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251.

Obtain mirror equation for the real image obtained by concave mirror.

Answer»

Solution :
An object AB is perpendicular to principal axis away from C of a concave mirror.
AM ray from point A incidents on mirror at M and reflected ray passes through principal FOCUS F.
AP ray from point A incidents on pole P and reflects back in form of PA.
These both reflected rays intersect at A., hence A. is real image of A.
A.B. is image of object AB due to reflection of rays from mirror.
Let FP = focal length f
CP = radius of curvature R
BP = object distance u
B.P= image distance v
For paraxial rays, MP can be considered to be a straight LINE perpendicular to CP.
The two right-angled triangles A.B.F and MPF are similar.
`therefore(B.A.)/(PM)=(B.F)/(FP)`
But PM = BA
`therefore (B.A.)/(BA)=(B.F)/(FP)`....(1)
Since `angleAPB = angleA.PB.`, the right angled triangles A.B.P and ABP are also similar. Therefore,
`therefore(B.A.)/(BA)=(B.P)/(BP)`...(2)
Comparing Equations (1) and (2),
`(B.F)/(FP)=(B.P)/(BP)`
but B.F = B.P - FP
`therefore (B.P-FP)/(FP)=(B.P)/(BP)` ....(3)
Now, ACCORDING to sign CONVENTION, B.P = - v, FP = - f, BP = - u
From equation (3),
`(-v+f)/(-f)=(-v)/(-u)`
`therefore (v-f)/(f)=v/u`
`therefore v/f-1=v/u`
`therefore v/f=1+v/u`
Now,dividing by v,
`therefore 1/f=1/v+1/u`
Which is mirror equation and it is called Gaussian equation as it was given by scientist Gauss.
Discussion
252.

Gren house effect is

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BLACK radiations from both the TYPES of SOURCES at high and low temperature
transmit radiations from both the types of source
transmit radiations from a source at low source at high temperature
transmit radiates from a source at high temperature and block radiations from a source at low temperature

Answer :D
Discussion
253.

Three identical objects each of mass M move along circle of radius R under the action of their mutual gravitational attraction, the speed of each is :

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`((GM)/( R ))^(1//2)`
`((GM)/( R ))^(1//2)`
`((GM)/(sqrt(3)R))^(1//2)`
`((GM)/(sqrt(2)R))^(1//2)`

ANSWER :C
Discussion
254.

Mention the way of producing induced emf.

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Solution :Induced EMF can be produced by changing MAGNETIC flux in any of the FOLLOWING ways: (i) By changing the magnetic FIELD B
(ii) By changing the area of the coil and
(iii) By changing the relative orientation of the coil with magnetic field
Discussion
255.

An stronmical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

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SMALL FOCAL LENGTH and LARGE diameter
large focal length and small diameter
small focal length and small diameter
large focal length and large diameter

Answer :D
Discussion
256.

If the work function of a certain metal is 3.2xx10^(-19)J and is it illuminated with light of frequency 8xx10^(14) Hz, the maximum kinetic energy of photo electrons would be (h=6.6xx10^(-34)Js)

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`2.1xx10^(-19)J`
`8.5xx10^(-19)J`
`5.3xx10^(-19)J`
`3.2xx10^(-19)J`

ANSWER :A
Discussion
257.

A circular coil of radius 8cm, 400 turns and resistance 2Omega is placed with its plane perpendicular to the horizantal component of the earth's magnetic fiedl. It is rotated about its vertical diameter through 180^(@) in 0.30 s. Horizontal component of earth magnitude of current induced in the coil is approximately

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`4xx10^(-2)A`
`8xx10^(-4)A`
`8xx10^(-2)A`
`1.92xx10^(-3)A`

Solution :Initial flux through the coil
`phi_(i)=BAcos theta`
`=3xx10^(-5)xxpixx(8xx10^(-2))xxcos 0^(@)`
`=192pixx10^(-9)Wb`
FINAL flux after the rotation
`phi_(F)=3xx10^(-5)xxpi(8xx10^(-2))^(2)cos 180^(@)`
`=-192pixx10Wb`
`:.` TEH magnitude of induced emf is
`epsi=N|(dphi|)/(dt)=(N|phi_(f)phi_(i)|)/(dt)`
`=(400xxx(384pxx10^(-9)))/(0.30)=1.6xx10^(-3)V`
Current `I=(epsi)/(R)=(1.6xx10^(-3))/(2)=8xx10^(-4)A`
Discussion
258.

