45406 + Interview Questions in Physics in Class 12 Page 60 InterviewSolution

2951.	The frequency of gamma-rays, X-rays and UV rays are a,b and respectively. Then
Answer» `a GT B gt C` `a LT b lt c` `a=b=c` `a gt c gt b` ANSWER :A

Discussion

2952.	The spped of light in the medium is :
Answer» maximum on the axis of the BEAM minimum on the axis of the beam the same everywhere in the beam directly proportional to the intensity 1. Solution :Since intensity is decreasing with radius, `mu (I) = mu_(1) + mu_(2)I` So is maximum along the axis of CYLINDER i.e. Refractive index `mu` is maximum along the axis. `RIGHTARROW` velocity is minimum along the axis of beam.

Discussion

2953.	A plane EM wave travelling in vacuum along z - direction is given by vec(E )=E_(0)sin (kz - omega t)hat(i)andvec(B)=B_(0)sin(kz - omega t)hat(j).Evaluate int vec(B).vec(d)s over the surface bounded by loop 1234.
Answer» Solution :Let SQUARE 1234 is made up of very large no. of small strip ds, let area of one strip = ds = h dz `THEREFORE int vec(B ).vec(d)s=int B ds COS 0^(@)` `= int B ds "" [because cos 0^(@)=1]` `= int_(z_(1))^(z_(2))B_(0)sin (kz- omega t)h dz "" [because ds = h dz]` `=-(B_(0)h)/(k)` `[cos(kz_(2)-omega t)-cos (kz_(1)-wt)] ""`....(2)

Discussion

2954.	A20 cm long cylindrical test tube is inverted and pushed verticallydown into water. When the closed end is at the water surface, how high has the water risen inside the tube ?
Answer» SOLUTION :`0.374 CM ]`

Discussion

2955.	A light having wavelength 300 nm falls on a metal surface work function of metal is 2.54 eV. What is stopping potential ?
Answer» 1.4 V 2.59 V 1.60 V 1.29 V Solution :`eV_(0) = H UPSILON - W_(0) = 2eV - 0.6 eV = 1.4 eV` `therefore""V_(0) = (1.4 eV)/(e) = 1.4 V1`

Discussion

2956.	A particle of mass m is moving along a circle of radius R such that its tangential acceleration a_tvaries with distance coveredx as a_t =ax^2 where alpha is a constant. the kinetic energy , K of the particle varies with the distance as K= beta x^c , where beta and c are constants. The values of beta and c are
Answer» `beta =(malpha)/3,c=3` `beta =(malpha)/4 ,c=4` `beta = (m alpha)/2, c=4` `beta = (malpha)/2 , c=3` Solution :Given mass of a particle = m tangential acceleration of particle , `a_t =ax^2 ` and KINETIC energy, `k= beta x^c` Here as we KNOW that TORQUE , `tau = FR = I alpha` `rArr FR = I (a_t)/R (because alpha =a/R) ` `rArrF*R = mR^2 (ax^2)/R (becauseI = MR^2)` ` rArr F =max^2 ` .........(i) Hence, the WORK done in displacement of dx, `W = int f. dx =int max^2 dx ` `W = 1/3 max^3 ` ...... (iii) From work -energy theorem , `W = Delta KE` here ,w = work done and `Delta KE ` = change in kinetic Energy , `(max^3)/3 = beta x^c ( because K= beta x ^c)` Now , from the above equation we get, `beta = (ma)/3 and c =3`

Discussion

2957.	Three charges q, q_(2) and q_(3) are placed in such a way that the distance q_(1) and q_(2)" is "r_(12), q_(2) and q_(3) is r_(23) and q_(2) is r_(31). a. What is the potential energy of the system? b. In the above question if there are 'N' point charges, then what is thenet potential energy of the system? Give the expression.
Answer» SOLUTION :`U=U_(1)+U_(2)+U_(3)=(1)/(4pi epsi_(0)) {(q_(1)q_(2))/(r_(12))+(q_(2)q_(3))/(r_(23))+(q_(1)q_(3))/(r_(13))}` b. Algebrac sum of PE of all attained pairs i.e, `U=1/2 UNDERSET(i=1)overset(n)sum underset(j=1)overset(n)sum (1)/(4pi epsi_(0)) (q_(1)q_(j))/(r_(ij))" with i "ne j`

