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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. |
Compute the binding energy of ""_(2)^(4)"He" nucleus using the following data: Atomic mass of Helium atom, M_(A) (He) = 4.00260 u and that of hydrogen atom, m_(H) = 1.00785 u. |
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Answer» Solution :BINDING ENERGY BE = `[Zm_(H) + Nm_(N) - M_(A)]c^(2)` For helium nucleus, `Z = 2, N = A -Z = 4 - 2 = 2` Mass defect `DELTA m = [(2 xx 1.00785u )] + (2 xx 1.008665u ) - 4.00260 u] Delta m = 0.03038 u` `B.E = 0.03038u xx c^(2)` `B.E = 0.03038 xx 931 MeV = 28 MeV` `[therefore 1 uc^(2) = 931 MeV]` The binding energy of the `""_(2)^(4)"He"` nucleus is 28 MeV. |
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| 2. |
(A) According to classical theory, the proposed path of an electron in Rutherford atom model will be parabolic. (R ) According to electromagnetic theory an accelerated particle continuously emits radiation. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT explanation of 'A'. |
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| 3. |
10 m/s के प्रारम्भिक वेग से एक पिण्ड गति करता है। यदि यह 2 सेकण्ड में 20m की दूरीतय करता है तब पिण्ड का त्वरण होगा ? (m/s^2 में ) |
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Answer» 0 |
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| 4. |
In A.C. circuit having only capacitor, the current ...... |
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Answer» lags behind the VOLTAGE by `PI/2` in PHASE |
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| 6. |
{:(overset(*)(A),overset(*)(S_(1)),overset(*)(B),overset(*)(S_(2)),overset(*)(C)):}In the figure shown, S_(1) and S_(2)represents two stationary sources of sound having equal frequency, one observer is moving form A toward C with velocity V_(0)then - |
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Answer» BEATS for three position A, B and C will be heard |
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| 7. |
A 12 pF capacitor is connected to a 50 V battery . How much electrostatic energy is stored in the capacitor ? If another capacitor of 6 pF is connected in series with it with the same battery connected across the combination , find the charge stored and potential difference across each capacitor. |
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Answer» Solution :Here , `C_1 = 12 pF = 12 xx 10^(-12) F, V = 50 V ` and `C_(2) = 6 pF = 6 xx 10^(-12) F` `therefore` Electrostatic energy in first capacitor `U_(1) = (1)/(2) C_(1) V^2 = (1)/(2) xx (12 xx 10^(-12)) xx (50)^2 = 15 xx 10^(-9) J = 15 nJ` When the two capacitors are joined in series across same battery , then resultant CAPACITANCE C is given by `1/C = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(12) + (1)/(6) = (1)/(4) implies C = 4 pF` `therefore` CHARGE stored on 1st capacitor = Charge stored on 2nd capacitor `Q = CV = (4 pF) xx 50 V = 200 pC` `therefore` Potential across 1st capacitor `V_1 = (Q)/(C_1) = (200 pC)/(12 pF) = 16.67 V` and potential difference across 2nd capacitor `V_2= (Q)/(C_(2)) (200 pC)/(6pF) = 33.33 V` |
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| 8. |
A 220 V input is supplied to a transformer. The output circuit draws a current of 2 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is |
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Answer» 5.0 A |
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| 9. |
Photoelectric effect in photocell convert…. |
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Answer» CHARGE in intensity of LIGHT into change in photo CURRENT |
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| 10. |
which of the following digital form is in electrical series circuit |
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Answer» AND |
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| 11. |
An LCR circuit in resonance . The capacitance is decrease to 1/4 of its original value. What should be the new inductance so that the circuit remains in resonace ? |
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Answer» increase 2 times |
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| 12. |
What do you mean by electron emission ? Explain briefly various methods of electron emission. |
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Answer» Solution :Electron emission : Free electrons posses some kinetic energy and this energy is DIFFERENT for different electrons. The kinetic energy of the free electrons is not sufficient to overcome the surface barrier. Whenever an additional energy is given to the free electrons, they will have sufficient energy to cross the surface barrier. And they escape from the metallic surface. The liberation of electrons from any surface of a substance is called electron emission. There are mainly four types of electron emission which are given below. (i) Thermionic emission : When a metal is HEATED to a high temperature, the free electrons on the surface of the metal get sufficient energy in the form of thermal energy so that they are emitted from the metallic surface. This type of emission is known as thermionic emission.  