According to the Bohr-Sommerfeld postulate the periodic motion of a particle in a potential field must satisfy the following qyantization rule: ointp dq= 2piħn, where q and p are generalized coordinate and momenum of the particle, n are integers. Making use of this rule, find the permitted values of energy for a particle of mass m moving (a) In a uniimensional rectangular potential well of width l (b) along a circule of radius r, (c ) in a unidimentional potential field U=alphax^(2)//2, where alpha is a positive constant: (d) along a round orbit in a central field, where the potential enargy of the particle is equal to U= -alpha//r(alpha is a positve constant)

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Solution :(a) if we measure energy from the bottom of the well, then `V(x) = 0` inside the walls. Then the quantization condition reads `oint p d x = 2 l p = 2 pi nħ`
or `p = pi ħ//l`
Hence`E_(n) = (p^(2))/(2 m) = (pi^(2) n^(2) ħ)/(2 m l)`.
`oint p d x = 2 l p` because we have to consider the INTEGRAL form `- (1)/(2)` to `(1)/(2)` and then back to `-(1)/(2)`.
(b) Here, `oint p d x = 2 pi r p = 2 pi nħ`
or `p = (n ħ)/(r )`
Hence`E_(n) = (n^(2)ħ^(2))/(2 m r^(2))`
(c ) By energy CONSERVATION`(p^(2))/(2 m) + (1)/(2) alpha x^(2) = E`
so`p = sqrt(2m E - m alpha x^(2))`
Then `oint p d x = oint sqrt(2m E - m alpha x^(2) dx)`
`= 2sqrt(m alpha) int_(-(sqrt( 2E))/(alpha))^(sqrt(2E)/(alpha)) sqrt((2 E)/(alpha) - x^(2)) dx`
The integral is`int_(-a)^(a)sqrt(a^(2) - x^(2)) dx = a^(2) int_(-x//2)^(x//2) cos^(2) theta d theta`
`= (a^(2))/(2) int_(-x//2)^(x//2)(1 + cos 2 theta) d theta = a^(2)(pi)/(2)`.
Thus `oint p d x = pi sqrt(m a). (2 E)/(alpha) = E.2 pi sqrt((m)/(alpha)) = 2 pi n ħ`
Hence `E_(n) = n ħsqrt((alpha)/(m))`.
( b) It is required to find the energy levels of the CIRCULAR orbit for the rotential
`U( r) = -(alpha)/(r )`
In a circular orbit, the particle only has TANGIBLE velocity and the qunatization condition reads `ointp d x = m v. 2 pi r = 2 pi nħ`
so `m v r = M = n ħ`
The energy of the particle is
`E = (n^(2) ħ^(2))/(2 m r^(2)) - (alpha)/(r )`
Equilibrium requires that the energy as a function of `r` be minimum. Thus
`(n^(2) ħ^(2))/(mr^(3))=(alpha)/(r^(2)) or r=(n^(2) ħ^(2))/(m alpha)`
Hence `E_(n)= -(malpha^(2))/(2n^(2) ħ^(2))`
Discussion
259.

The displacement-time graphs of two moving particles makes anlges of 30^(@) and 45^(@)with the X-axis.The ratio of their velocities is

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`SQRT(3:2)`
`1:1`
`1:2`
`:sqrt(3)`

Solution :`(V_A)/(V_B)=(TAN30^(@))/(TAN45^(@))=(1)((sqrt(3))/(1))=(1)/(sqrt(3))`
Discussion
260.

A ray of light travelling in a medium of refractive index mu is incident at an angle theta ona composite transparent plate consisting of 50 plates of R.I. 1.0mu, 1.02mu, 1.03mu,"……",1.50mu. The ray emerges from the composite plate into a medium of refractive index 1.6 mu at angle 'x'. Then

Answer»

`sin x (1.01/1.5)^(50) sin THETA`
`SINX = 5/8 sin theta`
`sinx= (8)/(5) sin theta`
`sinx = (1.5/1.01)^(50) sin theta`

Solution :
ACCORDINGTO SNELL's law, `musintheta= 1.6 mu sin x`
`:. sin x = 5/8 sin theta`
Discussion
261.

The rate of emission of heat energy per unit area of an iron ball of radius 10 cm is 10J//m^2s s, then rate ·of emission of heat energy per unit area by a copper ball of radius 5 cm at same temperature will be ( emissivity of both the balls is same)

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`2J//m^2s`
`2.5J//m^2s`
`2cal//m^2s`
NONE of these

Answer :B
Discussion
262.

