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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. |
For an elctron in circular orbit around a nucleus the ratio of P.E. to K.E. is: |
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Answer» `1/4` |
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| 2. |
A 4muF capacitor is charged to a potential 10V using a battery. The battery is then removed and then this capacitor is connected parallel to an unchanged capacitor of capacity 6muF. What is the common potential? |
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Answer» Solution :DATA SUPPLIED, `C_(1)=4muF, V_(0)=10V, C_(2)=6muF` Common potential =V By the law of conservation of charges, `C_(1)V_(0)=C_(1)V+C_(2)V=(C_(1)+C_(2))V` `V=(C_(1)V_(0))/(C_(1)+C_(2))=(4 XX 10^(-6) xx 10)/((4+6) 10^(-6))=4V` |
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| 3. |
Two lenses of power - 15 D and + 5 D are in contact with each other. The focal length of the combination is |
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Answer» Solution :`P = P_(1) + P_(2) = - 15 + 5 = - 10D` `F = (1)/(P) = (1)/(-10) m = - 10 CM` |
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| 4. |
In Fig. 27-40, R_1= 100 Omega, R_2=R_3= 50.00Omega, R_4 = 75.0 Omega, and the ideal battery has emf epsi= 12.0 V. (a) What is the equivalent resistance? What is i in (b) resistance 1, (c) resistance 2,(d) resistance 3, and (e) resistance 4? |
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Answer» |
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| 5. |
The phenomenon of perfect diamagnetism in super conductors is called the ___________ . |
| Answer» SOLUTION :MEISSNER EFFECT | |
| 6. |
A particle of mass 10 ^(-3)kg and charge5 mu Ccenters into a uniform electric field of2 xx 10^(5)N C^(-1)moving with a velocity of 20 ms^(-1)in a direction opposite to that of the field. Calculate the distance it would travel before coming to rest. |
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Answer» Solution :Here mass of particle m `= 10 ^(-3) kg , `charge `q= 5 muC= 5 xx 10^(-6)C, ` initial velocity of particle `=u =20 MS ^(-1) , ` FINAL velocity v= 0 and electric field E = `-2xx 10 ^(5)N C^(-1)`. (The electric field has been taken -ve because its direction is opposite to that of direction of motion of charged particle. ) ` THEREFORE ` Acceleration or particle ` a= (F)/(m)=(qE)/(m)= ( (5XX 10^(-6))xx (-2xx 10^(5)))/(10^(-3)) =-10 ^(3) ms ^(-2) ` From equation `v^(2) -u^(2) =2as`, the distance travelled by charged particle. ` "" s=( v^(2) -u^(2))/(2a) =((0)^(2) -(20)^(2))/(2(-10^(3)) =0.2 m ` |
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| 7. |
The capacitor A shown in fig. has a capacitance C_(1) = 3 mu F. The dielectric filled in it has a breakdown voltage of 40 V and it has a resistance of 3 MOmega. The capacitor B has a capacitance of C_(2) = 2 mu F and dielectric in it has a resistance of 2 MOmega. Breakdown voltage for B is 50 V. The switch is closed at t = 0. Will there be breakdown of any capacitor after the switch is closed ? If yes, which will breakdown first and at what time? |
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Answer» |
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| 8. |
Why did the authour accept his approaching death? |
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Answer» PIRATES had ATTACKED the ship |
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| 9. |
Suppose y=f(x)andy=g(x)are two functions whose graph intersect at three points (0.4), (2,2) and (4.0), with f(x)gtg(x)" for "0ltxlt2 and f(x)ltg(x)" for " 2ltxlt4.if int_(0)^(4)(f(x)-g(x))dx=10and int_(2)^(4)[g(x)-f(x)]dx=5.Ifareas between two curves for0ltxlt2isA_(1) and area between two curves for0ltxlt4 isA_(2), then - |
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Answer» `A_(1)=5` |
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| 10. |
(i) State Bohr's quantisation condition for defining stationary orbits. How does de-Broglie hypothesis explain the stationary orbits ? (ii) Find the relation between the three wavelengths, lambda_(1),lambda_(2) and lambda_(3)from the energy level diagram shown here. |
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Answer» Solution :(i) See ShortAnswer Question Number 29. (ii)LET energy of electron corresponding to LEVELS A, B and C by `E_(A)` , `E_(B)` and `E_(C)` RESPECTIVELY . Then `E_(C) - E_(B) = (HC)/(lambda_(1)) ""…(i) E_(B) =E_(A) = (hc)/(lambda_(2)) ""......(ii) and E_(C) -E_(A) =(hc)/(lambda_(3))"".......(iii) ` On adding(i) and (ii), we get . `E_(C) = E_(A) = (hc)/(lambda_(1)) + (hc)/(lambda_(2))"".......(iv)` Now comparing(iii) and (iv) , we findthat . `(1)/(lambda_(3)) = (1)/(lambda_(1)) + (1)/(lambda_(2))` 