Discussion

2958.	Twelve equal resistance are connected as shown in the diagram. Find the effective resistance between A and B
Answer» SOLUTION :MNO and POQ are two triangle with equal resistance between the vertices. So each is equivalent to `(2R)/(3)` between M and N or P and Q. So three resistors are in parallel between A and B. Whose resistance are `(R+(2R)/(3)+R),2R and (R+(2R)/(3)+R)` respectively `therefore (1)/(R_("EFF"))=(3)/(8R)+(1)/(2R)+(3)/(8R)=(10)/(8R)RARR R_("eff")=(4R)/(5)`

Discussion

2959.	An inclined plane making an angle beta with horizontal. A projectile projected from the bottom of the plane with a speed u at angle alpha with horizontal, then its maximum range R_("max") is
Answer» `R_("MAX") = (u^2)/(G(1 - sin beta))` `R_("max") = (u^2)/(g(1 + sin beta))` `R_("max") = (u^2)/(g(1 - sin beta))` `R_("max") = (u^2)/(g(1 + sin beta))` ANSWER :B

Discussion

2960.	If R is the radius of the meniscus of liquid in capillary and r is the radius of capillary. The angle of contact will be
Answer» `cos^-1(R/R)` `cos^-1(R/r)` `cos^-1(RR)` `cos^-1(R)` ANSWER :A

Discussion

2961.	Find the number of half lives elapsed, before which, 93.75% of a radioactive sample has decayed.
Answer» SOLUTION :FRACTION of atoms left undecayed is 100-93.75% 6.25% i.e., `N/N_0=6.25/100=1/16` or `N/N_0=(1/2)^4` or , `N=N_0(1/2)^4` COMPARING with `N=N_0(1/2)^n` , yields n=4 .

Discussion

2962.	(i) Explain total internal reflection from law of refraction. (ii) A ray of light is incident at a small angle theta on a rectangular glass slab of thickness t. If the refractive index of glass is mu . Show that the perpendicular distance between the emergent ray from the slab and the incident ray is t theta(mu - 1) // mu .
Answer» ANSWER :(i) N/A (II) N/A

Discussion

2963.	A container is filled with water (mu=1.33) upto a height of 33.25cm. A convex mirror I s placed 15cm above the water level and image of an object placed at the bottom is formed 25cm below the water level. Focal length of the mirror is
Answer» 15cm 20cm `-18.31cm` 10cm Solution :The image `I^(')` for first refraction (i.e., when the ray comes out of liquid ) is at a depth of `=(33.25)/(1.33)=25CM` `[ :' "Apparent depth" =("Real depth")/(mu)]` Now, reflection will occur at concave MIRROR. For this `i^(')` behaves as an object `:. u=-(15+25)-40cm, F=f, v=?` Using mirror formula `(1)/(f)=(1)/(v)+(1)/(u)=(1)/(v)-(1)/(u)-(1)/(40)rArr v=(40F)/(40+f)`(i) But `v =-[15+(25)/(1.33)]` (ii) where `(25)/(1.33)` is the real depth of the image Form Eqs. (i) and (ii) `(40f)/(40+f)=-[15+(25)/(1.33)]=-33.79` `rArr f=-18.31cm`

Discussion

2964.	A sample of Ammonium phosphate (NH_(4))_(3)PO_(4) contains 3.18 moles of oxygen atoms. The number of moles of oxygen atom in the sample is
Answer» 0.265 0.795 1.06 3.18 Solution :`because` 12 MOLES of HYDROGEN in the sample `therefore` 1 moles of hydrogen=`4/12` mole of oxygen `therefore` 3.18 mole of hydrogen=`4/12xx3.18` =1.06 moles of oxygen

Discussion

2965.	The most commonly used system of modulation for telegraphy is :
Answer» on-off keying frequency SHIFT keying analog modulation pulse CODE modulation. Answer :A

Discussion

2966.	Two infinitely long parallelsheets having surface charge densities +sigma and -sigmarespectively are separated by a small distance. The electric field in the region between the plates is
Answer» `0` `(SIGMA)/( 2 in_0) ` ` (sigma)/(in_0) ` ` (2sigma)/( in_0) ` ANSWER :C