A massless elastic cord (that obeys Hooke's law) break if the tension in the cord exceeds T_("max"). One end of the cord is attached to a fixed point, the other is attached to an object of mass 3m as shown in the figure. If a second, smaller object ofmass m moving at an initial speed v_(0) strikes the larger mass and the two stick together the cord will stretch and break but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one dimensional collision the cord will still break and the larger mass will move off with a final speed of v_(f). All motion occurs on a horizontal frictionless surface. (Assume that Hooke's law is obeyed throughout untill the cord breaks) Find (v_(f))/(v_(0))

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`(1)/(sqrt(12))`
`(1)/(sqrt(2))`
`(1)/(sqrt(6))`
`(1)/(sqrt(3))`

ANSWER :C
Discussion
263.

A massless elastic cord (that obeys Hooke's law) break if the tension in the cord exceeds T_("max"). One end of the cord is attached to a fixed point, the other is attached to an object of mass 3m as shown in the figure. If a second, smaller object ofmass m moving at an initial speed v_(0) strikes the larger mass and the two stick together the cord will stretch and break but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one dimensional collision the cord will still break and the larger mass will move off with a final speed of v_(f). All motion occurs on a horizontal frictionless surface. (Assume that Hooke's law is obeyed throughout untill the cord breaks) Find the ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision and the cord has broken to the initial knetic energy of the smaller mass prior to the collision

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`1//4`
`1//3`
`1//2`
`3//4`

ANSWER :D
Discussion
264.

Maximum value of photo electrons V_max = ____.

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SOLUTION :2hc(`lambda_0 - LAMBDA`)/`mlambda lambda_0`
Discussion
265.

The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length L_A=1.5m and one open end, contains a small sound source that sets up the standing wave with the second lowest resonant frequency of that pipe. Sound from pipe A sets up resonance in pipe B, which has both ends open. The resonance is at the second lowest resonant frequency of pipe B. The length of the pipe B is

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Solution :(a)Only odd HARMONICS can exist in a pipe with one OPEN end. The resonant frequencies are given by f = nv/4L for harmonic numbers n = 1,3,5,....
Because resonant frequency increases with increasing n, the second lowest resonant frequency CORRESPONDS to n = 3, the second lowest CHOICE of n.
(b) Any harmonic can exist in a pipe with two open ends. The resonant frequencies are given by f = n/2L for harmonic numbers n = 1,2,3,...
Because resonant frequency increases with increasing n, the second lowest resonant frequency corresponds to n = 2, the second lowest choice of n. Therefore, we can write
`f= (2v)/(2L_B)`
Solving for `L_B` and substituting known data yield
`L_B = v/f = (1522 m//s)/(761 Hz) = 2.00 m `
Discussion
266.

Rutherford's model suggested that electron revolves round the nucleus in ?

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SOLUTION :CIRCULAR ORBITS
Discussion
267.

A uniform wire of length L, diameter D ad densityP is stretched under a tension T. The correct relation between its fundamental freqency 'f', the length L and the diameter D is :

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`F prop (L)/(D^(2))`
`f prop (1)/(D^(2))`
`f prop (1)/(LD)`
`f prop (1)/(L sqrt(D))`

Solution :V= `(1)/(2l ) sqrt((T)/(m))`
Now mass per unit length m `rho . A= rho. (pi D^(2))/(4 ) `
v `= (1)/(2l) sqrt((T)/((pi D^(2))/(4))) = (1)/(l) sqrt((T)/(pi D^(2))) `
For CONSTANT tension T, v `prop (1)/(l)`
So correct choice is c.
Discussion
268.

A convex lens of power 0.04 dioptre produces an image which is double the size of object placed in front of it. Find the position of the object.

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Solution :DATA supplied,
P = 0.04 dioptre,`"" ` m = 2,U = ?
`F = (1)/(p) = (1)/(0.04) = (100)/(4) = 2.5 ` m
` m = (f)/(u + f)2 =(25)/(u + 25)2u + 50 = 252u = - 25u = - 12.5 m `
Discussion
269.

STATEMENT-1: It is not possible for a charged particle to move in a circular path around a long straight conductor carrying current. STATEMENT-2: The electromagntic force on a moving particle is normal to its plane of rotation.

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Statement-1 is TRUE, Statement-2 is True, Statement-2 is a CORRECT EXPLANATION, for Statement-1.
Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is FALSE.
Statement-1 is a False, Statement-2 is True.

Answer :C
Discussion
270.

