45406 + Interview Questions in Physics in Class 12 Page 758 InterviewSolution

37851.	Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint : The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Answer» Solution :As charge on each DEUTERON `q = 1.6 xx 10^(-19)` C and SEPARATION between their centres`= 2r=2xx2fm=4XX10^(-15)`m, hence their potential energy is : `U=1/(4piepsilon_(0)).((q)(q))/(2r)=1/(4piepsilon_(0)).q^(2)/(2r)=(9xx10^(9)xx(1.6xx10^(-19))^(2)/(4xx10^(-15))eV)=360keV`. This is the height of the potential barrier between the two deuterons.

Discussion

37852.	Considering the case of a parallel plate capacitor being charged, show how one is required to generalise Ampere.s circuital law to include the term due to displacement current.
Answer» Solution :Consider charging of a parallel plate capacitor by a time-varying current iw Let us FIND the magnetic field at a point P in a region outside the capacitor. For this we consider a plane CIRCULAR loop of radius r whose plane is perpendicular to the direction of the current carrying wire and which is centred symmetrically with respect to the wire [Fig]. Using symmetry condition and applying Ampere.s circuital law, we have `B(2pir)=mu_(0)i_((t))"....(i)"` However, if we consider a different surface, having the same boundary [either as SHOWN in Fig. or Fir. and apply Ampere.s circuital law as before, we find `B(2pir)=mu_(0)(0)=0"....(ii)"` It is because nocurrent passes through the surface of Figs. and 8.04. It causes a contradiction because we get a finite value of magnetic field B by doing calculation in one way and zero value of B by doing calculation in another way. To remove this contradiction Maxwell introduced the CONCEPT of displacement current `i_(D)=in_(0)(dphi_(E))/(DT),` where `(dphi_(E))/(dt)` is the rate of change of electric flux between the plates of given capacitor. Now, the Ampere - Maxwell.s circuital law is expressed as `oint vecB.vecdl=mu_(0)[i_((t))+i_(D)]` Thus, Ampere - Maxwell law successfully explains the flow the current through a capacitor when it is being charged (or discharged) by a battery.

Discussion

37853.	Focal lengths of two lenses are 10cm and -20cm and dispersive powers of their materials are 0.03 and w. To form achromatic combination from these, (a) what is the value of omega ? (b) What is the resulting focal length of the combination ?
Answer» ANSWER :`(a) omega=00.06`; `(B) f=20cm` (CONVEX LENS)

Discussion

37854.	The equation of a progressive wave is y=0.1 sin(8pi/7)(0.1t - x/20)m .the velocity of propagation of the wave is
Answer» 2 m/s 15 m/s 20 m/s 40 m/s Answer :A

Discussion

37855.	A body is projected at an angle of 60° with the horizontal with momentum p. At its highest point the magnitude of the momentum is :
Answer» `SQRT(3)/2P` `2/sqrt(3)p` `p` `p/2` Solution :At highest PT. `v_(y)=0,:. P_(y)=0` `p_(x)=mu_(x)=mucos60^@=(mu)/2=p/2`

Discussion

37856.	What is optical density ?
Answer» Solution :It is a MEASURE having a relatively high VALUE of RI. It.s LOW for a medium with low RI.

Discussion

37857.	The deflection in galvanometer falls to ((1)/(4))^(th) when it is shunted by 3Omega. If additional shunt of 2Omegais connected to earlier shunt in parallel, the deflection in galvanometer falls to
Answer» ANSWER :8.5

Discussion

37858.	Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is :
Answer» `sqrt((Gm)/(4R))` `sqrt((Gm)/(3R))` `sqrt((Gm)/(2R))` `1/2sqrt((Gm)/(R ))` Solution :`(Gm^(2))/((2R)^(2))=m omega^(2)R` `(Gm)/(4R^(3))=omega^(2)` `omega=sqrt((Gm)/(4R^(3)))` `v=omega R` `v= sqrt((Gm)/(4R^(3)))xxR=sqrt((Gm)/(4R))` So, correct CHOICE is (a).

Discussion

37859.	Which of the following assertions are correct?
Answer» A NEUTRON can decay to a proton only INSIDE a NUCLEUS A proton can change to a neutron only inside a nucleus An ISOLATED proton can change into a neutron An isolated neutron can change into a proton Solution :Inside a NUCLEAUS, a proton can change to a neutron. Out side the nucleus, an isolated neutron can change into a proton.