Discussion

2967.	Assertion :Static crashes are heard on radio, when lightning flash occurs in the sky. Reason:Electromagnetic waves having frequency of radiowave range, interfere with radiowaves.
Answer» If both the Asseration and Reason are true and reason explains the ASSERTION : If both the Assertion and Reason are true but reason does OT explain the Assertion : If Assertionis true but reason false If Assertionis false but reason is true Answer :A

Discussion

2968.	From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is
Answer» `(4MR^(2))/(3sqrt(3)PI)` `(MR^(2))/(32sqrt(2)pi)` `(MR^(2))/(16sqrt(2)pi)` `(4MR^(2))/(9sqrt(3)pi)` ANSWER :D

Discussion

2969.	When are two objects just resolved ? Explain. How can the resolving power of a compound microscope be incresed ? Use relevant formula to support your answer.
Answer» Solution :Two objects are said to be just resolved when central diffraction MAXIMUM of one object coincides with first diffraction MINIMA of the other object and vice versa. Resolving power of a compound microscope `=(2nsina)/(1.22lamda)`, where n=refractive index of medium present between the object and objective LENS, 2D is the angle subtended by the objective lens at the object and `lamda` the wavelength of light used to ILLUMINATE the object. generally the term n sin `alpha` is known as numerical aperture (N.A.) of the microscope. thus, to increase the resolving power of a compound microscope (i) its numerical aperrture should be increased, and (ii) the wavelength of illuminating light should be reduced.

Discussion

2970.	A plane EM wave travelling in vacuum along z - direction is given by vec(E )=E_(0)sin (kz - omega t)hat(i)andvec(B)=B_(0)sin(kz - omega t)hat(j).Use equation int vec(E ).vec(d)l =-(d phi_(B))/(dt) to prove (E_(0))/(B_(0))=c.
Answer» Solution :Now `oint vec(E ).vec(d)l = -(d)/(dt)oint vec(B).vec(d )s` `thereofre` From EQUATION (1) and (2), `E_(0)h[sin (kz_(2)-omega t)-sin (kz_(1)-omega t)] =-(d)/(dt)` `[(B_(0)h)/(K){COS (kz_(2)-omega t)-cos(kz_(1)-wt)}]` `therefore E_(0)h[sin(kz_(2)-omega t)-sin (kz_(1)-omega t)]` `=(B_(0)h)/(k)(omega)[sin (kz_(2)-omega t)-sin (kz_(1)-omega t)]` `therefore E_(0)h=R_(0)h((omega)/(k))` `therefore E_(0)=B_(0)c "" [because (omega)/(k)=c]` `therefore (E_(0))/(B_(0))=c`

Discussion

2971.	The electric flux through a closed Gaussian surface depends upon
Answer» net charge enclosed and permittivity of the MEDIUM. net charge enclosed, permittivity of the medium and the size of the Gaussian surface. et charge enclosed only. permittivity of the medium only. Solution :For a closed Gaussian surface in a GIVEN medium `phi_(E)=(sumq)/(in)`

Discussion

2972.	The unit of permittivity is ..................... .
Answer» ANSWER :`C^(2)N^(-1)m^(-2)`

Discussion

2973.	Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.
Answer» Solution :Let `I_(0)` be the intensity of polarised light after passing through the first polariser`P_(1)`. Then the intensity of light after passing through second polariser `P_(2)` will be `I=I_(0)cos^(2)theta`, where `theta` is the angle between pass AXES of `P_(1)` and `P_(2)`. SInce `P_(1)` and `P_(3)` are crossed the angle between the pass axes of `P_(2)` and `P_(3)` will be`((pi)/2-theta)`.HENCE the intensity of light emerging from `P_(3)` will be `I=I_(0)cos^(2)theta cos^(2)((pi)/2-theta)=I_(0)cos^(2) SIN^(2) theta=((I_(0))/4)sin^(2) 2 theta` Therefore the transmitted intensity will be maximum when `theta=(pi)/4`