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| 11. |
How field lines depend on area or on solid angle made by area ? |
Answer» Solution :Figure shows a set of field lines.  Imagine two equal and small elements of area placed at points R and S NORMAL to the field lines. The number of field lines in our picture cutting the area elements is proportional to the magnitude of field at these points. The picture shows that the field at R is STRONGER than at S. The angle subtended by `Deltal` at O can be approximated as `Deltatheta =(Deltal)/r` Likewise, in three-dimensions, the solid angle `DeltaOmega =(DeltaS)/r^(2)`. In a given solid angle the number of radial field lines is the same. For two points `P_(1)`and `P_2` at distances `r_(1)` and `r_2` from the charge, the element of area SUBTENDING the solid angle `DeltaOmega`is `r_(1)^(2) DeltaOmega` at `P_1` and an element of area `r_(2)^(2) DeltaOmega` at `P_(2)`. The number of lines n cutting these area elements are the same. The number of field lines, cutting unit area element is therefore `n/(r_(1)^(2)DeltaOmega)` at `P_(1)` and `n/(r_(2)^(2)DeltaOmega)` at `P_(2)` , respectively. Since n and `DeltaOmega` are COMMON, the strength of the field `1/r^(2)`.DEPENDS on |
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| 12. |
The velocity time graph of a body moving along a straight line is shown in Fig. The acceleration of the body during OA will be: |
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Answer» Zero `=(20)/(2)=10m//s^(2)` |
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| 13. |
A point charge q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is |
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Answer» `(q)/(epsilon_(0))` |
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| 15. |
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure .(##ARH_EGN_PRG_PHY_C17_E02_019_Q01.png" width="80%"> The change in internal energy of the gas during the transition is |
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Answer» 20kJ  The change in internal energy of gas in the transition from A to B is`Delta U=nC_(V)dT` `= n ((5R)/(2))(T_(B)-T_(A))=nR(5)/(2)((p_(B)V_(B))/(nR)-(p_(A)V_(A))/(nR))` =`(5)/(2)(2xx10^(3)xx6-5xx10^(3)xx4)` = `(5)/(2)(-8xx10^(3))=-(4xx10^(4))/(2)=-20kJ` |
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| 16. |
Two circular coils are made from a uniform copper wire. Radii of circular coils is in the ratio 3:4 and number of turns in the ratio 3:5. If they are connected in series across a battery Statement (A) : Ratio between magnetic field inductions at their centers is 4:5 Statement (B): Ratio between effective magnetic moments of the two coils is 16:15 |
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Answer» Both statements are WRONG |
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| 17. |
S.T. average power over a complete cycle in a pure inductor connected to ac is zero. |
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Answer» SOLUTION :The instantaneous power supplied to an INDUCTOR `P_(L)=iv` `=i_(m)sin(omegat-pi//2)(v_(m)SINOMEGAT)` `=-i_(m)cosomegatv_(m)sinomegat` i.e., `P_(L)=-(1)/(2)i_(m)v_(m)(sin2omegat)` where `sin2omegat=2sinomegatcosomegat` The AVERAGE power over a complete cycle is `P_(L)=(:-(i_(m)v_(m))/(2)sin2omegat:)` `=-(i_(m)v_(m))/(2)(:sin2omegat:)` = 0 `therefore (:sin2omegat:)=(:cos2omegat:)=0` Thus average power supplied to an inductor over one complete cycle is zero. |
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| 18. |
A wave represented by the equation y = a cos(kx - wt) is superposed with another wave to form astationary 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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| 19. |
In the following question a statement of assertion (A) is followed by a statement of reason (R ) A : When two conductors charged to different potentials are connected to each other , the negative charge always flows from lower potential to higher potential. R : In the charging process, there is always a flow of electrons only. |
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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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| 20. |
A magnetic needle lying parallel to a magnetic field requires w units of workto turn it through 60^(@)thetorque needed to maintain the needle in this positionwill be |
