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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. |
Match List - I(Electromagnetic wave type) with List - II (Its association / application) and select the correct option from the choices given below the lists.{:(,"List - I",,"List - II"),((a),"Infrared waves",(i),"To treat muscular strain"),((b),"Radiowaves",(ii),"For broadcasting"),((c ),"X - rays",(iii),"To detect fracture of bone"),((d),"Ultraviolet rays",(iv),"Absorbed by the ozone layer of the atmosphere"):} |
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Answer» (A) - (iii), (B)-(II), (C )-(i), (D)-(iv) |
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| 2. |
The capacitive time constant of the RC circuit shown in the figure is |
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Answer» Zero |
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| 3. |
A 2.0 - kg block executes SHM while attached to a horizontal spring of spring constant 200 N//m. The maximum speed of the block as it slides on a horizontal frictionless surface is 3.0 m//s. What is the amplitude of the block's motion? |
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Answer» 0.15 m |
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| 4. |
Draw a labelled diagram showing image formation by a simple microscope when the image is being formed at least distance of its distinct vision. Find an expression for its angular magnification (or the magnifying power. |
Answer» Solution :A simple microscope is a convex lens of small focal length f. The lens is HELD near the object AB at a small distance w(zz  `therefore` Angular magnification `m=beta/alpha =(ANGLEA.CB.)/(angleA_(0)CB.) = (angleACB)/(angleA_(0)CB.)` As values of angles `alpha` and `beta` are extremely small, hence `alpha = tan alpha` and `beta= tan beta`. `therefore m=beta/alpha =(tan beta)/(tan alpha) = (tan angleACB)/(tan angleA_(0)CB) =(AB//BC)/(AB//B.C) =(B.C)/(BC) =D/u` [since `A_(0)B. = AB`] From lens formula, we have : `1/(-D) -1/(-u) =1/f` `rArr 1/u =1/f + 1/D = (D+f)/(Df)` `therefore m=Du =D[(D+f)/(Df)] =(D+f)/f=1+D/f` Thus, magnifying power of microscope will be more if focal length of microscope lens is less, |
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| 5. |
A current of 4 mA is maintained in a circular loop of single tunn of 1 m circumference. The plane of the loop is parallel to a magneticfield of 0.2 T. (i) Calculate the magnetic moment of the loop. (ii) Calculate the magnitude of torque experienced by the loop. |
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Answer» |
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| 6. |
The critical angle of diamond is 30^@. What is its refractive index ? |
| Answer» SOLUTION :`N= 1/sin C=1/sin 30 = 1/1/2 =2. n=2` | |
| 7. |
A smooth sphere A is moving on a frinctionless horizontal plane with angular speed omega and centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction every where. After collision, their angular speeds are omega_(A) and omega_(B) respectively. Then : |
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Answer» `omega_(A)gtomega_(B)` |
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| 8. |
A 2.0 - kg block executes SHM while attached to a horizontal spring of spring constant 200 N//m. The maximum speed of the block as it slides on a horizontal frictionless surface is 3.0 m//s. What is the magnitude of its maximum and minimum accelerations? |
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Answer» `30 m//s^(2), 15 m//s^(2)` |
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| 9. |
The laws of Robotics are ........... |
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Answer» a robot MAY not injure a human being |
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| 10. |
A ray of light is incident at an angle of 60^@ on one face of a prism which has an angle of 30^@. The ray emerging out of the prism makes an angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the prism. |
| Answer» SOLUTION :`MU = SQRT(3)` | |
| 11. |
According to Bohr's model the orbit of electron revolving around the nucleus is ...... |
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Answer» elliptical |
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| 12. |
An electric heater of resistance 10Omega connected to 220 V power supply is immersed in the water of 1 kg. How long the electrical heater has to be switched on to increase its temperature from 30^(@)C to 60^(@)C. (The specific heat of water is s=4200J kg^(-1)/C) |