The intesity of the thermionic emission (the number of electrons emitted) depends on the metal used and its temperature.  Examples : cathode ray tubes, electron microscopes, X-ray tubes etc. (ii) Field emission : Electric field emission occurs when a very strong electric field is applied across the metal. This strong field pulls the free electrons and HELPS them to overcome the surface barrier of the metal.  Examples : Field emission SCANNING electron microscopes, Field-emission display etc. (iii) Photo electric emission : When an electromagnetic radiation of suitable frequency is incident on the surface of the metal, the energy is trasferred from the radiation to the free electrons. Hence, the free electrons get sufficient energy to cross the surface barrier and the photo electric emission takes place. The number of electrons emitted depends on the intensity of the incident radiation.  Examples : Photo diodes, photo electric cells etc. (iv) Secondary emission : When a beam of FAST moving electrons strikes the surface of the metal, the kinetic energy of the striking electrons is transferred to the free electrons on the metal surface. Thus the free electrons get sufficient kinetic energy so that the secondary emission of electron occurs.  Examples : Image intensifiers, photo multiplier tubes etc. |
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| 13. |
In Young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm. Number of fringe observed in the same segment of the screen is given by |
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Answer» 18 |
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| 14. |
Freezing is reverse of ………… |
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Answer» Condensation |
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| 15. |
A sample of of paramagnetic salt contains2.0xx10^(24) atomic dipoles each of dipole moment 1.5xx10^(-23)JT^(-1). The sample is placed under a homogeneous magnetic field of 0.64 T , cooled to a temperature of 4.2 K . The degree of magnetic saturation achieved is equal to 15% . What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K ? (Assume Curie's law) |
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Answer» SOLUTION :`N=2.0xx10^(24) ""m=1.5xx10^(-23)JT^(-1) ""B_("ex")=0.84 T, T = 4.2 K` % saturation = 15% , `sumNm` = ? `=B=0.98T,T =2.8K` Under the initial conditions , the total MAGNETIC moment `=1.5xx10^(-23)xx2.0xx10^(24)xx15/100=4.5JT^(-1)` By Curie law, `M=C.B_0/T "" because m/V=C.(B_0)/T""(m.)/V=C.(B)/(T.)` `:.(m.)/m=B/(T.)xxT/B_0"":.m.=(4.5xx0.98xx4.2)/(2.8xx0.84)=7.9JT^(-1)` |
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| 16. |
The volume of air decreased to one third of its original value as a result of shock compression. How many times did the pressure of air and its temperature increase? Compare with the variation of these quantities as a result of quasistatic adiabatic compression. |
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Answer» `p//p_0 = 4.67 ; T//T_0 = 1.56` (adiabatic compression for a QUASISTATIC process) |
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| 17. |
(a) State briefly, with what purpose was Davisson and Germer experiment performed and what inference was drawn from this. (b) Obtain an expression for the ratio of the accelerating potentials required to accelerate a proton and an alpha-particle to have the samede-Broglie wavelength associated with them. |
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Answer» SOLUTION :(a) Davisson and Germer experiment was performed to check the wave nature of electrons and to VERIFY de-Broglie.s formula for wavelength of electron waves. Davisson and Germer were able to demonstrate diffraction of electron waves and measured their wavelength. the experimentally determined VALUE of electron.s wavelength was found to be in close agreement with the ANSWER obtained as per de-Broglie relation. hence, wave nature of electrons was conclusively proved. (b) In terms of accelerating potential V, the de-Broglie wavelength of a particle of charge q and mass m is given as `LAMDA=(h)/(sqrt(2mqV))` As per question `lamda_("proton")=lamda_(alpha-"particle")` `implies m_(p)q_(p)V_(p)=m_(alpha)q_(alpha)V_(alpha)` `implies (V_(p))/(V_(alpha))=(m_(alpha))/(m_(p))*(q_(alpha))/(q_(p))=4xx2=8implies V_(p)=8V_(alpha)` |
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| 18. |
A solenoid is of length 50cm and has a radius of 2cm. It has 500 turns. Around its central section of coil of 50 turns is wound. Calculate the mutual inductance of the system. |
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Answer» Solution :`N_(P)= 500, N_(S)= 50` `A= PI xx 0.02 xx 0.02m^(2)` `mu_(0) = 4pi xx 10^(-7) HM^(-1), l= 50CM= 0.5m` Now, `M= (mu_(0)N_(P)N_(S)A)/(l)` `=(4pi xx 10^(-7) xx 500 xx 50 xx pi xx (0.02)^(2))/(0.5)H` `= 789.8 xx 10^(-7)H= 78.98 mu H` |
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| 19. |
E.M.F has dimensions of |
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Answer» `M^(1)L^(2)T^(-3)A^(-1)` |
