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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 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ""_(92)^(235)U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ""_(92)^(235)U and that this nuclide is consumed only by the fission process. |
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Answer» Solution :Energy generated PER GRAM of `""_(92)^(235)U = (6 xx 10^(23) xx 200 xx 1.6 xx 10^(-13))/(235) J g^(-1)` The amount of `""_(92)^(235)U` CONSUMED in 5y with 80% on - time. `=(5 xx 0.8 xx 3.154 xx 10^(16) xx 235)/(1.2 xx 1.6 xx 10^(13)) g = 1544 g` The initial amount of `""_(92)^(235)U = 3088 KG. ` |
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| 2. |
Rational number between 2 and 3 - |
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Answer» 2.25 |
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| 3. |
Determine the current through the galvanometer in the circuit. Given :P=2 Omega, Q = 4 Omega, R =8 Omega S =4 Omega G =10 Omega, E =5 and r =0. |
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Answer» Solution :By applying Kirchhoff.s Voltage law we get following EQUATIONS: `2I _(1) + 10 I _(g) - 8I _(2) =0` `4I_(1) - 18 I _(g) - 4 I _(2) =0` `12I _(1) + 10 I _(g) - 8 I _(2) =0` By solving the above equations `I _(g) = 0.12 A` |
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| 4. |
An electron in an atom is in a state with l=1. The minimum angle (in degree) between vecL and the z axis is |
| Answer» ANSWER :A | |
| 5. |
A long horizontal pipe is fitted with a piston of mass 10 kg which is connected to another mass 10.5 kg by a string passing over a frictionless pulley. A source of sound of frequency 512 Hz is placed in front of the piston. Initially the piston is almost in touch with the source and it moves away from the source when the hanging mass is released. Find the time/s when maximum sound is heard. Assume the string horizontal between pulley and piston. There is no friction. Velocity of sound = 340 m/s. |
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Answer» |
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| 6. |
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 ms^(-1) at right angles to the horizontal component of the Earth's magnetic field 0.30 xx10^(-4) Wb " "m^(-2). (c ) Which end of the wire is at the higher electrical potential ? |
| Answer» Solution :The EAST END will be at a HIGHER electrical POTENTIAL (-ve potential). The conventional CURRENT flowes from `+ve underset(("West"))` end to `-ve underset(("East"))` end of the wire. | |
| 7. |
If a cell of e.m.f. 1.5V is applied to a resistance R=4 Omega and a reactance of 3 Omega. What is the power factor of the a.c. circuit? |
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Answer» zero 
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| 8. |
Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces. |
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Answer» SOLUTION :A plot of potential energy of a pair of nucleons as a function of their separation ‘R’ has been shown in Fig. 13.06. Two important conclusions drawn from this are : (1) The potential energy is a minimum at a distance `r_(0)cong0.8` fm. It means that the nuclear force between two nucleons is attractive for distances larger than 0.8 fm and repulsive if they are separated by distances less than 0.8 fm. (2) For distances greater than 3 fm potential energy and hence nuclear force is practically zero. THUS, nuclear force is an extremely SHORT RANGE force. |
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| 9. |
Express Ohm.s law in terms of electrical resistivity and intensity of electric field |
| Answer» SOLUTION :`(J= sigmaE)` | |
| 10. |
A TV tower has a height of 110 m. How much population is covered by the TV broadcast of the average population density around the tower is 1000 km^-2? Given that radius of earth=6.4xx10^6m. |
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Answer» SOLUTION :RADIUS of the AREA covered by TV broadcast is `d = SQRT(2Rh)` Here, `R=6.4xx10^6m,h=110m` `d=sqrt(2xx6.4xx10^6xx110)=37500m=37.5km` POPULATION covered = `Population densityxxpid^2=1000xxpi37.5)^2=4.4xx10^6` |
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| 11. |
A spring of force constant k is cut into three equal parts. The force constant of each part would be |
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Answer» k/3 Here, n=3 THEREFORE, the force constant of each part would be 3k. |
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| 12. |