Discussion

37860.	The plane of the coil of a T.G. is placed perpendicular to the magnetic meridian and an electric current is passed through it. The deflection of the needle is
Answer» `0^@` only `0^@ or 180^@` `180^@` only `90^@` ANSWER :B

Discussion

37861.	The resultant magnetic moment for the following arrangement is
Answer» M 2M 3M 4M Solution :`M_(R)=sqrt(M_(1)^(2)+M_(2)^(2)+2M_(1)M_(2)COSTHETA)`

Discussion

37862.	A converging glass lens with a refractive index of 1.5 has a focal length of f in air. When it is completely immersed in a liquid of refractive index 2, its focal length and nature will be respectively
Answer» 2F and coverging 3f and converging 2f and diverging 3f and diverging Answer :C

Discussion

37863.	Statement I: Photoemission from a photosensitive metal is possible only if the incident radiation has a frequency above the threshold frequency. Statement II : In photoelectric emission, the maximum energy of photoelectrons increases with increasing intensity of incident light.
Answer» A. Statement-I is FALSE, statement-II is TRUE. Statement-I is true, statement-II is false. Statement-I is true, statement-II is true. Statement-II is CORRECT EXPLANATION of statement-I Statement-I is true, statement-II is true and statement - II is not correct explanation of statement-I. Answer :B

Discussion

37864.	The charge per unit area of a large fiat sheet of charge is 3.0muC//m^2.. Calculate the electric field intensity at a point just above the surface of the sheet, measured from, it’s midpoint
Answer» Solution :Surface CHARGE density `sigma=3.0xx10^-6C//m^2` ELECTRIC INTENSITY `=E=sigma/(2epsilon_0)=(3.0xx10^-6)/(2xx8.85xx10^-12)=1.7xx10^5NC^-1`

Discussion

37865.	The magnetic field in plane e.m. wave is given by B_(y) = 2 xx 10^(7) sin (0.5 xx 10^(3)x+1.5 xx 10^(11)t) Tesla. Wavelength of e.m. wave is
Answer» 1.26 cm 10.2 cm 12.6 cm 31.4 cm Answer :A

Discussion

37866.	A ball is projected from the point O with velocity 20m//s at an angle of 60^(@) with horizontal as shown in figure. At highest point of its trajectory it strikes a smooth plane of inclination 30^(@) at point A. The collision is perfectly inelastic. The maximum height from the ground attained by the ball is 75/k meter. Find the value of k? (g=10m//s^(2))
Answer» SOLUTION :`[4]`

Discussion

37867.	Which from the following represents displacement current ?
Answer» `epsilon_(0)(DB)/(DT)` `epsilon_(0)(d phi_(E ))/(dt)` `epsilon_(0)(d phi_(B))/(dt)` `epsilon_(0)(dI)/(dt)` Answer :B

Discussion

37868.	The intensities of magnetic field at two points on the axis of a bar magnet at distances 0.1 m and 0.2 m from its middle point are in the ratio 12.5 : 1 . Calculate the distance between the poles of the magnetic .
Answer» ANSWER :`0.1` m

Discussion

37869.	Two charges q_(1) and q_(2) are placed at (0,0,d) and (0,0,-d) respectively. Find icons of points where the potential is zero.
Answer» Solution :In fig, we haveshown two charges `q_(1) (0,0,d) and q_(2) (0,0,d)`. For potential to be zero at (x,y,Z) we should have`(q_(1))/(4pi in_(0) sqrt(x^(2) + y^(2) + (z - d)^(2))) + (q_(2))/(4pi in_(0) sqrt(x^(2) + y^(2) + (z + d^(2)))= 0` `(q_(1))/(sqrt(x^(2) + y^(2) + (z - d^(2)))) = (-q_(2))/(sqrt(x^(2) + y^(2) + (z + d^(2))))`...(i) Clearly, total potential can be zero when `q_(1), q_(2)` have OPPOSITE signs. Sqaring both sides of (i), we GET `q_(1)^(2) [x^(2) + y^(2) + (z + d)^(2)] = q_(2)^(2) [x^(2) + y^(2) + (z - d)^(2)]` on simlifyingwe get`x^(2) + y^(2) + z^(2) + [((q_(1)//q_(2))^(2) + 1)/((q_(1)// q_(2))^(2) - 1)] (2zd) + d^(2) = 0`. This is the equaction of a sphere will center at `[0,0, -2d ((q_(1)^(2) + q_(2)^(2))/(q_(1)^(2) - q_(2)^(2)))]`. If `q_(1) = -q_(2)`, then `z = 0`. Therefore, locus of pointswhere potential is zero is the plane through MID pointof the two charges.