Discussion

2974.	The direction of flow of negative charge what we call ?
Answer» SOLUTION :ELECTRIC CURRENT

Discussion

2975.	S_1 and S_2 are two sources of light separated by a distance d. A detector can move along S_2 P perpendicular to S_1S_2. What should be the minimum and maximum path difference at the detector?
Answer» Solution :MINIMUM path DIFFERENCE is zero (when p is at infinity MAXIMUM path difference =d

Discussion

2976.	In the arrangement shown in the figure, friction exists only between the two blocks, A and B. The coefficient of static friction mu_(s)=0.6 and coefficient of kinetic friction mu_(k)=0.4, the masses of the blocks A and B are m_(1)=20 kg and m_(2)=30kg, respectively. Find the acceleration (in ms^(-2)) of m_(1), if a force F=150N is applied, as shown in the figure. [Assume that string and pulleys are massless]
Answer» ANSWER :1.5

Discussion

2977.	Give the theory of interference of light by considering waves of equal amplitude and hence arrive at the conditions for constructive and destructive interference in terms of path difference.
Answer» Solution :Theory of interference : Consider TWO harmonic light waves of same amplitude. Let `phi` be the constant phase difference between them. The displacement of these waves can be represented as. `y_1 =a cos omega t""rarr(1)` `y_2 = a cos (omega t+ phi)""rarr(2)` According superposition principle `y=y_1 +y_2` `y= a cos omega t + a cos (omega t+phi)` `y = [cos omega t + cos (omega t +phi )]` We know that `cos C+cos D=2cos((C+D)/(2))cos((C-D)/(2))` So, `y=a[2cos((omega t+omega t+phi)/(2))cos((omega t-omega t-phi)/(2))]` `y=2a cos(omega t+phi/2)cos(- phi/2)` `y=[2acos(phi/2)]cos(omega t+phi/2)` `y=A cos(omega t+phi/2)` Where `A = 2a cos(phi/2)` called amplitude of resultant wave. So intensity of resultant wave, `I = A^2 = 4a^2 cos^2(phi/2)` Condition for constructive interference For MAXIMUM intensity `cos^2(phi/2) =+-1` `cos(phi/2)=+-1=cos(2n pi)` where `n=0,1,2,3,....,phi = 0, +- 2pi, +- 4pi, +-6pi.....` `:.` Path difference `delta=n lambda` `delta=0,lambda,2lambda,3lambda,......` Condition for destructive interference For MINIMUM intensity `cos(phi/2)=0` so, `phi = +-pi,+-3PI,+-5PI......` `:.` Path difference `delta= (2n+1) (lambda)/(2)` `delta = (lambda)/(2), (3lambda)/(2), (5lambda)/(2) ....`

Discussion

2978.	The dimensions of electric susceptibility are :
Answer» `M^(@)L^(@)T^(@)A^(@)` `M^(@)L^(-2)T^(@)A^(2)` `MLT^(-2)A` `M^(@)L^(-1)T^(@)A` Solution :ELECTRIC SUSCEPTIBILITY `X=(1)/(K)` `:.x=(1)/(M^(0)L^(0)T^(0))=[M^(0)L^(0)T^(0)]` Thus `(a)` is CORRECT choice.

Discussion

2979.	A coil of number of turns N, area A, is rotated at a constant angular speed omega, in a uniform magnetic field B, and connected to a resistor R. Deduce expressions for : (i) maximum emf induced in the coil, (ii) power dissipated in the coil.
Answer» Solution :(i) As SHOWN in Short Answer Question Number 14, induced emf `varepsilon = NBA omega SIN omegat.` `therefore` Maximum emf induced in the coil `varepsilon_(max) = N B A omega` [when `sin omegat = 1]` (ii) Istantaneous power dissipation in the coil Pinsi `P_(inst).= varepsilon_(2)/R = (N^(2)B^(2)A^(2)omega^(2))/R sin^(2)omegat` For one complete cycle mean VALUE of `sin^(2) omegat =1/2`. Hence, the mean value of power dissipation in the coil is given by `P("average") = (N^(2) B^2)A^(2) omega^(2))/R.(1/2)=(N^(2)B^(2)A^(2)omega^(2))/(2R)`.