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Answer» w `tau =MB sin 60^(@)=2w.sqrt(3)/(2)=sqrt(3)W` |
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| 21. |
An image of an object is formed by using a convex lens. If the orientation of the lens is reversed, will the position of the image change ? |
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Answer» Solution :As `(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))` On reversing the ORIENTATION of the lens, `R_(1) and R_(2)` will get interchanged. At the same time signs of `R_(1) and R_(2)` also get interchanged, hence focal length remains UNAFFECTED. POSITION of image is also unaffected. |
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| 22. |
Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36 m. During the collision at the bottom of the elevator shaft, a 90 kg passenger is stopped in 5.0 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)? |
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| 23. |
(a) How dpes an unpolarized light incident on a polaroid get polarized? Describe briefly with the help of a necessary diagram the polarization of light by reflection from a transparent medium. (b) Two polaroids 'A' and 'B' are kept in crossed position . How should a third polaroid 'C' be placed between them so that the intensity of polarized light transmitted by polaroid B reduces to 1//8^(th) of the intensity of unpolarized light incident on A ? |
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Answer» Solution :(a) When an unpolarised light falls on a polaroid, it lets only those of its ELECTRIC vectors that are oscillating along a direction perpendicular to its aligned molecules to pass through it. The incident light thus gets linearly polarised. Alternatively,  WHENEVER unpolarised light is incident on a transparent surface, the reflected light gets partially or completely polarized/the reflected light gets completely polarized when the reflected and refracted light are perpendicular to each other. (b) Let `theta` be the angle between the pass axis of A and C Intensity of light passing through `A=(I_(0))/(2)` Intensity of light passing through`C=((I_(0))/(2))cos^(2)theta` Intensity of light passing through B `=((I_(0))/(2))cos^(2)theta.[cos^(2)(90-theta)]` `implies(I_(0))/(2)xx(SIN^(2)theta)/(4)=(I_(0))/(8)` (Given) `:.sin2theta=1` `2theta=90^(@)` The third polaroid is placed at `theta=45^(@)` |
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| 24. |
A transformer with 80% efficiency works at 4 kW and 200 V. If the secondary voltage is 1000 V then the primary and secondary currents are respectively |
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Answer» 20 A, 3.2 A As output VOLTAGE is 1000 V, hence output current `I_(s) = 3200/1000 = 3.2 A` |
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| 25. |
The length of the conductor in the diagram shown in Fig. 30.5 is l= 20 cm, its speed is v=1 m/s and the resistance of the bulb is R =1 ohm. A magnetic field with induction B=0.5 T is set up perpendicular to the plane of the diagram. What force should be applied to the conductor to make it move at the speed specified? |
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| 26. |
If Bohar 's quantisationpostulate( angularmomentum = nh//2pi) is a basic law of nature , it should be equalbe equallyvalid forthe caseof planetarymotionalso . Whythen dowe never speak of quatisation of orbitof planets around the Sun ? |
| Answer» SOLUTION :If we tryto applyBohr.s quantum conditionsto PLANETARY motion then on calculatingangularmomentum OFPLANET we find the VALUEOF angularmomentum is incomparably large relativeto Planck.s constant h. For earth , it issuchthat quantum numbern `=10^(70)` .For such large values ofn,the difference in thesuccessive energies and angularmomenta of thequantised levels of theBohr modelare sosmall that of all purposewe may considerthe angularmomentum aswellas ENERGY levels continuous. | |
| 27. |
A man running on the horizontal road at8 km h^(-1) find the rain appears to be falling vertically. He incresases his speed to 12 km h^(-1) and find that the drops make angle30^2 with the vertical. Fin dthe speed and direction of the rain with respedt to the road. |
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Answer» `4 sqrt7km//h` |
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| 28. |
The no. of lines of force passing normally through a given unit area is called______. |
| Answer» SOLUTION :MAGNETIC INDUCTION | |
| 29. |
In an induction coil, the secondary e.m.f. is |
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Answer» ZERO during BREAK of the CIRCUIT |