Discussion

37870.	Two charges -q each are fixed separated by distance 2d. A third charge q of mass m placed at the midpoint is displaced slightly by x (x < < d) perpendicular to the line joining the two fixed charged as shown in figure. Show thatq will perform simpleharmonic oscillation of time period. T = [(8pi^(3) epsilon_(0)md^(3))/q^(2)]^(1//2)
Answer» <P> Solution :Suppose charge at A and B are - q and O is mid point of AB and PO is X. `therefore AB = AO + OB` =d+d =2d `x lt d` and `angleAPO =theta` m is mass of charge q. Attractive force by each charge A and B on charge at P, `F = (k(q)(q))/r^(2)` where r = AP = BP `F sin theta`COMPONENTS of force are of same magnitude but in opposite directions hence, their resultant is zero and Fcos6 components are in same direction, `F. = 2F cos theta` `=(2kq^(2))/r^(2) cos theta` But from figure, `r = sqrt(d^(2) + x^(2))` and `cos theta = x/r` `therefore F. = (2kq^(2))/(d^(2) + x^(2))^(2).x/(d^(2) + x^(2))^(1//2)` `=(2kq^(2)x)/(d^(2) + x^(2))^(3//2)` If `x lt lt d`, then x can be neglected, `F. = (2kq^(2)x)/(d^(3))`..........(1) `F. = Kx` where `K = (2kq^(2))/d^(3)` is constant `therefore F. prop x` Hence, q can perform SIMPLE harmonic motion. Here, force is DIRECTLY proportional to x and towards point O. `omega = sqrt(K/m)` `therefore (2pi)/T = sqrt(K/m)` `therefore T = 2pi sqrt(m/(2kq^(2)//a^(3))` By taking `k = 1/(4pi epsilon_(0))` `therefore T=[(8pi^(3)epsilon_(0)md^(3))/q^(2)]^(1//2)`

Discussion

37871.	Which of the following is a suitable material for making electromagnet ?
Answer» Tungsten Bismuth Copper SOFT IRON Solution :Soft iron

Discussion

37872.	A circular current loop of magnetic moment M is in an arbitrary orientation in an external uniform magnetic field vecB. The work done to rotate the loop by 30^(@) about an axis perpendicular to its plane is
Answer» `(MB )/(2)` `MB ` ZERO `sqrt(3) (MB)/(2)` Solution :`W=0`

Discussion

37873.	A radioactive element ._92X^238 emits one alpha-particle and one beta^+particle in succession. Wht is the mass number of the new elementformed ?
Answer» SOLUTION :`._92X^238 overset(ALPHA -"decay")to._90X^234 , ._92X^234 overset(BETA^+ "decay")to ._89Y^234 + ._(+1)^(e^0)` MASS number new elementthe is 234.

Discussion

37874.	In yound's double slit experiment, the separation between the slits is halved and distance between the slits and screen is doubled. The figure width is
Answer» unchanged halved doubled quadrupled Solution :`BETA=(lambdaD)/(d),beta=(lambdaxx2D)/(d//2)=4beta`

Discussion

37875.	A point mass M is attached to one end of a masslesa rigid non-conducting rod of length L. Another equal point mass is attached to the other end of the rod. The two particles carry charges +q and -q respectively. This arrangement is held I the region of uniform electric field E such that the rod makes a small theta (say about 5^(@)) with field direction. Moment of inertia of the rod is I. Now answer the question Time period for the rod to become parallel to E is
Answer» `2pi sqrt(ML)/(QE))` `(2pi)/3sqrt((ML)/(qE))` `(pi)/3sqrt((ML)/(2QE))` `(pi)/4sqrt(ML)/(2qE))` Answer :C