Discussion

2980.	In 1924, the French physicist Louis de-Broglie, suggested that moving particles of matter should display wave like properties under suitable conditions. His reasoning was that nature was symmetrical and so the two basic physical entities, matter and energy, must have symmetrical character. if radiation shows dual character so should matter. he proposed that the wavelength lamda associated with a particle of momentum p is given as: lamda=(h)/(p)=(h)/(mv) Dual aspect of matter is evident in this relation. whereas wavelength lamda is an attribute of a wave, the momentum p is a typical attribute of a particle. the Planck's constant 'h' relates the two attributes. The de-Broglie's hypothesis of matter waves has been basic to the development of medern quantum mechanics. it also led to the field of electron optics. the wave properties of electrons have been utilised in the design of electron microscope which is a great improvement, with higher resolution, over the optical microscope. Q. An electron an a proton both are accelerated by same voltage. which of the two will hae greater value of de-Broglie wavelength and why?
Answer» SOLUTION :As PER relation `lamda=(h)/(sqrt((2mqV))` for de-Broglie wavelength of a charged particle accelerated through a voltage V, we find that for same VALUE of V and charge `q, lamdaprop(1)/(sqrtm)`. As mass of electron is LESS as compared to that of a proton, the de-broglie wavelength of electron is more as compared to a proton.

Discussion

2981.	Assertion Mutual inductance of two coils depends on the distance between the coils and their orientation. Reason It does not depend on the magnetic material filled between the coils.
Answer» If both assertion and reason are true and reason is the corrent explanation of assertion. If both assertion and reason are true but reason is not the correct explanation of assertion. If assertion is true but reason is false. Solution :The MUTUAL inducrtance in case of a medium of ralative permeability `mu_(R)` present is `M=mu_(r)mu_(0)n_(1)n_(2)pi r_(1)^(2)l`

Discussion

2982.	A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the plane of the coil. If the current in the coil is 3.0 A, calculate the : (i) Total troque on the coil, (ii) total force on the coil, (iii) average force on each electron on each electron in the coil, due to the magnetic field. Assume, the area of cross-section of the wire to be 10^(-5) m^(2) and the free electron density as 10^(29) m^(-3).
Answer» Solution :Here`N = 200`, radius of coil `r = 10 cm = 0.1 m` and hence AREA of circular coil `A = pi r^2 = pi (0.1)^(2) = 0.01 pi m^2`, uniform magnetic field `B = 0.5 T, theta = 0^@` and current I = 3.0 A (i) TORQUE on the coil `tau = NAI B sin theta = 200 xx 0.01 pi xx 3.0 xx 0.5 xx sin 0^@ = 0` (ii) Total force on a current carrying coil placed in a uniform magnetic field is ALWAYS zero. (iii) Average force on each electron in the coil, `F= B.e.v_d = Be 1/(n eA.)`, where n = free electron dnesity `10^(29) m^(-3)` and A. = area of cross-section of wire = `10^(-5) m^(2)` `:. F = (0.5 xx 1.6 xx 10^(-19) xx 3.0)/(10^(29) xx 1.6 xx 10^(-19) xx 10^(-5)) = 1.5 xx 10^(-24) N`.

Discussion

2983.	In 1924, the French physicist Louis de-Broglie, suggested that moving particles of matter should display wave like properties under suitable conditions. His reasoning was that nature was symmetrical and so the two basic physical entities, matter and energy, must have symmetrical character. if radiation shows dual character so should matter. he proposed that the wavelength lamda associated with a particle of momentum p is given as: lamda=(h)/(p)=(h)/(mv) Dual aspect of matter is evident in this relation. whereas wavelength lamda is an attribute of a wave, the momentum p is a typical attribute of a particle. the Planck's constant 'h' relates the two attributes. The de-Broglie's hypothesis of matter waves has been basic to the development of medern quantum mechanics. it also led to the field of electron optics. the wave properties of electrons have been utilised in the design of electron microscope which is a great improvement, with higher resolution, over the optical microscope. Q. In an electron microscope which property of electron beam is utilised and what is its significance ?
Answer» Solution :In electron microscope we make use of wave property of electrons. In fact, electron waves are used as a PROBE to OBSERVE the structure at atomic and subatomic LEVEL, because RESOLVING power of an electron microscope is MILLIONS times higher than that of an optical microscope.