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| 30. |
A : When two lenses in contact form an achromatic doublet, then the materials of the two lenses are always different. R : The dispersive powers of the materials of the two lenses are of opposite sign. |
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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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| 31. |
An equiconvex lens of focal length 10 cm (in air) and R.I. 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves M_(1)andM_(2) and placed at the end of the tube. M_(1)andM_(2) are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axis of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5xx10^(14)Hz. The light reflected from M_(1)andM_(2) forms interference pattern on the left end EF of the tube. O is an opaque substance to cover the hole left by M_(1)andM_(2). Width of the fringes on EF is (x xx10)mum. Find the value of x. |
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Answer» 5 m |
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| 32. |
Twomonochromaticcoherent point sources S_(1) andS_(2) are separated by a distance L , Each source emitslight of wavelengths lambdawhereL gt gt lambda. The line S_(1)S_(2)when extended meets a screenperpendicular to it at point A . Which of the following is correct ? |
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Answer» The interference frings are circular in SHAPE |
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| 33. |
A point source of light is placed at the bottom of a water lake. If the area of the illuminated circle on the surface is equal to 3 times the square of the depth of the lake. The refractive index of water is |
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Answer» `SQRT(PI+1)` |
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| 34. |
A block of mass 1kg is dropped on a spring - mass system as shown in the figure. The block traves 100 meters in the air before strinking the 3 kg mass. Calculate maximum compression in the spring, if both the blocks move together after the collision. Spring constant of the string k=1.25xx10^(6). |
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Answer» 2 cm |
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| 35. |
Define the terms : Stopping potential. |
| Answer» Solution :STOPPING POTENTIAL is the MINIMUM negative (retarding) potential of anode for which phtocurrent stops or becomes ZERO. | |
| 36. |
Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see whatis wrong ? In what way is the formula to be modified ? |
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Answer» SOLUTION :V = 10 MV = `10^(7) V , v = sqrt(e/m xx2v)= sqrt(1.76 xx10^(11) xx 2 xx10^(7)) :. v = 1.8762 xx10^(9) `m/s This SPEED is greater than speed of light , which is not possible . As `v`approaches to c, then mass m = ` (m_(0))/(sqrt(1-(v^(2))/(c^(2))))` |
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| 37. |
A smooth circular table is surrounded by a rim whose interior is vertical . A ball is projected along the table from a point on the rim in a direction making an angle theta to the radius through the point and returns to the point of projection after two impacts . If e be the coefficient of restitution, then |
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Answer» `tan THETA = SQRT(((E^(3))/( 1+ e + e^(2))))` |
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| 38. |
A series LCR circuit is connected to an ac source of frequency v and a voltage V. At this frequency, reactance of the capacitor is 350 Omegawhile the resistance of the circuit is 180Omega. Current in the circuit leads the voltage by 54^@and power dissipated in the circuit is 140 W. Then the voltage V is |
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Answer» 250V |
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| 39. |
0.07200 मे सार्थक अंकों की संख्या है- |
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Answer» 2 |
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| 40. |
If radius of earth is reduced by 1%, the escape velocitywill (If mass of earth remains same): |
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Answer» INCREASE by 0.5% % INCREASEIN ESCAPE velocity `=(V_(e)^(.)-V_(e ))/(V_(e )) xx 100` `=(sqrt((200)/(99)-(GM)/(R ))-sqrt((2GM)/(R )))/(sqrt((2GM)/(R )))xx100=( sqrt((100)/(99))-1)xx100` `=0.5%` So, correct choice is (a). |
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| 41. |