Discussion

37876.	Draw V-I characteristicsof a p-n junction diode. Answer the following questions, giving reasons: (i)Why is the current under reverse bias almost independent of the applied potential upto a critical voltage ? (ii) Why does the reverse current show a sudden increase at the critical voltage ? Name any semiconductor device which operates under the reverse bias in the breakdown region.
Answer» Solution :The V-I characteristics of a p-n junction diode in forward bias and REVERSE bias arrangement have been shown in . (i) In reverse bias the junctionwidth increases. The higher junction potential restricts the flow of majority charge carriers. HOWEVER, such a field favours flow of minority carriers. Thus, reverse bias current is due to flow of minority carriers only. Since the NUMBER of minority carriers very small, the current is small and ALMOST independent of the applied potential upto a critical (before breakdown) voltage. (ii)At a critical (breakdown) voltage the reverse bias current shows a sudden increase. Under high reverse bias, the high junction field may strip an electron from the valence band, which can tunnel to the n-side through the thin depletion layer. This mechanism of emission of electrons after a critical applied voltage LEADS to a high reverse (breakdown) current. A zener diode operates under the reverse bias in the breakdown region.

Discussion

37877.	The size of the nucleus is_____.
Answer» SOLUTION :`10^(-15)m`

Discussion

37878.	What are eddy currents ? How are these produced ? In what sense are eddy currents considered undesirable in a transformer and how are these reduced in such a device?
Answer» Solution :For eddy currents and their production, see Point Number 12 under the heading "Chapter At A Glance". The eddy currents are considered undesirable in a transformer as they cause heating of iron core of transformer. On one hand, it means loss of electrical energy i.e., decrease in efficiency of the transformer, on the other hand, due to EXCESSIVE heating the insulation of transformer windings MAY be damaged. Eddy currents in a transformer are reduced by taking the iron core to be a laminated iron core. In such an arrangement instead of a solid iron core, we take a large number of THIN iron laminas and these are joined together with layers of insulating materials (SAY lacquer) in between so as to minimise eddy currents.

Discussion

37879.	In a-p-n junction, the thickness of the depletion layer is 110^-6m. If the potential difference across it is 0.2 V, then the electric field set up across the junction is
Answer» `10^6V//m` `2xx10^5V//m` `10^6V//m` `10^5V//m` ANSWER :B

Discussion

37880.	The moment of inertia of a cylinder of radius R, length L and mass M about an axis passing through its centre of mass and normal to its length is
Answer» `(ML^(2))/(12)` `(MR^(2))/(4)` `M[(L^(2))/(12)+(R^(2))/(4)]` `M[(L^(2))/(12)+(R^2)/(2)]` Answer :C

Discussion

37881.	The quantity having negative dimensionsin mass is
Answer» GRAVITATIONAL Potential Gravitational CONSTANT Acceleration due to gravity none of the above ANSWER :B

Discussion

37882.	Write the symbol, truth table, function and Boolean equation for ORgate.
Answer» `Y=barA` `Y=A+B` `Y=barA+barB` `Y=A*B` ANSWER :B

Discussion

37883.	A proton of mass 1.6xx10^-(27) kg, revolves in a circular path of radius 0.1 m. Calculate the angular velocity of the proton if it is acted upon by a centripetal force of 2.5xx10^(-12) N.
Answer» Solution :`F=(mv^2)/r=mrw^2` `therefore w^2=F/(MR)=(4xx10^(-13))/(1.6xx10^(-27)xx10^(-1)` `w^2 =25xx10^14 therefore w=5xx10^7 (rad)/s`

Discussion

37884.	(a) Determine the electrostatic potential energy of a system consisting of two charges 7 µC and –2 µC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? (c) Suppose that the same system of charges is now placed in an external electric field E = A(1//r^(2)):A = 9 xx 10^(5) NC^(-1) . What would the electrostatic energy of the configuration be?
Answer» Solution :(a) `U = 1/(4pi epsilon_(0)) . (q_(1)q_(2))/( r) = 9 xx 10^(9) xx (7 xx -2) xx 10^(-12)/0.18 = 0.7 J` (B) `W = U_(2)-U_(1) =0-U = 0- (-0.7) = 0.7 J` ( c) The mutual interaction ENERGY of the two charges remains UNCHANGED. In addition, there is the energy of interaction of the two charges with the EXTERNAL electric field. We find, `q_(1) V(r_(1)) + q_(2)V(r_(2)) = A(7 muC)/(0.09 m) + A(-2 muC)/(0.09 m)` and the net electrostatic energy is `q_(1) V(r_(1)) + q_(2)V(r_(2)) + (q_(1)q_(2))/(4pi epsilonr_(12)) = A(7muC)/(0.09 m) + A(-2 muC)/(0.09 m) -0.7 J = 70-20-0.7 = 49.3 J`