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Answer» Solution :According to Joule.s heating law `H=I^(2)Rt` The current PASSED through the ELECTRICAL heater `=(220V)/(10Omega)=22A` The heat produced in one second by the electrical heater `=(220V)/(10Omega)=22A` The heat produced in one second by the electrical heater`H=I^(2)R` The heat produced inone second `H=(22)^(2)xx10=4840J=4.84kJ`. In FACT the power rating of this electrical heater is 4.84 k W. The amount of energy to increase the temperatureof 1 kg water from `30^(@)C` to `60^(@)C` is `Q=ms DeltaT` Here `m=1kg, s=4200J kg^(-1), DeltaT=30` so `Q=1xx4200xx30=126kJ` The time required to producethis heat energy `t=(Q)/(I^(2)R)=(126xx10^(3))/(4840)=26.03s` |
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| 13. |
Assume that clouds are distributed around the entire earth at a height of 3000 m above the ground. The atmosphere can be modeled as a spherical capacitor with the earth as one plate and the cloud as other. When the electric field between the earth and the cloud becomes large, the air begins to conduct and the phenomena is called lightning. On a typical day 4 xx 10^(5) C of positive charge is spread over the surface of the earth and equal amount of negative charge is there on the cloud. Resistivity of the air is rho = 3 xx 10^(13) Omega m and radius of the earth = 6000 km. (a) Find the resistance of the air gap between the earth’s surface and cloud. (b) Estimate the potential difference between the surface of the earth and the cloud (c) In how much time the capacitor formed between the earth and the cloud will lose 63 % of the charge ? |
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Answer» (b) `3 XX 10^(5)V` (C) `165.3 s` |
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| 14. |
A 2.0 - kg block executes SHM while attached to a horizontal spring of spring constant 200 N//m. The maximum speed of the block as it slides on a horizontal frictionless surface is 3.0 m//s. How long does the block take to complete 7.0 cycles of its motion? |
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Answer» 4.0 s |
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| 15. |
If the energy of a photon corresponding to a wavelength of 6000Å is 3.32xx10^(-19)J, the photon energy in J) for a wavelength of 4000Å will be : |
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Answer» `:.(E_(2))/(E_(1))=(lambda_(1))/(lambda_(2)) rArr E_(2)=(lambda_(1))/(lambda_(2))E_(1)` |
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| 16. |
A sphere of mass m and radius R is kept on a trolley of mass Mas shown in the figure. The coefficient of static and kinetic friction between the sphere and the trolley are mu_(s) and mu_(k) respectively. The maximum horizontal force F that can be applied to the trolley for which the solid sphere does not slip is |
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Answer» `mu_(s)G(+(7)/(2)M)` |
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| 17. |
Four charges of +q, +q, +q and +q are placed at the corners A, B, C and D of a square of side a. Find the resultant force on the charge at D. |
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Answer» Solution :Let SIDE = `a, BD= SQRT(2)a` Along AD force `F_1= 1/(4 pi epsilon_0)cdot (q^2)/(a^2)` Along CD force `F_2 = 1/(4 pi epsilon_0) cdot (a^2) :. F_1 = F_2` Along AD force `F_1 = 1/(4 pi epsilon_0) cdot (q^2)/(a^2) "" :. F_1 = F_2` RESULTANT of `F_1` and `F_2 = sqrt((F_1^2) + F_2^2)) = sqrt(2) .F_1` Force along BD `F_3 = 1/(4 pi epsilon_0) cdot (q^2)/((sqrt(2)a)^2) = 1/2 1/(4 pi epsilon_0) (q^2)/(a^2)` Resultant force at `D = sqrt(2) F_1 + F_3` `= sqrt(2) 1/(4 pi epsilon_0) cdot (q^2)/(a^2) + 1/2 1/(4 pi epsilon_0) cdot (q^2)/(a^2)` `= 1/(4 pi epsilon_0) (q^2)/(a^2) [ sqrt(2) + ½] = (q^2)/(8 pi epsilon_0 a^2) cdot [1 + 2sqrt(2)]` 
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| 18. |
Draw a ray diagram showing the path of a ray of light entering through a triangular glass prism. Deduce the expression for the refractive index of glass prism in terms of the angle of minimum deviation and angle of the prism. |
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Answer» Solution :Refraction of light rays through a prism has been shown in FIG. 9.77. Here `anglei` is the incidence angle at FIRST surface of prism and `anglee`is the angle of emergence from the second surface. `angler_(1)`and `angler_(2)`are the respective refraction angles at the two FACES and `angleinfty`is the angle of deviation. These angles are corelated as : `anglerr_(1) + angler_(2) = angleA`..........(i) and `anglei + anglee = angleA + angledelta`........(ii)  A graph showing variation in angle of deviation (8) with variation in angle of incidence (z) is shown in Fig. 9.78. As minimum deviation position corresponds to only one angle of incidence, it means that for minimum deviation position `anglei = anglee`and also `angler_(1) = angler_(2)`Thus, in minimum deviation position `=(A+D_(m))/2` and `2angler_(1) =A` or `angler_(1) = A/2` `therefore` Refractive INDEX of prism `n =(sin i)/(SINI r_(1)) = ((sin(A+D_(m))/2)/(sin A/2))` 
