation in which e or m occuring ALONE. As a result of it, we study `e//m` of electron and do not talk of e and m separately far an electron. (c) At ordinary pressures a few positive ions and electrons produced by the ionisatation of the gas molecules by energetic rays (like X-rays, `gamma`-ray, cosmic rays etc. coming from outer space and entering the earth's atmosphere) are not able to reach their respectively electrods, even at high VOLTAGES, due to their frequent collisions with gas molecules and recombinations. That is why the gasses at ordinary pressures are insulators. At low pressures, the density of the gas decreases, the mean free path of the gas molecules becomes larages. Now under the effect of external high voltage, the ions aquire sufficient energy before they collide with molecules causing further ionisation. Due to it, the number of ions in the gas increases and it becomes a conductor. (d) By work function of a metal , we mean the minimum energy required for the electron in the highest level of conduction band to get out of the metal. Since all the electrons in the metal do not belong to that level but they occupy a continous band of levels, therefore, for the given incident radiation, electrons knocked off from different levels come out with different energies. (e) De-Broglie wavelength associated with the moving particle is `lambda=h/p or p=h/lambda` Energy of the wave E is `E=hv=(hc)/lambda` , Energy of moving particle `=1/2 (p^(2))/m = 1/2 ((h//lambda)^(2))/m= 1/2 (h^(2))/(lambda^(2)m)` For the relations of E and p, we note that ` lambda` is physically sufficient but v has no direct physical significance. |
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| 40. |
An open organ pipe and closed pipe have same length. The ratio of frequencies of their n^(th) over tone is |
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Answer» `(N+1)/(2N+1)` |
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| 42. |
A closely wound solenoid of 800 turns and area of cross section 2.5xx10^(-4)m^2 carries a current of 3.0 A . Explain te sense in which the solenoid acts like a bar magnet . What is its associated magnetic moment ? |
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Answer» Solution :`N = 800, A = 2.5 xx10^(-4) m^2, I = 3.0 A, m = ? ` ` m = ANI = 2.5 xx10^(-4) xx800xx3.0=7.5 xx8xx10^(-2) =0.600 Am^2` The field of SOLENOID is UNIFORM along the axis , and for short solenoid, it is similar to that of a BAR magnet and the polarity depends on the direction of circulation of current . |
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| 43. |
A wire of density 9 xx 10^(3) kg//m^(3) is stretched between two clamps l m apart and is subjected to an extension of 4.9 xx 10^(-4) m. What will be the lowest frequency of transverse vibrations in the wire ? (Y=9xx10^(10)N//m^(2)). |
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Answer» 25Hz |
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| 44. |
The graph between potential energy U and displacement X in the state of stable equilibrium will be– |
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Answer»
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| 45. |
The photo electric threshold wavelength for a metal is 2750A^(@), The minimum energy of a photon that can produce photoelectric effect in the metal is |
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Answer» 0.045eV |
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| 46. |
A certain radioactive substance has a half life of 5 years. Thus for a nucleus in a sample of the element, probability of decay in 10 years is |
| Answer» ANSWER :B | |
| 47. |
In the depletion region of the an unbaised p-n junction diode, there are |
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Answer» only free ELECTRONS. |
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| 48. |
What does the term E = - ( dV)/(dx)signify ? |
| Answer» Solution :Electric FIELD is in the direction along which the electric POTENTIAL DECREASES SHARPLY. | |
| 49. |
The number of electric lines of force crossing through a given area is |
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Answer» |
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| 50. |
A transmitter radiates 12 KW power at 80 % modulation index. The power in carrier is nearly: |
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Answer» 8.12 KW |
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