Discussion

37885.	(a) Derive an expression for unit vector along reflected ray (hat r) if unit vectors hat i and hat n represents unit vectors along incident light ray and normal (at point of reflection and outward from surface) respectively. (b) If vector along the incident ray on a mirror is -2 hat i+ 3 hat j + 4 hat k. Considering the x-axis to be along the normal. Then, find the unit vector along the reflected ray.
Answer» SOLUTION :VECTOR ALONG the reflected ray `rArr 2 HAT (i)+ 3 hat (j) + 4 hat (k)` `:.` it's unit vector `= (2 hat(i)+3 hat(j)+4 hat(k))/(SQRT(29))`.

Discussion

37886.	Name them :
Answer» SOLUTION :DIA, FERRO, PARA

Discussion

37887.	In Young's double slit experiment, when light of wavelength 4000 A^(@) is used 90 fringes are seen on the screen. When light of 3000 A is used, the number of fringes seen is
Answer» 70 120 140 68 Answer :B

Discussion

37888.	A source is moving across a circle given by the equation x^2 + y^2 = R^2 , with constant speed v = (330 pi)/(6sqrt3) m//s , in anti-clockwise sense. A detector is at rest at point (2R, 0) w.r.t the centre of the circle. If the frequency emitted by the source is f and the speed of sound, C = 330m/s. Then
Answer» The POSITION of the SOURCE when the detector records the maximum FREQUENCY `( + (sqrt3)/(2) R, - R/2)` The co-ordinate of the source when the detector records maximum frequency is (0, R) Themaximum frequency RECORDED by the detector is`(6 sqrt3)/(PI + 6 sqrt3) f ` The minimum frequency recorded by the detector is `(6 sqrt3)/(6 sqrt3 - pi) f ` Answer :A::B::C::D

Discussion

37889.	Write down the advantages and limitations of frequency modulation (FM)? Advantages of FM
Answer» Solution :(i) Large decrease in noise. This LEADS to an increase in signal-noise ratio. (ii) The operating range is quite large. (iii) The transmission efficiency is very high as all the transmitted power is useful (iv) FM bandwidth COVERS the entire frequency range which humans can hear. Due to u FM radio has better quality compared to AM radio. LIMITATIONS of FM (i) FM requires a much WIDER channel (ii) FM transmitters and receivers are more complex and costly (iii) In FM reception, less area is covered compared to AM.

Discussion

37890.	A Camot engine working between450 K and 600 K has a work output of 300 J//"cycle". What is the amount of heat energy supplied to the engine from the source in each cycle ?
Answer» 400 J 800 J 1200 J 1600 J Answer :C

Discussion

37891.	Uniform circular motion is,
Answer» PERIODIC motion Oscillatory motion Vibratory motion None of these Answer :A

Discussion

37892.	Carbon dating is best suited for determining the age of fossils of their age in years is of the order of
Answer» `10^3` `10^4` `10^5` `10^6` Solution :CARBON dating is best suited for DETERMINING the age of FOSSILS, if their age is of the order of `10^4` years.