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| 20. |
Let bar(v)(t) be the velocity of a particle al lime t. Then : |
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Answer» `|dbar(V)(t)//DT|and d|bar(v)(t)|//dt|"are ALWAYS equal"` |
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| 21. |
Suppose there is a circull consisting of only resistances and batteries. Suppose one is to double (or increase it to n-times) all voltages and all resistances. Show that currents are unaltered. |
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Answer» Solution :LET equivalent internal resistance of GIVEN cells be `R_(eq). ` Equivalent voltage be `V_(eq)` EXTERNAL resistance be R. From Ohm.s law, `I_(1) = (V_(eq))/(R_(eq) + R) "" `.... (1) When value of resistors and all batteries MADE n times initial value then, `therefore V_(eq) = nV_(eq), R_(eq) = nR_(eq)` and new external resistor R= NR Current in new (modified ) circuit, `I_(1) = (n V_(eq))/(nR_(eq) + nR) = (n V_(eq))/(n (R_(eq) + R)) ` ` = (V_(eq))/(R_(eq)+ R)` `therefore ` Comparing (1) and (2) `I_(1) = I_(1)` thus current do not change. |
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| 22. |
Calculate the force experienced by a moving electron which is entering into a condenser having its plates 0.1m apart and potential difference of 300V. The direction of the electric field is perpendicular to the motion of the electron. |
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Answer» `5.8xx10^(-16)N` |
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| 23. |
If the temperature of a hot body is raised would increased by |
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Answer» 0.16 `(R_(2)-R_(1))/(R_(1))=0.16=16%`. |
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| 24. |
Which is more advantageous, amplitude modulations or frequency modulation ? |
| Answer» SOLUTION :AMPLITUDE MODULATION. | |
| 25. |
If the vectors vec(a) and vec(b) are such that |vec(a) + vec(b)|=|vec(a)| then does it imply that: |
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Answer» `vec(b)=0` |
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| 26. |
Stationary waves of frequency 200 Hz are formed in air. If velocity of the wave is 360 m/s, the shortest distance between two antinodes is : |
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Answer» 1.8 m |
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| 27. |
In short wave communications, waves of which of the following frequencies will be reflected back by the ionosphere layer having electron density 10^(11-3)m? |
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Answer» `2MHundersetz` |
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| 28. |
A relation between time t and distance x ist= alpha x^2 + beta x where alpha and beta are constants. The retardation is |
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Answer» `-2 ALPHA v^3` |
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| 29. |
A vibrating tuning fork generates a wave given by y = 0.1 sin pi (0.1 x- 2t) , Where x and y are in metre and t is in second. The distance travelled by the wave while the fork completes 30 vibrations is : |
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Answer» 600 m COMPARING it with y = R sin `(kx - omega t)`. `rArr "" k = 0.1 pi and omega= 2pi ` `(2pi)/(lambda) = 0.1 pi "" rArr "" lambda ` = 20 m, `therefore` DISTANCE travelled by wave when TUNNING fork completes 30 vibrations = 30 `lambda` = 600 m Correct choice is a. |
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| 30. |
Draw a ray diagram of an astronomical telescope in the normal adjustment position. State two drawbacks of this type of telescope |
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Answer» Solution :A refractive TYPE astronomical telescope suffers with the following drawbacks : (i) The IMAGE formed has both chromatic as well as SPHERICAL aberration. (ii) It is extremely difficult to design and MAINTAIN the mechanical support of the telescope. |
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| 31. |
In reflecting type telescope ......... |
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Answer» `f_0=f_(E),D_0=D_e` |
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| 32. |
Let a small block of ice of 0^(@)C fall from a certain height into a water k. It at 0^(@)C, we find that (1)/(8) thof molts when it reaches the ground. The height of the fall is: |
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Answer» 2100 m `therefore` Work done W= mgh. Heat produced `Q=(W)/(J)=(mgh)/(J)` Now heat required to MELT `(1)/(8)`th block `Q.=(m)/(8) xxL` `therefore (mgh)/(J) =(m)/(8)xxL" "RARR h=(JL)/(8_(g))` `h=(4 CDOT 2xx80xx10^(3))/(8xx10)=4200 m`. THUS, correct choice is (b). |
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| 33. |