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| 19. |
A compound microscope consists of two lenses_____and ______. |
| Answer» SOLUTION :OBJECTIVE LENS, EYE lens | |
| 20. |
S_(1) and S_(2)in list I represent coherent point sources ,S represents a point source lambda = wavelength of light emitted bythe sources .Fringes pattern is observed on the screen . Exclude the position of source while detecting the fringe pattern. |
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Answer» |
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| 21. |
Find V_(AB) |
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Answer» 10 V  `impliesV_(AB)=30/((10+5))xx5=10V` |
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| 22. |
Given the uses of Foucault current. |
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Answer» Solution :it is useful in some other case. A few of them are a. Induction stove b. Electromagnetic damping c. Eddy current brake d. Eddy current testing a. Induction stove : i. Induction stove is used to cook the food quickly and safely with less energy cosumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone. When the stove is switched on, an alternating current flowing in the coil produces HIGH frequency alternating magnetic field which induces very strong eddy currents in the pan produce so much of heat due to joule heating which is used to cook the food. NOTE: The frequency of the domestic AC supply is increased from 50-60 Hz to around 20-40 KHz before giving it to the alternating magnetic field. b. Electromagnetic damping i. This eddy current braking system trainsand roller coasters. Strong the rails. ii. To stop the train, electromagnetes are these magnets induces eddy currents in the rails which oppose or resist the current linear brake. iii. In some cases, the circular disc, conneted to the wheel of the train through a common shaft, is made to ROTATE in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stops the train. This is Eddy current circular brake. c. Eddy current brake i. It is one of the simple non-destructive testing methods to find defects like a specimen. ii. A coil of insulated wire is given an it produces an alternating magnetic field. iii. When this coil is brought near the test surface, eddy current is induced in the test surface. iv. The PRESENCE of defects causes the changh in phase and amplitude of the eddy current that be detected by someother means. In this way, the defects present in the specimen are identified. Eddy current testing i. The armature of the galavanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the armature to rest immediately and then galvanometer shows a steady deflection. This is CALLED electromagnetic damping. |
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| 23. |
For a chosen non zero value of voltage, there can be more than one value of current is |
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Answer» COPPER wire |
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| 24. |
A source of sound attached to the bob of a simple pendulum execute S.H.M. The difference between the apparent frequency of sound as received at the mean position of the S.H.M. is 2% of the natural frequency of the source. The velocity of the source at the mean position is: (velocity of sound in the air is 340 m/s). [ Assume velocity of sound source |
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Answer» 1.4 m/s |
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| 25. |
Circuit for the measurement of resistance by potentiometer is shown in figure. The galvanometer is first connected at point A and zero deflection is observed at length PJ = 10 cm. In second case, it is connected at point C and zero deflection is observed at a length 30 cm from P, then the unknown resistance X is . |
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Answer» 2 R a potentiometer wire be K.  For zero deflection, the current will flow independently in TWO circles : `IR = K xx 10` .......(i) ` I R + I X = K xx 30` .......(II) Subtracting Eq. (ii) from Eq. (i), we GET `IX = k xx 20 ` ......(III) Dividing Eq. 9i) by Eq. (ii), we have `R/X = 1/2` . |
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| 26. |
A small ball strikes at one end of a stationary uniform frictionless rod of mass m and length l which is free to rotate, in gravity-free space. The impact elastic. Instantaneous axis of rotation of the rod will pass through |
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Answer» its center of MASS |