Discussion

2984.	A charge is distributed with a linear density lambda over the length L along radius vector drawn from the point where a point charge q is located. The distance between q and the nearest point on linear charge is R. The electrical force experienced by the linear charge due to q is
Answer» `(q lambdaL)/(4piepsilon_0R^2)` `(qlambdaL)/(4piepsilon_0R(R+L))` `(qlambdaL)/(4piepsilon_0 RL)` `(qlambdaL)/(4piepsilon_0L^2)` SOLUTION :As Coulomb.s force is an action reaction PAIR, so force EXPERIENCED by the linear charge is equal and OPPOSITE to the force experienced by point charge q. Here, we are computing electric force experienced by q due to the charge. Consider a small element dr on line charge as shown, then force experienced by q due to this element is, `dF=(qlambdadr)/(4piepsilon_0r^2)` `THEREFORE F=int dF= int_R^(R+L) (qlambdadr)/(4piepsilon_0r^2)` `=(qlambdaL)/(4piepsilon_0R(R+L))`

Discussion

2985.	An AC voltage V = 5 cos (1000 t) V is applied to a L-R series circuit of inductance 3 mH and resistance 4Omega . The value of maximum current in the circuit is …….. A.
Answer» 1 `5/sqrt7` `5/7` 0.8 Solution :By comparing , V=5 cos (100T)V with `V=V_m cos (omegat)V` `omega`=1000 rad/s , `V_m` = 5 V Impedence of SERIES connectionof L-R, `\|Z\|=sqrt((omegaL)^2 + R^2)` `=sqrt((1000xx3xx10^(-3))^2+(4)^2)` `=sqrt(9+16)` `=sqrt25` `= 5Omega` `THEREFORE I_m=V_m/"\|Z\|"=5/5` = 1 A

Discussion

2986.	A body in motion along a straight line. From position time graph Calculate itsinitial velocity
Answer» Solution :(i) At `t = 0` , the curve OA has a non ZERO SLOPE. The body possesses some INITIAL velocity which is GIVEN by `u = 20/10=2m/s`

Discussion

2987.	A: The relation between magnetic moment and angular momentum is true for every finite size body. R: Ratio of magnetic dipole moment and angular momentum is just dependent on specific charge of the body.
Answer» If both Assertion & Reason are TRUE and the reason is the CORRECT explanation of the assertion then mark (1) If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark 2. If assertion is true statement but Reason is false, then mark (3) If both Assertion and Reason are false STATEMENTS, then mark (4) ANSWER :A

Discussion

2988.	Velocity of light in air is 3 xx 10^8 m//s and in glass its velocity is 2 xx 10^(8) m//s so for glass angle of polarisation is ......
Answer» `56.50^(@)` `56.30^(@)` `56.1^(@)` `56^(@)` Solution :Refractive index of glass `mu= tan theta_(p)` but `mu=(C)/(v)=(3XX10^(8))/(2XX10^(8))=1.5` `:.theta_(p)=tan^(-1)(1.5)` `:.theta_(p)=56.30^(@)`

Discussion

2989.	In photoelectric effect the electrons are assumed to be ___ and minimum energy ___ to the work function is required to free them .
Answer» SOLUTION :BOUND , EQUAL

Discussion

2990.	Two magnets of magnetic moments M and 2M are placed in a vibration magnetometre, with the identical poles in the same direction . The time period of viberation is T. If the magnets are placed with opposite poles together and vibrate with time period T_(2) then
Answer» `T_(2)` is INFINITE `T_(2)=T_(1)` `T_(2) gt T_(1)` `T_(2) lt T_(1)` SOLUTION :(C) `T_(1)=2pisqrt((I_(1)+I_(2))/((M+2M)H))=2pisqrt((I)/(3MH))` `T_(2)=2pisqrt((I_(1)+I_(2))/((2M-M)H))=2pisqrt((I)/(MH))` Obviously, `T_(2) gt T_(1)`

Discussion

2991.	The highest frequency of transmitted wave that is returned to the earth by the layer of ionosphere after having been sent at some angle to normal is called
Answer» MAXIMUM USABLE frequency critical frequency LAUNCHING frequency threshold frequency Answer :A

Discussion

2992.	Displacement current is given by,
Answer» Solution :The CURRENT due to CHANGING ELECTRIC field (TIME varying electric field) is called displacement current. To remove the inconsistency of Ampere.s circuital law.