The diagram shows the basic setup for the production of X-rays. A_(1) and A_(2) are two ammeters, reading 2.55A and 2.566A respectively. F is a filament which is alos the cathode. The potential difference applied between P and Q is 50000 V. Assume that all X-rays photons have the maximum possible energy and that one X-ray photon is emitted for every 100 electron incident on the target. You may assume that the kinetic energy ofother electrons reappear as heat in the tube. The rate at which heat is produced in the X-ray tube is approximately |
| Answer» Answer :D | |
| 34. |
Show that in the free oscillationsof an LC circuit , the sum of energies stored in the capacitor and the inductor is constant in time . |
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Answer» SOLUTION :Let `q_0` be the initial charge on a capacitor . Let the charged capacitor be connected to an inductorof inductance L. this LC circuit will sustain an oscillation with frequency ` omega ( = 2PI v= (1)/(IsqrtLC) )` . At an instant t, charge Q on the capacitor and the current I are given by `q(t)= q_0 cos omega t` `I(t) = - q_0 omega sin omega t` Energy stored in the capacitor at time t is `U_E = 1/2 CV^2 = 1/2 (q^2)/(C ) = (q_0^2)/(2C) cos^2 (omega t)` Energy stored in the inductor at time t is `U_M = 1/2 Li^2` ` = 1/2 Lq_0^2 omega^2 sin^2 (omega t) = (q_0^2)/(2C) sin^2 (omega t) ( because omega^2= (1)/(sqrtLC) )` sum of ENERGIES `U_E+ U_M = (q_0^2)/(2C) cos ^2 omega t + sin omega t = (q_0^2)/(2C)` this sum is constant in time as `q_0` and C, both are time independent . Is equal to the initial energy of the capacitor . |
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| 35. |
The diagram shows the basic setup for the production of X-rays. A_(1) and A_(2) are two ammeters, reading 2.55A and 2.566A respectively. F is a filament which is alos the cathode. The potential difference applied between P and Q is 50000 V. Assume that all X-rays photons have the maximum possible energy and that one X-ray photon is emitted for every 100 electron incident on the target. You may assume that the kinetic energy ofother electrons reappear as heat in the tube. The number of X-ray photons produced per second is approximately |
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Answer» `10^(12)` |
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| 36. |
Answer the followingquestion , whichhelp you understand the differencebetween Thomson'smodel and Rutherford's model better. Is theprobailityof backward scattering(i.e.,scattering of alpha - particle at angles greaterthan 90^(@)) perdictedby Thomsom's model much less , aboutthe same , or muchgreater than the predicted by Thomson'n modelmuch less ,aboutthe same ,or muchgreater than thatpredicted by Rutheford's model ? |
| Answer» SOLUTION :The probability of backward scattering (i.e., scattering of `alpha` - PARTICLES at ANGLE greater than `90^(@)`) predictedby Thomson.smodel ismuchless than that predicted by Rutherford.s ATOM model. | |
| 37. |
The diagram shows the basic setup for the production of X-rays. A_(1) and A_(2) are two ammeters, reading 2.55A and 2.566A respectively. F is a filament which is alos the cathode. The potential difference applied between P and Q is 50000 V. Assume that all X-rays photons have the maximum possible energy and that one X-ray photon is emitted for every 100 electron incident on the target. You may assume that the kinetic energy ofother electrons reappear as heat in the tube. The momentum of each X-ray photon is approximately |
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Answer» `3 XX 10^(-17) kg ms^(-1)` |
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| 38. |
A short bar magnet of amgnetic moment 25 JT^(-1)is placed with its axis perpendicular to earththe resultant field is inclined at 45^(@)with earth fieldif H=0.4xx10^(-4)T |
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Answer» 5m so B=H `therefore (mu_(0))/(4pi)(2M)/(r^(3))=H` `therefore r^(3)=(mu_(0))/(4pi)(2M)/(H)=0.5 m` |
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| 39. |
Four spheres of radius r each and mass m are placed with their centres on the four corners of the square of side 'a'. The moment of inertia of system about an axis along one of sides of square is : |
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Answer» `(4)/(5)MR^(2)+2ma^(2)` |
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| 40. |
Three deminsions. Three point particles are fixed in place in a xyz coordinate system. Particle A, at the origin, has mass m_A. Particle B, at xyz coordinates (2.00d,1.00d. 200), has mass 2.00m_A, and particle C, at coordinates (-1.00d,2.00d,-3.00d), has mass 3.00m_A. A fourth particle D,with mass 4.00 m_A,is to be placed near the other particles. In terms of distance d, at what (a)x, (b)y, and (c)z coordinate should D be placed so that the net gravitational force on A from B, C, and D is zero ? |
| Answer» SOLUTION :`(a) -1.88d, (b) -3.90d, (C ) 0.489d`. In this way, we are ABLE to deduce that (X,y,Z) = (1.88d,3.90d, 0.489d). | |
| 41. |
Answer the followingquestion , whichhelp you understand the differencebetween Thomson'smodel and Rutherford's model better. Is the average angle of deflection of a particles by a thin gold foil predicted by Thomson's model much less, about the same, or much greater than that predicted by Rutherford's model ? |