Figure shows roughly how the force F between two adjacent atoms in a solid varies with inter atomic separation r. Which of the following statements are correct ? |
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Answer» OQ is the equilibrium separation. `(dF)/(DR)|_(P^(+))=(dF)/(dr)|_(P^(-))=(dF)/(dr)|_(P)=-alpha` and F (at P) =0 so Hooke's law valid near POINT P. Energy required to separate the atoms =`|DeltaU|=|-intvec(F).vec(dr)|=`|Area enclosed between curve and r-axis| |
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| 13. |
y(x,t)= 0.8/[((4x + 5t)^(2) +5)] represents a moving pulse, where x and y are in metres and .t. is in seconds, then |
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Answer» PULSE is moving in +ve x direofibn |
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| 14. |
What was the impact of the pool incident |
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Answer» DEVELOPED fear |
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| 15. |
The nuclear radius of ""_(13)^(27)Al is 3.6 fermi. Find the nuclear radius of ""_(29)^(64)Cu. |
| Answer» SOLUTION :`(R_(Cu))/(R_(Al)) =(R_(Cu))/(3.6f) =((64)/(27))^(1/3) =4/3 RARR R_(Cu)=3.6 XX 4/3 =4.8 f` | |
| 16. |
A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with xgt0 is now bent so that it now lies in the y-z plane. |
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Answer» The magnitude of magnetic moment now diminishes. `M=IA=I(pir^(2))`  When ring is bent, dipole moment for each part, `M.-Ipi(r/2)^(2)" "[becausem=IA]` `M.=(IPIR^(2))/4` Resultant magnetic moment for both circular ring, `M_("net")=sqrt((M.)^(2)+(M.)^(2))=sqrt2M.=sqrt2(Ipir^(2))/4` Thus, `M_("net")ltM` Thus, resultant moment will DECREASE. |
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| 17. |
Linear magnification produced by a concave mirror may be : |
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Answer» LESS than 1 or EQUAL to 1 |
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| 18. |
which of the following is simple harmonic motion? |
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Answer» Particle MOVING in a CIRCLE with UNIFORM speed |
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| 19. |
A ball is projected from a point in a horizontal plane so as to strike a vertical wall at right angle and afterrebounding from the wall and once from the horizontal plane it returns to the point of projection. Then the co-efficient of restitution (e) for the collisions is : (Assuming it to be same for both the collisions). |
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Answer» `1//2` `V_(1_(x))=eucostheta(larr)`, `V_(1_(y))=sqrt(2gh)=usin theta(DARR)` `V_(2_(x))=V_(1_(x))=eucostheta(larr)`, `V_(2_(y))=eu sintheta(UARR)` From `BtoC`, equation of trajectory gives, `-h=O-(gx_(1)^(2))/(2(eV)^(2))impliesx_(1)=(eu^(2)sinthetacostheta)/(g)"and"x_(2)=(2V_(2x)V_(2y))/(g)=(2e^(2)u^(2)sin theta cos theta)/(g)` `because""x_(1)+x_(2)=x""implies""e+2e^(2)=1""e=(1)/(2)` 
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| 20. |
Figure shows two, identical narow slits S_(1) and S_(2). A very small completely strip is placed at distacne y from the point C. C is the point on the screen equidistant from S_(1) and S_(2). Assume lamda lt lt d lt lt D, where lamda, d and D have usual meaning. When S_(2) is covered the force due to light acting on strip is f and when both slits are opened the force acting on strip is 2f. Find the minimum positive y(lt lt D) coordinate (in cm) of the strip if (lamdaD)/d=4cm |
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Answer» When waves due to both the sources superpose at `P`,the net intensity at `P` is `I=4I_(0)"COS"^(2) (piy)/(beta)`, where `beta=(lamdaD)/d` Since FORCE on strip at `P` is twice due to individual source Hence `I=2I_(0)` `:."cos"^(2) (piy)/2=1/2` or `(piy)/(beta)=(PI)/4` or `y=(beta)/4=(lamda D)/(4d)` 
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| 21. |
A parallel electron beam accelerated in an electric field with a potential difference of 15 V falls on a narrow rectangular diaphragm 0.08 mm wide. Find the width of the principal diffraction maximum on a screen placed 60 cm away from the diaphrugm. |
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Answer» `x=2l tantheta=2llamda//D` |
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| 22. |
Minimum excitation potential of Bohr's first orbit in hydrogen atom is ...... eV |
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Answer» 13.6 Minimum excitation energy `=(13.6)/(2^(2))-(-(13.6)/(1^(2)))` `=-(13.6)/(4)+13.6` =10.2 eV |