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| 27. |
A: If the amplitude of a message signal is half of the amplitude of the carrier wave, then the index of modulation is 50%. R: In amplitude modulation the amplitude of the carrier wave varies in accordance with the signal voltage of the message signal. |
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Answer» |
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| 28. |
12 identical wires, each with resistance Rare interconnected to form a cube. Find equivalent resistance of this cube across its diagonally opposite points. |
| Answer» SOLUTION :`R. = (5R)/(6)` | |
| 29. |
A helicopter is ascending vertically with a speed of 8.0 ms^(-1) .At a height of 12 m above the earth,a package is droped from a window .Ho much time does it take for the package to reach the fround? |
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Answer» 1.23 s |
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| 30. |
(i) State the principle of working of a meter bridge. (ii) In a meter bridge balance point is found at a distance I_(1) with resistances R and S as shown in the figure. When an unknown resistance X is connected in parallel with the resistance S, the balance point shifts to a distance I_(2). Find the expression for X in terms of l_(1), l_(2) and S. |
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Answer» Solution :(i) A SLIDE wire BRIDGE is KNOWN as Metre bridge. It is constructed on the principle of balanced wheatstone bridge. When a Wheaststone bridge is balanced then, `P/Q=l/(100-l)`  (ii) When resistace R and S are connected : Since, balance point is found at a distance `l_(1)` from the zero and `:. R/S=l_(1)/(100-l_(1))` ...(i) When unknown resistance X is connected in parallel to S. `:.` Total resistance in the right hand GAP is `S_(1)=(SX)/(S+X)""[ :] 1/R=1/R_(1)+1/R_(2) implies R=(R_(1)R_(2))/(R_(1)+R_(2))]` Since, balance point is obtained at a distance `l_(2)` from the zero end `:. R/S_(1)=l_(2)/(100-l_(2))` Putting the value of `S_(1)` we get `R/((SX)/(S+X))=l_(2)/(100-l_(2))` `(R(S+X))/(SX)=l_(2)/(100-l_(2))` ...(ii) Dividing (ii) by (i), we get `(R(S+X))/(SX).S/R=l_(2)/(100-l_(2))xx(100-l_(1))/l_(1)"",""(S+X)/X=(l_(2)(100-l_(1)))/(l_(1)(100-l_(2)))` `S/X+X/X= (l_(2)(100-l_(1)))/(l_(1)(100-l_(2)))""implies""S/X+1=(l_(2)(100-l_(1)))/(l_(1)(100-l_(2)))` `S/X+X/X=(l_(2)(100-l_(1)))/(l_(1)(100-l_(2)))""implies""S/X+1=(l_(2)(100-l_(1)))/(l_(1)(100-l_(2)))` `S/X=(l_(2)(100-l_(1)))/(l_(1)(100-l_(2)))-1` `=(100 l_(2)-l_(1)l_(2)-100 l_(1)+l_(1)l_(2))/(l_(1)(100-l_(2))),""S/X=(100(l_(2)-l_(1)))/(l_(1)(100-l_(2)))` Hence, `X=(l_(1)(100-l_(2)))/(100(l_(2)-l_(1))).S"":.""X=(100l_(1)s-l_(1)l_(2)s)/(100(l_(2)-l_(1)))` |
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| 31. |
Three capacitors C_1,C_2,andC_3 of capacitance 1muF, 2muF, and 3muF, respectively, are charged separately as shown in the figure. Now these charged capacitors are connected to a battery of epsilon = 20 V and an uncharged capacitor of C = 2muF as shown in figure. , The charge oncapacitor marked C is |
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Answer» `3.75 muC`  The circuit can be redrawn as Conserving CHARGES on LOWER plates, we get `-[(x+20)1+(x+2)2+(x+20)3+2x]=-60` or `x=-(15)/(2)V` `|Q_(1muF)|=12.5muC` `|Q_(2muF)|=25muC` `|Q_(3muF)|=37.5muC` `|Q_(2muF)|=15muC` |
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| 32. |
Steel wireof length l hasa magnetic moment M it is htne bentinto a semi circualr arc the new magnetic moment is |
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Answer» ML |
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| 33. |
Three capacitors C_1,C_2,andC_3 of capacitance 1muF, 2muF, and 3muF, respectively, are charged separately as shown in the figure. Now these charged capacitors are connected to a battery of epsilon = 20 V and an uncharged capacitor of C = 2muF as shown in figure. , The potential of point P is |
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Answer» `12.5V`  The circuit can be REDRAWN as Conserving CHARGES on lower plates, we get `-[(X+20)1+(x+2)2+(x+20)3+2x]=-60` or `x=-(15)/(2)V` `|Q_(1muF)|=12.5muC` `|Q_(2MUF)|=25muC` `|Q_(3muF)|=37.5muC` `|Q_(2muF)|=15muC` |
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| 34. |
Draw the output signals C_(1) and C_(2) in the given combination of gates as shown in figure. |
Answer» SOLUTION :For figure (3),  Here `C_(1)=barY= barbar(Y_(1)*Y_(2))=Y_(1)*Y_(2)=barA*barB=bar(A+B)` `rArr ` Given logic circuit BEHAVES like NOR GATE. Its OUTPUT `C_(1)` can be shown as follows:  Here `C_(2)=bar(Y_(1)+Y_(2))=bar(barA+barB)=barbar(A*B)=A*B` Given logic circuit behaves like AND gate. Its truth table and output will be as shown below. 