| Answer» SOLUTION :The average angleof deflection of `alpha` - PARTICLE BYA THIN glodthin GLOD foil predicted by Thomson.s model is aboutthe same as that predicted by Rutherford.s model. | |
| 42. |
What amount of energy sould be added to an electron to reduce its de Broglie wavelength lamda_1=550nm incident on it, causing the ejection of photoelectrons for which the stopping potential is V_(s1)=0.19V. If the radiation of wavelength lamda_2=190nm is now incident on the surface, (a) calculate the stopping potential V_(S2), (b) the work function of the surface, and (c) the threshold frequency for the surface. |
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Answer» Solution :We know that, de Broglie wavelength `lamda=(h)/(mv)` and`R=(1)/(2)mv^2` `lamda=(h)/(SQRT(2mE))` In first case, `100xx10^(-12)=(h)/(sqrt(2mE_(1))` …(i) In second case, `50xx10^(-12)=(h)/(sqrt(2mE_(2))`..(ii) Dividing Eqs. (i) by (ii), we get `2=sqrt(((E_2)/(E_1)))` or `E_2=4E_1` So, energy to be ADDED `=4E_1-E_1=3E_1` Now, `(h)/(sqrt(2mE_1))=100xx10^(-12)` or `sqrt(2mE_1)=(6.625xx10^(-34))/(10^(-10))` or `sqrt(2mE_1)=6.625xx10^(-24)` or `E_1=((6.625xx10^(-24))^(2))/(2XX(9.1xx10^(-31)))=150eV` There, energy added `=3E_1=450eV` |
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| 43. |
Gibbon is related to .............. |
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Answer» Greece |
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| 44. |
If the tube length of astronomical telescope is 1.5 cm and magnifying power is 20 for normal setting, calculate the focal length of the objective. |
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Answer» 100 cm |
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| 45. |
Define inductance, give its units and write factors on which its value depends. |
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Answer» Solution :An electric current can be induced in a coil by flux change produced by another coil in its vicinity or flux change produced by the same coil. However, in both the cases, the flux through a coil is proportional to the current. That is, `Phi_B prop I`. For a closely wound coil of N turns, the same magnetic flux is linked with all the turns. So, total flux linkage is EQUAL to `NPhi_B`, which can be given by `NPhi_B prop I` The constant of proportionality in this relation is called inductance. Inductance depends only on the geometry of the coil and intrinsic material properties. (This aspect is similar to capacitance which for a parallel plate capacitor depends on the plate area and plate separation (geometry) and the DIELECTRIC constant K of the intervening medium (intrinsic material property). Inductance is a scalar quantity. It has the dimensions of `[ML^2 T^(-2)A^(-2)]` . The SI unit of inductance is henry and is denoted by H. It is named in honour of JOSEPH Henry who discovered electromagnetic induction in USA, independently of Faraday in England. There are two type of inductance :(1) MUTUAL inductance (2) Self inductance. |
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| 46. |
The third band in colour coded resistance represents |
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Answer» THIRD digit of resistance |
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| 47. |
In the figure shown consider the first reflection at the plane mirror and second at the convex mirror. AB is object. |
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Answer» the SECOND image is real, inverted of 1/5th magnification |
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| 48. |
A battery of negligible internal resistance is connected with 10m long wire. A standard cell gets balanced on 6m length of this wire.On increasing the length of potentiometer wire by 2m,the null point with the same standard cell in the secondary will be (no series resistance in primary) |
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Answer» INCREASED by 2m |
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| 49. |
A mass 'M' is suspended from a spring of negligible mass. The spring is pulled a little and then released. It execuutes S.H. oscillations of period T. When mass is increased by 'm', the period becomes (5)/(4) T, the ratio (m)/(M) is : |
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Answer» `(9)/(16)` and `(5T)/(4)=2pi sqrt((M+m)/(k))""…(ii)` DIVIDING (ii) by (i) `:.""(5)/(4)=sqrt((M+m)/(M))""IMPLIES""(25)/(16)=(M+m)/(M)` `:.""(m)/(M)=(9)/(16)` Hence correct choice is (a). |
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| 50. |
When a 10muCcharge is enclosed by a closed surface, the flux passing through the surface is phi. Now another -10muCcharge is placed inside the closed surface, then the flux passing through the surface is......... |
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Answer» `2phi` `therefore sumq = q_(1) + q_(2) =10-10=0` `therefore` FLUX `phi =(sumq)/epsilon_(0)`, in this equation `sumq=0` `therefore phi=0` |
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