Discussion

2993.	A conducting loop of resistance R is pulled in a uniform magnetic field vecB, with uniform velocity vecv (see fig.). During this process it is found that …….
Answer» the TEMPERATURE of the LOOP remains constant the temperature of the loop decreases the temperature of the loop increases none of these above. Solution :When a loop moves the flux associated with changes and hence induced current produced and DUE to the resistance of a loop `I^2Rt` HEAT ENERGY produces, so the temperature of the coil increases.

Discussion

2994.	The substance which are transparent to thermal radiation are
Answer» athermanous diathermanous thermo electric radioactive Answer :B

Discussion

2995.	A ray of light is incident on a glass slab making on angle of 30^@ with the surface. If the angle of refraction in glass is 33^@ , the angle of deviation of the ray during its passage through the glass slab is :
Answer» `30^@` `27^@` `40^@` `20^@` ANSWER :B

Discussion

2996.	Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids ?
Answer» <P> Solution :Let `I_(0)` be the intensity of polarised light after passing through the first polariser `P_1` Then the intensity of light after passing through second polariser `P_2` will be `I =I_(0)cos^(2)e,` where `theta` is the angle between pass axes of `P_1 and P_2.` SINCE `P_1 and P_3` are crossed the angle between the pass axes of P, and `P_3` will be `(pi//2 -theta)`. Hence the intensityof light EMERGING from `P_3` will be `I=I_(0)cos^T(2)thetacos^(2)((pi)/(2)-theta)` `=I_(0)cos^(2) thetasin^(2)theta=(I_(0)//4)SIN^(2)2theta` Therefore, the transmitted intensity will be maximum when `theta= lambda//4.`

Discussion

2997.	In a liquid with density 900 kg//m^3,longitudinal waves with frequency 250 Hz are found to have wavelength 8.0 m. Calculate the bulk modulus of the liquid.
Answer» SOLUTION :`3.6 XX 10^(9) PA`

Discussion

2998.	In the figure shown, pulley and spring are ideal and strings are lilght and inextensible. Intially all the bodies are at rest when string connecting A and B is cut. Find initial acceleration of C.
Answer» SOLUTION :`[0]`

Discussion

2999.	A : Two gases with the same average translational kinetic: energy have same temperature even if one has greater rotational energy as compared to other. R:Onlyaveragetranslationalkineticenergyofa gas contributes to its temperature.
Answer» If bothAssertion & Reason are true and the reason is the correct explanation of the ASSERTION, then Mark (1) If both Assertion & Reason are true but the reasonis not the correct explanation of the assertion, then mark (2) If Assertion is true statement but Reason is false, then mark (3) If both Assertion and Reason are false STATEMENTS, then mark (4). Answer :A

Discussion

3000.	Pick the correct options for maximum value of equivalent resistance .
Answer» IV II R III iv S II ii Q I iv S Answer :D

Discussion

Explore topic-wise InterviewSolutions in .

The frequency of gamma-rays, X-rays and UV rays are a,b and respectively. Then

The spped of light in the medium is :

A plane EM wave travelling in vacuum along z - direction is given by vec(E )=E_(0)sin (kz - omega t)hat(i)andvec(B)=B_(0)sin(kz - omega t)hat(j).Evaluate int vec(B).vec(d)s over the surface bounded by loop 1234.

A20 cm long cylindrical test tube is inverted and pushed verticallydown into water. When the closed end is at the water surface, how high has the water risen inside the tube ?

A light having wavelength 300 nm falls on a metal surface work function of metal is 2.54 eV. What is stopping potential ?

Twelve equal resistance are connected as shown in the diagram. Find the effective resistance between A and B

An inclined plane making an angle beta with horizontal. A projectile projected from the bottom of the plane with a speed u at angle alpha with horizontal, then its maximum range R_("max") is

If R is the radius of the meniscus of liquid in capillary and r is the radius of capillary. The angle of contact will be

Find the number of half lives elapsed, before which, 93.75% of a radioactive sample has decayed.