A ball of mass m is thrown vertically upward from the ground and reaches a height h before momentarily coming to rest. If g is acceleration due to gravity, the impulse received by the ball due to gravity force during its flight is

Answer»

`SQRT(2 m^(2) gh)`
`sqrt(4 m^(2) gh)`
`sqrt(8 m^(2) gh)`
`4 sqrt(m^(2) gh)`

Answer :C
Discussion
271.

Two shells are fired from a canon successively with speed u each at angles of projection alphaandbeta, respectively. If the time interval between the firing of shells is dt and they collide in mid-air after a time t from the firing of the first shell. Then

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`tcosalpha=(t-dt)COSBETA`
`ALPHAGTBETA`
`(t-dt)cosalpha=dtcosbeta`
`(usinalpha)t-1/2gdt^(2)=(usinbeta)t-dt-1/2g(t-dt)^(2)`

ANSWER :A::B::D
Discussion
272.

A variable frequency alternating voltage of constant magnitude is applied across a capacitor. Which of the following graph shows the variation of current set up in the circuit with frequency v?

Answer»




ANSWER :A
Discussion
273.

An error of 3% is committed in the measurement of volume of a cube. What is the percentage of error committed in the measurement of each of its sides.

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Solution :The volume of a cube `V=L^3`
l=length of each side
so `long(V) = 3 log l`
Differentiating both sides `DeltaV/V=3(Deltal)/l` so, `(Deltal)/l=l//3xx(DeltaV)/V=1//3xx3%=1%`
Discussion
274.

A capacitor of capacitance 10mu F is charged to a potential of 50V with a battery. The battery is disconnected and an additional charge 200 mu C is given to positive plate of the capacitor. The potential difference across the capacitor will be

Answer»

50V
80 V
70 V
60 V

Answer :D
Discussion
275.

Two plane wavespropagate in a homogeneous elasticmedium, one along the x axis and the other along the y axis : xi _(1) = a cos ( omega t - kx), xi _(2) = a cos ( omega t - ky ) .Find the motion patternfo particles in the plane xy if both waves. (a) are tansverse and their oscillation directions coincide, (b) are longitudinal.

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Solution :`(a)` Equation of the resultant wave,
`xi=xi_(1)+xi_(2)=2 a cos k ((y-x)/(2))XOA {omegat-(k(x+y))/(2)}, `
`=a'cos { omegat-(k(x+y))/2}`,where `a^(') = 2 a cos k^(') ((y-x)/2)`
Now, the equation of wave pattern is,
`x+ y=k`, (aConst.)
For sought plots see the answer `-` sheet of the problem book.
For antinodes, i.e. MAXIMUM intensity
`cos((k(y-x))/(2))=+-1= cos n pi`
or, `+-(x-y)=(2npi)/(k) = n lambda`
or `y=x+- nlambda, n=0,1,2,.....`
Hence, the particles of the medium at the POINTS, lying ono the solide straight lines `(y=x+-n lambda)` , oscillate with maximum amplitude.ltbr. For nodes, i.e. minimum intensity,
`cos ((k(y-x))/(2))=0`
or `+- (k(y-x))/( 2)=(2n+1) ( pi)/(2)`
or , `y=x+- ( 2n +1) lambda//2`,
and hence the particles at the points, lying on dotted lines do not oscillate.
`(b)` When the waves are longitudinal,
For sought plots see the answer `-` sheet of the problem book.
`k(y-x)=cos^(-1)((xi_(1))/(a))- cos^(-1)((xi_(2))/( a))`
or, `(xi_(1))/( a)= cos { k(y-x)+cos ^(-1) ((xi_(2))/(a))}`
`=(xi_(2))/( a) cos k(y-x) - SIN k y( y-x) sin(cos ^(-1)((xi_(2))/(a)))`
`=( xi_(2))/(a) cos k ( y-x)- sin k ( y-x) SQRT(1-(xi_(2)^(2))/( a^(2)))...(1)`
from `(1)`,
if `sin k ( y-x) = 0 =cos ( 2n +1) ( pi)/(2)`
`(xi_(1)^(2))/(a)=1-xi_(2)^(2)//a^(2)`, acircle.
Thus the particles, at the points, where `y= x+-(n+- 1//4) lambda` , will oscillate along circles,
In general, all other particles will move along ellipse.
Discussion
276.

Which of the following determine shortest and longest wavelengths in hydrogen atom spectrum ?