Discussion

37893.	Transformation ratio of one transformer is 1:2. If Leclanche cell having emf 1.5 V is connected to primary coil of a transformer then voltage obtained across its secondary is ………..
Answer» 3 V 1.5 V 0.75 V 0 Solution :Because Leclanche cell is a D.C. supply and so it gives constant voltage to primary which does not change its MAGNETIC FLUX and so no EMF is INDUCED across secondary because `epsilon_2=-M(dI_1)/(dt)=0 "" (because (dI_1)/(dt)=0)`

Discussion

37894.	Which one of the following force is conservative ?
Answer» GRAVITATIONAL force Frictional force Air resistance Viscous force. Answer :A

Discussion

37895.	If the refractive index of diamond is 2.4 find the velocity of light in diamond. (c= 3 xx10^8 m//s)
Answer» SOLUTION :`1.25 XX 10^8 m//s`

Discussion

37896.	Find the energy required to split ._(8)^(16)O nucleus into four alpha - particles. The mass of an alpha- particle is 4.002603u and that of oxygen is 15.994915u.
Answer» SOLUTION :`Deltam=m(4He_2^4)-m(O_8^16)` =4 x 4.002603 - 15.994915 =0.015497 amu HENCE `BE=Deltam XX 93.1` MEV= 14.435 MeV

Discussion

37897.	A convex lens of refractive index 1.5 has a focal length of 18 cm in air. Calculate the change inits focal length when it is immersed in water of refraction index 4/3.
Answer» Solution :Here, it is given that `N=1.5, f_("air") = 18 cm` and `n_(m) = 4/3` `therefore` Focal LENGTH of lens in WATER `f_("water") =((n-1)n_(m)f_("air"))/(n-n_(m)) = ((1.5 -1) xx (4/3) xx 18)/(1.5-4/3) = 72 cm` `therefore` Change in focal length `=f_(m) - f_("air") = 72 cm - 18 cm = 54 cm `

Discussion

37898.	(a) A monoenergetic electron beam with electron speed of 5.20 xx 10^(6) m s^(-1) is subject to a magnetic field of 1.30 xx10^(-4) T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 xx 10^(11)C kg^(-1) . (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
Answer» Solution :Here velocity `v=5.20xx10^(6)ms^(-1)` Magnetic field B`=1.30xx10^(-4)`T `(e )/(m)=1.76xx10^(11)C kg^(-1)` Kinetic energy of electron BEAM K=20 MEV Mass of electron m`=9.1xx10^(-31)Kg` (a) Centripetal fornce provided by magnetic field to electron, `(mv^(2))/(r)=bev` `therefore r=(mv)/(be)=(v)/(B((e)/(m)))` `therefore r=(5.20xx10^(6))/(1.30xx10^(-4)xx1.76xx10^(11))` `therefore r=2.2727xx10^(-1)` `therefore r~~0.227 m=22.7cm` (b)K=20 MeV `therefore (1)/(2)mv^(2)=20xx10^(6)xx1.6xx10^(-19)J` `therefore v=sqrt((2xx20xx10^(6)xx1.6xx10^(-19))/(m))` `therefore v=sqrt((64xx10^(-13))/(9.1xx10^(-31)))=sqrt(7.033xx10^(18))` `therefore v=2.65xx10^(9)m//s` No. `implies`Here speed of electron v`GT` speed of light c `therefore` This suggest that there is something incorrect in calculation .HEnce equation `r=(m_(0)v)/(eB)`,not valid.Hence we should use equation of RELATIVITY `r=(mv)/(eB)` but `m=(m_(0))/(sqrt(1-(v^(2))/(c^(2))))` `therefore` Equation `r=(m_(0)v)/(eBsqrt(1-(v^(2))/(c^(2))))` should be used.

Discussion

37899.	This poem is written by...
Answer» THOMAS Cambell William Wordsworth Robert Frost William Shakespeare Answer :C

Discussion

37900.	How is an electric circuit shielded from stray electric field by placing it inside a conductive box?
Answer» SOLUTION :We know that the INTERIOR of any METAL is always shielded from the electric fields outside. Thus, electric circuit is shielded from any stray electric field when kept INSIDE a conductive box.

Discussion

Explore topic-wise InterviewSolutions in .

Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint : The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

Considering the case of a parallel plate capacitor being charged, show how one is required to generalise Ampere.s circuital law to include the term due to displacement current.

Focal lengths of two lenses are 10cm and -20cm and dispersive powers of their materials are 0.03 and w. To form achromatic combination from these, (a) what is the value of omega ? (b) What is the resulting focal length of the combination ?

The equation of a progressive wave is y=0.1 sin(8pi/7)(0.1t - x/20)m .the velocity of propagation of the wave is

A body is projected at an angle of 60° with the horizontal with momentum p. At its highest point the magnitude of the momentum is :

What is optical density ?