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| 23. |
A coin placed on a rotating gramophone disc.remains at rest when it is at a distance of 8 cm from it's centre. The angular velocity of the disc is then doubled. At what distance from the centre the coin should be shifted,towards the centre so that it will just slip? |
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Answer» Solution :C.P. FORCE = force of friction. `mrw^2 = mumg``THEREFORE R=(mug)/w^2``therefore r ALPHA 1/w^2` `r_2/r_1=(w_1/w_2)^2=1/9``therefore r_2=9/9 = 1cm`. |
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| 24. |
The magnetic field is zero at mid point of AB and at P as shown |
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Answer» The current is equal in both wires and inside  Since magnetic field at `M` is ZERO. For this `i_(1)=i_(2)=i` and should in same direction. Let  `B_(P)=2B'cos theta` `=2(mu_(0)i)/(2pisqrt((a^(2))/4+x^(2))). x/(SQRT((a^(2))/4+x^(2)))` `=(mu_(0)ix)/(pi((a^(2))/(4) + x^(2)))` For `B_(P)` to be maximum `(dB_(P))/(sx)=0 implies ((a^(2))/4+x^(2)). 1-x. 2x=0` `x=+-a/2` Left of `M:` 
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| 25. |
The axes of the polariser and analyser are inclined to each other at 60^(@). If the amplitude of the polarised light emergent through analyser is A. The amplitude of unpolarised light incident on polariser is |
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Answer» `(A)/(2)` |
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| 26. |
Focal length of the plano-convex lens is 15cm. A small object is placed at A as shown in figure. The plane surface is silvered. The image will form at |
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Answer» 60cm to the left of lens `v=60cm UNDERSET(+"ve direction") rarr` i.e., FIRST image is formed at 60cm to the rigth of lens system. Reflection frm mirror: After reflection from the MIRRO, the second image will be formed at a distance 60cm to the left of lens sytem. Refraction from lens: `(1)/(v_(3))-(1)/(60)=(1)/(15)larr+"ve direction"` or `v_(3)=12cm` Therefore, the final image is formed at 12 cm to the left of the lens system. |
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| 27. |
Assertion A beam of electron can pass undeflected through a region of E and B. ReasonForce onmoving charged particle due to magnetic field may be zero in some cases. |
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Answer» |
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| 28. |
What is the number of significant figures in 0.310xx10^3 ? |
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Answer» 2 |
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| 29. |
The fission properties of ""_(94)^(239)Pu are very similar to those of ""_(92)^(235) U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ""_(94)^(239)Pu undergo fission? |
| Answer» SOLUTION :`4.536 XX 10^(26) MEV`. | |
| 30. |
The frequency of oscillations of a mass m connected horizontally by a spring of spring constant k is 4 HZ. When the spring is replaced by two identical spring as shown in figure. Then the effective frequency is |
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Answer» `4sqrt(2)` |
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| 31. |
An a.c. voltage V=Vm sin omega tis applied across a (i) series RC circuit in which the capacitive impedance is 'a' times the resistance in the circuit, (ii) series RL circuit in which the inductive impedance is 'b' times the resistance in the circuit. Calculate the value of the power factor of the circuit in each case. |
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Answer» SOLUTION :Power factor of an a.c. circuit is given by `cos phi = R/Z` <BR> (i) In a series RC circuit it is given that Z =aR `rArr` Power factor `=R/Z = R/(aR) = 1/a` (ii) In a series RL circuit it is given that Z= bR `rArr` Power factor `=R/Z = R/(bR) = 1/b` |
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| 32. |
The figure shows a block of mass M=2m having a spherical smooth cavity of radius R placed on a smooth horizontal surface .There is a small ball of mass m moving at an instant vertically downward with a velocity v with respect to the block .At this instant : |
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Answer» The normal reaction on the ball by the block is `(MV^(2))/R` For the block, from Newton's `II^(nd)` law, we have N=Ma=2ma For ball (with respect to the block), from Newton's `II^(nd)` law, we have `N+ma=(mv^(2))/R` SOLVE the two equations. 