Answer»

`(1)/(lamda)=R((1)/(1^(2))-(1)/(oo^(2))),(1)/(lamda)=R((1)/(5^(2))-(1)/(6^(2)))`
`(1)/(lamda)=R((1)/(2^(2))-(1)/(oo^(2))),(1)/(lamda)=R((1)/(5^(2))-(1)/(oo^(2)))`
`(1)/(lamda)=R((1)/(3^(2))-(1)/(4^(2))),(1)/(lamda)=R((1)/(4^(2))-(1)/(5^(2)))`
NONE of these

Solution :`(1)/(lamda)=R((1)/(1^(2))-(1)/(oo^(2))),(1)/(lamda)=R((1)/(5^(2))-(1)/(6^(2)))`
First equation of option (A) shows SERIES limit of Lyman series, which is in ultraviolet region. So `lamda` is minimum, second equation of option (A) shows first line of Pfund series, which is in INFRA RED region and shows maximum wavelength.
Discussion
277.

An equilateral prism has a refractive index equal to 1.56 for the light of wavelength 450 nm. Determine the minimum angle of deviation for light of 450 nm.

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SOLUTION :`42.52^(@)`
Discussion
278.

For a television network, 5 xx 10^(5) channels are granted. If the central frequency of the microwave link is 25 GHz and the alloted bandwidth for each channel is 2 KHz, then how much percentage of the link is used for the network ?

Answer»

`4%`
`10%`
`25%`
`5%`

Solution :Let `x%` of link is used for network. Given, central frequency = 25 GHz
`= 25 XX 10^(9) Hz`
Bandwidth of channels `= x%` of 25 GHz
`= (x)/(100) xx 25 xx 10^(9) = 25x xx 10^(7)`
Number of channel
`= ("Total Bandwidth")/("Bandwidth needed per channel")`
`5XX 10 ^(5) = (25 xx 10 ^(7))/(2 xx 10^(3))`
`10xx 10 ^(8) = 25 x xx 10 ^(7)`
`implies x = 4%`
Discussion
279.

The height of a T.V. tower is 100m. If radius of earth is 6400km, then what is the maximum distance of transmission from it ?

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`100KM`
`64sqrt(10)KM`
`6.4sqrt(10)km`
`8sqrt(20)km.`

ANSWER :D
Discussion
280.

Give the expression for the wavelength of a charged particle in terms of accelerating potential and explain the meaning of the symbols used.

Answer»

SOLUTION :`LAMDA = (h)/( sqrt(2ME V))`
Discussion
281.

Consider an ideal juction diode. Find the value of current flowing through AB is

Answer»

Solution :The barrier POTENTIAL of the diode is neglected as it is an ideal diode. The value of CURRENT flowing through AB can be OBTAINED by using Ohm.s law
`I=V/R = (3-(-7))/(1 xx 10^3)= 10/(10^3) = 10^(-2)A = 10 m A`
Discussion
282.

If 13.6 eV energy is required to 10ise the hydrogen atom, then energy required to remove an electron from n = 2 is

Answer»

`10.2` eV
`0` eV
`3.4` eV
`6.8` eV

Solution :`E_(N) = (136)/(n^(2))eV`
`therefore DELTA E = E_(INFTY - E_(2) = 0 + (13.6)/(2^(2)) = 3.4 eV`
Discussion
283.

A thermos bottle containing coffee is vigorously shaken. Consider coffee as a system: Has work been done on it ?

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SOLUTION :Yes, work has been DONE on COFFEE against VISCOUS FORCES
Discussion
284.

If the bar magnet in exercise 13 is turned around by 180^@ where will the new null points be located ?

Answer»

SOLUTION :On the equatorial line . LET d. be the corresponding distance . Since the two fields are equal to earth.s field
`10^(-7)xx(2M)/d^3=10^(-7)xxm/(d.^3)"":.d.^3=(d^3)/2`
`d.="d"xx2^(-1/3)=14xx2^(-1/3)=11.1cm`
Discussion
285.

Two particles having position vectors vec(r_1) = (3veci+5vecj)m and vec(r_2) =(-5veci +3vecj)m are moving with velocities vec(V_1) =(4veci -4vecj)ms^(-1) and vec(V_2) = (aveci – 3vecj)ms^(-1). If they collide after 2 seconds, the value of 'a' is

Answer»

2
4
6
8

Answer :D
Discussion
286.

Give the ratio of the number of holes and the number of conduction electrons in an intrinsic semiconductor.

Answer»

SOLUTION :1 (i.E., `n_(H) = n_(e) = n_(i)` ).
Discussion
287.