An inductor, a capacitor, and a resistor are connected in series with a frequency generator to complete a circuit. The frequency generator can supply alternating current at frequencies ranging from about 0 Hz to 100 MHz. As the frequency is increased from the lowest to the highest value of the generator there is a particular value for which the voltage across the resistor is maximum. At that frequency |
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Answer» The voltage across the inductor is zero at all TIMES |
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| 42. |
In Davisson-Germer experiment, an electron beam of 60eV energy falls normally to the surface of the crystal and maximum intensity is obtained at an angle of 60^@to the direction of incident beam. The inter- atomic distance in the lattice plane of the crystal is |
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Answer» 18Å |
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| 43. |
Nitrogen is a diatomic gas. Its molar specific heat a constant volume is very nearly: |
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Answer» `(5R)/(2)` `f=3" for "T le 70 k`. `f=5" for "250 k le T le 750 k` `f=7" for "T GE 750 k` Thus, CORRECT choice is (d). |
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| 44. |
An alternating-current emf device in a certain circuit has a smaller resistance than that of the resistive load in the circuit , to increase the transfer of energy from the device to the load , a transformer will be connected between the two . (a) Should N_s be greater than or less than N_p ?(b) Will that make it a step-up or step-down transformer ? |
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Answer» |
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| 45. |
A stone is dropped from a hill of height 180 m. Two seconds later another stone is dropped from a point P below the top of the hill . If the two stones reach the groud simultaneously, the height of P from the ground is (g = 10 ms^(-2)) |
| Answer» ANSWER :C | |
| 46. |
The plates or a parallel plate capacitor have an area of 90 cm^(2) each and are separated by 2.5 mm. The capacitor is charged by connecting It to a 400 V supply. (a) How much electrostatic energy is stored bythe capacitor ? (b) View this energy as stored in the electrostatic field between the plates, und obtain the energy per unit volume "· Hence arrive at a relation between u and the magnitude of electric field E between the plates. |
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Answer» Solution :(a) Here A = 90 `cm^(2) = 90xx10^(-4) m^(2) = 9xx10^(-3) m^(2)` d = 2.5 MM = `2.5 xx10^(-3)` m `in_(0) = 8.85 xx10^(-12) Fm^(-1)` V = 400 V Capacitance of PARALLEL of parallel of parallel plate capacitor `C= (in_(0)A)/(d)=(8.85xx10^(-12)xx9xx10^(-3))/(2.5xx10^(-3))` `:. C = 31.86xx10^(-12) F = 31.86 pF` Energy stored in capacitor `U = (1)/(2) CV^(2) =(1)/(2) xx31.86 xx10^(-12)XX(400)^(2)` `:. U = 254.88xx10^(-8) `J `:. U = 2.55 xx10^(-6) J = 2.55 mu J` (b) Enerby per unit of VOLUME or energy density in capacitor `rho_(E) = u = (U)/(Ad) = (2.55xx10^(-6))/(9xx10^(-3)xx2.5xx10^(-3))` `:. u =0.113 jm^(-3)` Relation between `rho_(E)(u)` and E `rho_(E) (U)/(Ad) =(1//2CV^(2))/(Ad)` `rho_(E)=(1)/(2)(in_(0)A)/(d) xx(V^(2))/(Ad)[because C = (in_(0)A)/(d)]` `:. rho_(E) = (1)/(2) in_(0)((V^(2))/(d^(2)))` `:. rho_(E)=(1)/(2)in_(0)E^(2)[because(V)/(d)=E]` |
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| 47. |
In a Rutherford scattering experiment when a projectile of charge Z_(1) and mass M_(1) approaches a target nucleus of charge Z_(1) and mass M_(2) , the distance of closest approach is r_(0) . The energy of the projectile is . |
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Answer» directlyproportional to `Mx_(1) xx M_(2)` |
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| 48. |
A setting sun appears to be at an altitude higher than it really is. This isbecause of |
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Answer» absorption of LIGHT |
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| 49. |
(A) : Mass, Volume and time may be taken as fundamental quantities in a system. (R) : Quantities which are independent of one another are called fundamental quantities. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 50. |
Assume that lasers are available whose wavelengths can be precisely ''tuned'' to anywhere in the visible range-that is, in the range 450nmltlambdalt650nm. If every television channel occupies a bandwidth of 10 MHz, how many channels can be accommodated within this wavelength range? |
| Answer» SOLUTION :`2XX10^(7)` CHANNELS | |