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| 33. |
Define kilowatt-hour unit of energy. |
| Answer» Solution :It is defined as the ELECTRICAL energy CONSUMED at the RATE of ONE kilowatt for one hour. | |
| 34. |
Ge and Si diode conduct at 0.3 V and 0.7respectively. In the following figure if Ge diode connection is reversed, the value of V_0 changes by |
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Answer» 0.2 V Equivalent VOLTAGE DROP cross the combination Ge and Si diode = 0.3 V `implies ` Current `i=(12-0.3)/(5kOmega)=2.34mA` `:.` Output voltage `V_0 =Ri = 5kOmega xx 2.34 mA = 11.7V` Now consider the case when diode connection are REVERSED. In this case voltage drop across the diode.s combination = 0.7 V `implies ` Current `i=(12-0.7)/(5kOmega) = 2.26mA` `:. V_0=iR=2.26mA xx5kOmega =11.3 V` HENCE CHARGE in the value of `V_0=11.7-11.3 = 0.4V` |
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| 35. |
Visiblelight of wavelength 6000 xx 10^(-8) cm fall normally on a single slit and produces diffraction pattern. It is found that the second diffraction minima is at 60^(@) from the centre maxima. If the first minima is produced at theta is close to |
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Answer» `20^(@)` `d sin theta= 2lambda` `:.sin theta=(2lambda)/(d)` `:. Sin 60^(@)=(2lambda)/(d)` `:. (sqrt(3))/(2)=(3lambda)/(d)` `:. (lambda)/(d)=(sqrt(3))/(4) ""...(1)` For first minimum, `d sin theta = lambda` `:. sin theta=(lambda)/(d)` From EQUATION (1), `:. sin theta=(sqrt(3))/(4)` `:. sin theta=0.4330` From table of sine, `theta=25.65(@)` `:. theta~~25^(@)` (NEARER value) |
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| 36. |
A receiver reconstructs the original message after propagation through the channel |
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Answer» MAY be TRUE |
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| 37. |
If the velocity of light c, Planck's constant h and gravitation constant G are adopted as fundamental units of a system, the dimensions of force in the system would be : |
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Answer» `G^(4)h^(-1)` |
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| 38. |
What is the ratio of heat generated in R and 2R |
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Answer» `2:1` |
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| 39. |
The materials suitable for making electromagnets should have |
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Answer» HIGH retentivity and high COERCIVITY. |
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| 40. |
Define Refractive Index. |
| Answer» Solution :REFRACTIVE Index OD 2nd medium with respect to 1ST medium is defined as `^1mu_2`=v_2/v_1 where `v_1` = VELOCITY of 1st medium.`v_2`= velocity of 2nd medium. | |
| 41. |
When Douglas tried to yell |
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Answer» everyone came to his rescue |
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| 42. |
The fig.12.15 showes energylevel diagram of hydrogenatom . (a) Find thetransition whichresults in theemissionof a photon of wavelenght496 nm . (b) Whichtransitioncorrespondsto the emission of radiationof maximum wavelenght?Justify your answer . |
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Answer» SOLUTION :(a) Energy of photon of wavelenght `lambda = 496` NM ` = 496 xx 10^(-9) m ` is . `E = (hc)/(lambda) I = (hc)/(elambda) EV = ((6.63xx 10^(-34))xx(3xx10^(8)))/((1.6 xx 10^(-19)) xx(496 xx 10^(-9))) = 2.51 eV` From the energylevel diagram we findthat aphoton of energy2.51 eV can beemitted onlywhentransitiontakes place from n = 4ton = 2 statebecausethen `DeltaE = E_(4)- E_(2)= - 0.85 - (-3.4 eV) = + 2.55 eV` (b) Radiationof maximum wavelenghtwill be emittedphoton is minimumandit is possiblefor TRANSITION from n = 4 ton = 3 state . |
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| 43. |
The height of a TV transmitting antenna is 20m. The telecast can cover a radius of |
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Answer» 8km |
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| 44. |
A 1m long metal wire moving perpendicular with speed 5m//s is a magnetic field of 0.1T. Then the induced emf between two ends of wire is……V |
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Answer» 1 `=0.1 times5 TIMES 1` =0.5V |
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| 45. |