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| 35. |
The Fahrenheit and kelvin scale of temperature will give the same reading at |
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Answer» -40 |
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| 36. |
Assertion: No interference pattern is detected when two coherent sources are infinitesimally close to each other. Reason: The fringe width is inversely proportional to the distance between the two slits. |
| Answer» SOLUTION :`beta=(lamdaD)/(d)`. If two coherent sources are infinitesimally close to each other then the WIDTH of interference fringe will be INFINITELY large. So, on the screen it will be central maxima only and no interference pattern is observed. | |
| 37. |
An organ pipe enclosed at one end has fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear, is : |
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Answer» 6 maximum frequency which a man can hear = 20, 000 Hz `therefore` In n is the number of overtones GENERATED then `V. = (2N + 1) v_(1) ` 20,000 = (2n + 1) `xx` 1500 `""rArr ` n= 6 Correct choice is (a). |
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| 38. |
A carbon resistor is coloured with four different bandsred, green, orange and silver respectively. Find the range of its probable resistance. Or, EMF of an electrical cell is 2 volt. A 10Omega resistance is joined at its two ends then potential difference is measured 1.6 volt. Find out the internal resistance and lost volt. |
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Answer» Solution :Red : 2, GREEN : 5, orange:3,zeroes, silver: 10% error `therefore` RESISTANCE = 25000`pm 10% Omega` ` 10% of 25000 Omega = 2500Omega` `therefore`Range of the resistance = (25000 - 2500) to (25000 + 2500) =22500`Omega` to 27500`Omega` Or, LOST VOLT = 2 - 1.6 = 0.4 V Current through this circuit , `I = (1.6 V)/(10Omega) = 0.16 A` Internal resistance, `r = (Ir)/(I) = (" lost volt")/(I) = (0.4V)/(0.16 A) = 2.5 Omega` |
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| 39. |
A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in, where t = 0 at the instant the ball is struck. The scaling on the vertical axis is set by v_(a)=19m//sandv_(b)=31m//s. (a) How far does the golf ball travel horizontally before returning to ground level? (b) What is the maximum height above ground level attained by the ball? |
| Answer» SOLUTION :(a) 95 m, (B) 31 m | |
| 40. |
Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m . Assume that the efficiency of the bulb is 2.5 % and it is a point source. |
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Answer» Solution :The bulb as a point source, radiates light in all directions uniformly. At a distance of 3 m, the surface area of the surrounding sphere is `A=4 pi r^(2) = 4pi(3)^(2)=113 m^(2)` The intensity at this distance is `I=("POWER")/("Area")=(100 Wxx2.5%)/(113 m^(2))=0.022 W//m^(2)` HALF of this intensity is provided by the electric field and half by the magnetic field. `(1)/(2)l=(1)/(2) (epsi_(0)E_("rms")^(2)C)` `=(1)/(2)(0.022 W//m^(2))` `E_("rms")=sqrt((0.022)/((8.85xx10^(-12))(3xx10^(8))))V//m=2.9 V//m` The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal , the peak electric field, `E_(0)` is `E_(0)=sqrt(2)E_("rms")=sqrt(2)xx2.9` V/m `=4.07 ` V/m Thus , you see that the electric field strength of the light that you for reading is fairly large. COMPARE it with electric field strength of TV or FM waves, which is of the order of a few microvolts per metre. Now, let us calculate the strength of the magnetic field . It is `B_("rms")=(E_("rms"))/(C)=(2.9 V m^(-1))/(3xx10^(8) ms^(-1))=9.6xx10^(-9) T`. Again, since the field in the light beam is sinusoidal, the peak magnetic field is `B_(0)=sqrt(2) B_("rms")=1.4 xx10^(-8)T`. Note that although the energy in the magneticfield is equal to the energy in the electric field , the magnetic field strength is evidently very WEAK. |
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| 41. |
A dielectric spheri- cal shell of radius R, having charge Q is rotating with angular speed w about its diameter. Calculate the magnetic dipole moment (M) of the shell. Write the ratio of M and angular momentum (L) of the rotating shell. This ratio is called gyro-magnetic ratio. Mass of shell is m. |
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Answer» |
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| 42. |
The earth's magnetic field is |
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Answer» `10^(-4)T` |
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| 43. |
When skin gets exposed UV radiation for a long time, it develops ….. |
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Answer» MANGANESE |
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| 44. |
Focal length of a thin convex lens is 30 cm. At a distance of 10 cm from the lens there is a plane refracting surface of refractive index 3/2. Where will the parallel rays incident on lens converge ? |
| Answer» SOLUTION :40 CM from LENS | |
| 45. |