The deflection in galvanometer falls to ((1)/(4))^(th) when it is shunted by 3Omega. If additional shunt of 2Omegais connected to earlier shunt in parallel, the deflection in galvanometer falls to

Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is :

Which of the following assertions are correct?

The plane of the coil of a T.G. is placed perpendicular to the magnetic meridian and an electric current is passed through it. The deflection of the needle is

The resultant magnetic moment for the following arrangement is

A converging glass lens with a refractive index of 1.5 has a focal length of f in air. When it is completely immersed in a liquid of refractive index 2, its focal length and nature will be respectively

Statement I: Photoemission from a photosensitive metal is possible only if the incident radiation has a frequency above the threshold frequency. Statement II : In photoelectric emission, the maximum energy of photoelectrons increases with increasing intensity of incident light.

The charge per unit area of a large fiat sheet of charge is 3.0muC//m^2.. Calculate the electric field intensity at a point just above the surface of the sheet, measured from, it’s midpoint

The magnetic field in plane e.m. wave is given by ﻿ B_(y) = 2 xx 10^(7) sin (0.5 xx 10^(3)x+1.5 xx 10^(11)t) Tesla.﻿ Wavelength of e.m. wave is﻿

Which from the following represents displacement current ?

The intensities of magnetic field at two points on the axis of a bar magnet at distances 0.1 m and 0.2 m from its middle point are in the ratio 12.5 : 1 . Calculate the distance between the poles of the magnetic .

Two charges q_(1) and q_(2) are placed at (0,0,d) and (0,0,-d) respectively. Find icons of points where the potential is zero.

Which of the following is a suitable material for making electromagnet ?

A circular current loop of magnetic moment M is in an arbitrary orientation in an external uniform magnetic field vecB. The work done to rotate the loop by 30^(@) about an axis perpendicular to its plane is

A radioactive element ._92X^238 emits one alpha-particle and one beta^+particle in succession. Wht is the mass number of the new elementformed ?

In yound's double slit experiment, the separation between the slits is halved and distance between the slits and screen is doubled. The figure width is

The size of the nucleus is_____.

What are eddy currents ? How are these produced ? In what sense are eddy currents considered undesirable in a transformer and how are these reduced in such a device?

In a-p-n junction, the thickness of the depletion layer is 110^-6m. If the potential difference across it is 0.2 V, then the electric field set up across the junction is

The moment of inertia of a cylinder of radius R, length L and mass M about an axis passing through its centre of mass and normal to its length is

The quantity having negative dimensionsin mass is

Write the symbol, truth table, function and Boolean equation for ORgate.

A proton of mass 1.6xx10^-(27) kg, revolves in a circular path of radius 0.1 m. Calculate the angular velocity of the proton if it is acted upon by a centripetal force of 2.5xx10^(-12) N.

Name them :

In Young's double slit experiment, when light of wavelength 4000 A^(@) is used 90 fringes are seen on the screen. When light of 3000 A is used, the number of fringes seen is

Write down the advantages and limitations of frequency modulation (FM)? Advantages of FM

A Camot engine working between450 K and 600 K has a work output of 300 J//"cycle". What is the amount of heat energy supplied to the engine from the source in each cycle ?

Uniform circular motion is,

Carbon dating is best suited for determining the age of fossils of their age in years is of the order of

Transformation ratio of one transformer is 1:2. If Leclanche cell having emf 1.5 V is connected to primary coil of a transformer then voltage obtained across its secondary is ………..

Which one of the following force is conservative ?

If the refractive index of diamond is 2.4 find the velocity of light in diamond. (c= 3 xx10^8 m//s)

Find the energy required to split ._(8)^(16)O nucleus into four alpha - particles. The mass of an alpha- particle is 4.002603u and that of oxygen is 15.994915u.

A convex lens of refractive index 1.5 has a focal length of 18 cm in air. Calculate the change inits focal length when it is immersed in water of refraction index 4/3.

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How is an electric circuit shielded from stray electric field by placing it inside a conductive box?

The magnetic field in plane e.m. wave is given by B_(y) = 2 xx 10^(7) sin (0.5 xx 10^(3)x+1.5 xx 10^(11)t) Tesla. Wavelength of e.m. wave is