An N-type silicon sample of width 4 xx10^(-3)m thickness 0.25mm and length 6 xx10^(-2) m carries a current of 4.8 mA when the voltage is applied across the length of the sample. What is the current density? If the free electron density is 10^(22) m^(-3) , then find how much time it takes for the electrons to travel the full length of the sample. |
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Answer» SOLUTION :The CURRENT density J is given by `J= (I )/(A )= ( 4.8 xx 10^(-3))/(( 4 x 10^(-3)) ( 25xx 10^(-5)) )= ( 4.8 xx 10^(-3))/( 10^(-6))` The drift velocity `v_d` is given by `v_d = (J )/( n e)= ( 4800)/( 10 ^(22) xx 1.6 xx 10^(-19))= 3m//s` The timetakent isgivenby `t=(L )/(u_d ) =( 6 xx 10^(-2))/( 3 ) = 0.02sec ` |
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| 46. |
Two opposite charaged particles oscillate about their mean equilibrium position in free space, with a frequency of 10^(9) Hz. The wavelength of the corresponding electromagnetic wave produced is …….. |
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Answer» 0.3 m `c=3xx10^(8)m//s` `f=1xx10^(9)Hz` `THEREFORE lambda =(3xx10^(8))/(10^(9))=0.3 m` |
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| 47. |
The number of geometric isomers in CH_(3)CH=CH-CH=CH-CH=CHCl is |
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Answer» 2 TOTAL no. of geomettical isomers `=2^(n)=2^(3)=8`. |
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| 48. |
Standing waves are established on a string of length L such that A is a node and B is an immediate anti node. Oscillation amplitude of point B is a_(0). Let a' be the oscillation amplitude for point C. Pick the suitable option (s) for correct value of a' and the possible equation (s) for standing waves on the string. |
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Answer» `a'=(a_(0))/2` `(KL)/3=npi` and `(kL)/2=(n+1/2)PI` `impliesn=1, k=(3pi)/L` Also: `a'=|a_(0)sin((3pi)/Lxx(5L)/6)|=a_(0)` |
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| 49. |
A metallic loop is placed in a uniform magnetic field vec(B) with the plane of the loop perpendicular to vec(B). Under which condition(s) given below an emf will be induced in the loop ?''If the loop is ……..'' |
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Answer» moved along the direction of `vec(B)`. So, emf can only be INDUCED in a loop if there is rate of change of flux across the loop. Now, magnetic flux with the loop remains unchanged when the loop is rotated about its own axis and moved along the direction of magnetic field `(vec(B))`. Now, when the loop remains unchanged when the loop is rotated about its own axis and moved along the direction of magnetic field `(vec(B))`. Now, when the loop is squeezed to a smaller area then DUE to change in area, flux linked with the loop is CHANGED, resulting in an e.m.f. will be induced in the loop. Again, when the loop is rotated about one of its diameters then an emf will be induced in the loop due to change in flux linked with the loop. More emf will be induced in the loop if the loop is rotated at faster rate. |
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| 50. |
Use Huygen's principle to show how a plane wavefront propagates from a denser to rarer medium. Hence, verify snell's law of refraction. |
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Answer» Solution :Consider a plane surface XY separating a denser medium of refractive index `n_(1)` from a rarer medium of refractive index `n_(2)`. Let `c_(1) and c_(2)` be the values of SPEED of light in the two media, where `c_(2)gtc_(1)`. AB is a plane wavefront incident on XY at an angle `i`. Let at a given instant the end A of the wavefront just strikes the surface XY but the other end B has still to COVER a path BC. if it takes time `t`, then `BC=c_(1)t`. Meanwhile, point A begins to emit secondary wavelets which cover a DISTANCE `c_(2)t` in second medium in time t. draw a circular arc with a as centre and `c_(2)t` as radius and draw a tangent CD from point C o this arc. then CD is the refracted wavefront in the rarer medium which advances in the DIRECTION of rays 1.2.. the refracted wavefront subtends an angle r from surface XY. Now in `DeltaABC,sini=(BC)/(AC)=(c_(1)t)/(AC)`  and in `DeltaADC, sinr=(AD)/(AC)=(c_(2)t)/(AC)` `therefore (sini)/(sinr)=(c_(1)t//AC)/(c_(2)t//AC)=(c_(1))/(c_(2))=`a constant`=(n_(2))/(n_(1))=n_(21)`. Which is snell.s law of refractive. |
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