A cyclotron is device for accelerating ions and charged particles. It was developed by Lawerence in 1932. The heart of the appritus consists of a split metal pillbox. Figure sHOws top and front views of the halves called dees.A rapidly oscilating potential difference is applied between the Dees. This produces an oscilating electric field in the gap between the dees, the region inside each dee being essentially free of electric field. The Dees are enclosed in an evaccutaed container, and the entire unit is placed in a uniform magnetic field B whose direction is normal to the plane Of Dees, A charged particle of mass 'm' and charge 'q' in the gap between the dees, it moves with constant speed in a semi-circle. The period of uniform circular motion is T=(2pim)/(qB) and is independant of speed. If the time-period of the oscilating elctric field is equal to this time, then the charged particle will be accelerated again and again Answer the following questions (consider the mass of particles remains constant during motion): A cyclotron has been adjusted to accelereted deutrons. It is now to be adjusted to accelerate other particles , for this following changes may be made: (a) In order to keep the frequency of oscillating electric field same, the magnetic field is halved for proton (b) If the magnetic field remain unchanged, the oscillation frequency of electric field should be halved for alpha particle (c ) If the magnetic field remain unchanged, the oscillation frequency of electric field should be double for proton (d) If the frequency of oscillating electric field is kept same, the magnetic field should be kept same for alpha particle |
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Answer» a,b and c |
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| 46. |
Wavelengths of light used in an optical instrument are lamda_1 = 4000 A and lamda_2 = 5000 A. What is the ratio of their respective resolving powers? |
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Answer» `4:5` |
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| 47. |
A cyclotron is device for accelerating ions and charged particles. It was developed by Lawerence in 1932. The heart of the appritus consists of a split metal pillbox. Figure sHOws top and front views of the halves called dees.A rapidly oscilating potential difference is applied between the Dees. This produces an oscilating electric field in the gap between the dees, the region inside each dee being essentially free of electric field. The Dees are enclosed in an evaccutaed container, and the entire unit is placed in a uniform magnetic field B whose direction is normal to the plane Of Dees, A charged particle of mass 'm' and charge 'q' in the gap between the dees, it moves with constant speed in a semi-circle. The period of uniform circular motion is T=(2pim)/(qB) and is independant of speed. If the time-period of the oscilating elctric field is equal to this time, then the charged particle will be accelerated again and again In a cyclotron resonance experiment, if we assume that maximum radius attained by any chared particles is equal to radius of dees, then minimum kinetics energy gained by the particles will be |
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Answer» ALPHA particle |
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| 48. |
When a piece of wire is held diametrically in a screw gauge [pitch = 1mm, number of divisons on the circular scale=100]. The reading obtained are as shown Now we have to measure the same with the help of vernier callipers [1 MSD= 1mm, 10 divisons of vernier coinciding with 9 divisions of main scale] having a negative zero error of 0.5 mm, then find which of the following figures correctly represents the reading - |
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Answer»
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| 49. |
The conduction current is the same as the displacement current when the source is |
| Answer» Answer :D | |
| 50. |
A cyclotron is device for accelerating ions and charged particles. It was developed by Lawerence in 1932. The heart of the appritus consists of a split metal pillbox. Figure sHOws top and front views of the halves called dees.A rapidly oscilating potential difference is applied between the Dees. This produces an oscilating electric field in the gap between the dees, the region inside each dee being essentially free of electric field. The Dees are enclosed in an evaccutaed container, and the entire unit is placed in a uniform magnetic field B whose direction is normal to the plane Of Dees, A charged particle of mass 'm' and charge 'q' in the gap between the dees, it moves with constant speed in a semi-circle. The period of uniform circular motion is T=(2pim)/(qB) and is independant of speed. If the time-period of the oscilating elctric field is equal to this time, then the charged particle will be accelerated again and again Answer the following questions (consider the mass of particles remains constant during motion): A cyclotron is accelerating deutrons deutrons having mass=2xx1.6xx10^(-27) kg and charge +e. If B=2T, then the required angular frequency of osciliating electric field is : |
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Answer» `10^(8)` radian`//`SEC |
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