A source of sound gives five beats per second, when sounded with another source of frequency 100 s^(-1). The second harmonic of the source, together with a source of frequency 205 s^(-1) gives five beats per second. What is the frequency of the source ?

Answer»

`105 s^(-1)`
205 `s^(-1)`
95 `s^(-1)`
100 `s^(-1)` .

Solution :The freq. of the SOURCE is `(100 pm 5) ` i.e., 105 or 95 Hz. The second harmonic is 210 or 190 . As is GIVES 5 beats / sec with 205 , THEREFORE, its freq. is 105 and harmonic 210 .
correct choice is (a).
correct choice is (c).
Discussion
288.

A current flowing through a galvanometer is 10(-6)A and produce a deflection of 1 scale devision. If the resistance of moving coil is 400Omega and galvanometer has 30 scale division, the voltage across the coil is :

Answer»

`4xx10^(-5) V`
`4xx10^(-4) V`
`4xx10^(-6) V`
`4xx10^(-3) V`

Answer :B
Discussion
289.

Figure shows a rectangular 20-trun coil of wire of dimensions 10 cm by 5.0 cm. It carries a current of 0.10 A and is hinged along one long side. It is mounted in the xy. Plane, at angle theta=30^(@) to the direction of a uniform magnetic field of magnitude 0.50 T. In unit-vector notation, what is the torque acting on the coil about the hinge line?

Answer»


ANSWER :`(-4.3xx10^(-3)N.m)HATJ`
Discussion
290.

Which of the following is an example of point to point communication mode ?

Answer»

RADIO
TELEVISION
telephony
All of these

Answer :C
Discussion
291.

A charged particle enters in a magnetic field B with its initial velocity making an angle of 45° with B. The path of the particle will be

Answer»

an ellipse
straight line
a circle
a heilical

Answer :D
Discussion
292.

An electric lamp designed for operation on 110 V AC is connected to a220 V AC supply, through a choke coil of inductance 2H, for proper operation. The angular frequency of the AC is 100sqrt(10) rad //s. If a capacitor is to be used in the place of the choke coil, its capacitance must be

Answer»

`1muF`
`2 MUF`
`5 muF`
`10 muF`

ANSWER :C
Discussion
293.

Three point-mases m_(1),m_(2)andm_(3) are located at the vertices of an equilateral triangle of side a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through m_(1)?

Answer»

`(m_(1)+m_(2))(a^(2))/(4)`
`(m_(1)+m_(2)+m_(3))(a^(2))/(4)`
`(m_(2)+m_(3))(a^(2))/(4)`
`(m_(1)+m_(3))(a^(2))/(4)`

Solution :`I_(AD)=m_(1)xx0^(2)+m_(2)xxBD^(2)+m_(3)xxCD^(2)`
`=0+m_(2)((a)/(2))^(2)+m_(3)((a)/(2))^(2)`
`I_(AD)=(m_(2)+m_(3))(a^(2))/(4)`
Discussion
294.

At a place horizontal component of the earths magnetic field is B and angle of dip at the place is 60°. What is the value of horizontal component of the earths magnetic field. (i) at Equator, (ii) at a place where dip angle is 30°

Answer»

SOLUTION :`2B, Bsqrt3`.
Discussion
295.

For a series L-C-R circuit the power loss at resonance is :

Answer»

`I^2omegaL`
`I_omegaC`
`I^2R`
`(E^2)/SQRT(R^2+(omegaL-1/(OMEGAC))^2`

ANSWER :C
Discussion
296.

Bodies in non-inertial frames disobey "___________".

Answer»

NEWTON's law
Theory of RELATIVITY
GENERAL Relativity
Einstein's theories

Answer :A
Discussion
297.

Match the following

Answer»

SOLUTION :
Discussion
298.

{:((i)"Quantisation of charges",(a) int_(infty)^(P)- vec(E).d vec(r) ),((ii)"Electric field",(b)"ne"),((iii)"Electric dipole moment",(c) (F)/(q)),("(iv) Electric potential",(d)2qa):}

Answer»


ANSWER :(i)`RARR` (d); (II)`rarr`(a); (iii) `rarr` (B); (iv) `rarr`(c)
Discussion
299.

What is the value of magnetic susceptibility for diamagnetic materials ?

Answer»

SOLUTION :SMALL and POSITIVE
Discussion
300.

Electromagnetic waves can be propagated throught conductor. In an electromagnatic wave, what is the phase difference between electric and magnetic field variation ?

Answer»

SOLUTION :They are in PHASE
Discussion