A container is filled with water (mu=1.33) upto a height of 33.25cm. A convex mirror I s placed 15cm above the water level and image of an object placed at the bottom is formed 25cm below the water level. Focal length of the mirror is

A sample of Ammonium phosphate (NH_(4))_(3)PO_(4) contains 3.18 moles of oxygen atoms. The number of moles of oxygen atom in the sample is

The most commonly used system of modulation for telegraphy is :

Two infinitely long parallelsheets having surface charge densities +sigma and -sigmarespectively are separated by a small distance. The electric field in the region between the plates is

Assertion :Static crashes are heard on radio, when lightning flash occurs in the sky. Reason:Electromagnetic waves having frequency of radiowave range, interfere with radiowaves.

From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is

When are two objects just resolved ? Explain. How can the resolving power of a compound microscope be incresed ? Use relevant formula to support your answer.

A plane EM wave travelling in vacuum along z - direction is given by vec(E )=E_(0)sin (kz - omega t)hat(i)andvec(B)=B_(0)sin(kz - omega t)hat(j).Use equation int vec(E ).vec(d)l =-(d phi_(B))/(dt) to prove (E_(0))/(B_(0))=c.

The electric flux through a closed Gaussian surface depends upon

The unit of permittivity is ..................... .

Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.

The direction of flow of negative charge what we call ?

S_1 and S_2 are two sources of light separated by a distance d. A detector can move along S_2 P perpendicular to S_1S_2. What should be the minimum and maximum path difference at the detector?

Give the theory of interference of light by considering waves of equal amplitude and hence arrive at the conditions for constructive and destructive interference in terms of path difference.

The dimensions of electric susceptibility are :

A coil of number of turns N, area A, is rotated at a constant angular speed omega, in a uniform magnetic field B, and connected to a resistor R. Deduce expressions for : (i) maximum emf induced in the coil, (ii) power dissipated in the coil.

Assertion Mutual inductance of two coils depends on the distance between the coils and their orientation. Reason It does not depend on the magnetic material filled between the coils.

A charge is distributed with a linear density lambda over the length L along radius vector drawn from the point where a point charge q is located. The distance between q and the nearest point on linear charge is R. The electrical force experienced by the linear charge due to q is

An AC voltage V = 5 cos (1000 t) V is applied to a L-R series circuit of inductance 3 mH and resistance 4Omega . The value of maximum current in the circuit is …….. A.

A body in motion along a straight line. From position time graph Calculate itsinitial velocity

A: The relation between magnetic moment and angular momentum is true for every finite size body. R: Ratio of magnetic dipole moment and angular momentum is just dependent on specific charge of the body.

Velocity of light in air is 3 xx 10^8 m//s and in glass its velocity is 2 xx 10^(8) m//s so for glass angle of polarisation is ......

In photoelectric effect the electrons are assumed to be ___ and minimum energy ___ to the work function is required to free them .

Two magnets of magnetic moments M and 2M are placed in a vibration magnetometre, with the identical poles in the same direction . The time period of viberation is T. If the magnets are placed with opposite poles together and vibrate with time period T_(2) then

The highest frequency of transmitted wave that is returned to the earth by the layer of ionosphere after having been sent at some angle to normal is called

Displacement current is given by,

A conducting loop of resistance R is pulled in a uniform magnetic field vecB, with uniform velocity vecv (see fig.). During this process it is found that …….

The substance which are transparent to thermal radiation are

A ray of light is incident on a glass slab making on angle of 30^@ with the surface. If the angle of refraction in glass is 33^@ , the angle of deviation of the ray during its passage through the glass slab is :

Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids ?

In a liquid with density 900 kg//m^3,longitudinal waves with frequency 250 Hz are found to have wavelength 8.0 m. Calculate the bulk modulus of the liquid.

In the figure shown, pulley and spring are ideal and strings are lilght and inextensible. Intially all the bodies are at rest when string connecting A and B is cut. Find initial acceleration of C.

A : Two gases with the same average translational kinetic: energy have same temperature even if one has greater rotational energy as compared to other. R:Onlyaveragetranslationalkineticenergyofa gas contributes to its temperature.

Pick the correct options for maximum value of equivalent resistance .

In photoelectric effect the electrons are assumed to be _ and minimum energy _ to the work function is required to free them .