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43401.

According to Maxwell a changing electric field gives rise to

Answer»

an cmf
electric current
magnetic field
pressure variation

Answer :C
Discussion
43402.

A charge Q is situated at the centre of a cube. The electric flux through one of the faces of the cube is

Answer»

`Q//epsilon_(0)`
`Q//2epsilpon_(0)`
`Q//4epsilon_(0)`
`Q//6epsilon_(0)`

ANSWER :D
Discussion
43403.

The writer decided to learn to swim when he was about

Answer»

TEN or ELEVEN YEARS old
fifteen or SIXTEEN years old
twenty years old
eighteen years old

Answer :A
Discussion
43404.

Give the expression for maximum induced emf in an ac generator.

Answer»

Solution :Let the call be rotating with angular velocityat any INSTANT .t. when the normal to the plane of the coil makes an angle `THETA`with the magnetic field. Hence magnetic flux,
`phi = NBA cos omega t`,There for induced EMF `(epsilon)`
`epsilon = -(d phi)/(d t)`
`rArr epsilon = NBA omega SIN omega t`
Induced emf will be maximum when `omega t = 90^(@)`
Hence, `epsilon_(max) = NBA omega`
Direction of induced emf can be determined using Flemming.s right hand rule.
[ALTERNATIVELY :Statement of the above rule.]
Discussion
43405.

The refractive index of the material of a prism is sqrt2 and the angle of the prism is 30^@. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is

Answer»

zero
`60^@`
`30^@`
`45^@`

SOLUTION :
`A=30^@,r_1=30^@,r_2=0^@`
`therefore A=r_1+r_2`
`30^@=r_1+r_2`
`therefore r_1=30^@`
From Snell.s law,
`n_1sini=n_2sinr_1`
(1)SINI=`sqrt2sin30^@`
`therefore sini=sqrt2xx1/2`
`therefore sini =1/2`
`therefore i=45^@`
Discussion
43406.

If alpha -particle (""_(2)He^4) is revolving in a circular orbit of radius 3.14 A with speed of 8 xx 10^6m/s. Then the equivalent current is

Answer»

`6. 4 XX 10^(-12) A `
`12.8 xx 10^(-4)A`
` 12 .810^(-16 )A`
`6.4 xx 10^(-9)A `

ANSWER :B
Discussion
43407.

What is magnetic susceptibility ?

Answer»

Solution :MAGNETIC SUSCEPTIBILITY is defined as the ratio of the intensity of magnetisation `(vecM)`induced in the MATERIAL due to the MAGNETISING field `(vecH)`
`X_m = abs((vecM)/(vecH))`
Discussion
43408.

A spring can be used to determine the mass m ofan object in two ways: (i) by measuring the extension in the spring due to the object, and (ii) by measuring the oscillation period for the given mass. Which of these methods can be used in a space-station orbiting Earth?

Answer»

Both
Only the extension METHOD
Only the oscillation method
Neither

Answer :C
Discussion
43409.

Is Ohm's law a fundamental law?

Answer»

Solution :Ohm.s law is not a fundamental law. Ohm.s law does not HOLD GOOD in case of gases, crystal rectifier, thermionic valves, TRANSISTOR etc., and it is applicable only at CONSTANT temperature.
Discussion
43410.

An ac emf is given by e= 220 sin 314t volt. Find the frequency, e= E_0 sin omegat.

Answer»

Solution :e=220 sin 314 volt.
`RARR``OMEGA`= 314 rad/sec.
`V = 314/(2pi) = 157/(PI) =50 Hz`
Discussion
43411.

In order to increase the sensitivity of a moving coil galvanometer one should decrease

Answer»

STRENGTH of the magnet
torsional CONSTANT of suspension
the NUMBER of TUMS of coil
area of the coil.

Answer :B
Discussion
43412.

Is it possible that a body be in accelerated motion under a force acting of the body, yet, no work is being done by the force ? Give example.

Answer»

SOLUTION :Yes , it is possible when a force is perpendicular to the direction of motion. For example, the MOON REVOLVES AROUND the earth under centripetal force of attraction of the earth but the earth does no work on the moon.
Discussion
43413.

(a) What is amplitude modulation? Draw a diagram showing an amplitude modulated wave obtained by modulation of a carrier sinusoidal wave on a modulating signal. (b) Define the terms (i) modulation index, and (ii) side bands. Mention the significance of side bands.

Answer»

SOLUTION :(b) (i) Modulation Index :
(ii) SIDE Bands : Side band is the portion of a modulated carrier wave that is either above or below the BASE band signal. `(w_(1)-w_(m))` is the lower side band `(w_(1)+w_(m))` is called the upper side band. Both side bands are used to carry a MESSAGE.
Discussion
43414.

What we call to the velocity of electron which is very small of the order of few millometer per second ?

Answer»

SOLUTION :DRIFT VELOCITY
Discussion
43415.

The difference in the speed of light in water and glass of refractive indices 1.33 and 1.5 respectively is :

Answer»

`2.5xx10^7 m//s`
`2.5xx10^5 m//s`
`3.5xx10^5 m//s`
`3.5xx10^5 m//s`

ANSWER :C
Discussion
43416.

Energy of a capacitor of capacitance C, when subjected to a pontential V, is given by

Answer»

`(1)/(2) CV^(2)`
`(1)/(2) C^(2)`V
`(1)/(2)`CV
`(1)/(2) (C)/(V)`

ANSWER :A::B::C
Discussion
43417.

Four positive point charges( +q) are kept at the four corners of a square of a side 'l'. The net electric field at the mid point of any one side of the square is, (take 1/(4 pi epsilon_0) = K)

Answer»

`(4Kq)/(1^2)`
`(16 KQ)/(5sqrt(5)I^2)` 
`(8 Kq)/(SQRT(5)1^2)`
`(Kq)/(1^2)`

ANSWER :B
Discussion
43418.

The dimensional formula of mobility is ... .

Answer»

`M^(-1) L^(1)T^(2) A^(1)`
`M^(1) L^(0) T^(-2) A^(-1)`
`M^(1)L^(-1) T^(-2) A^(-1)`
`M^(-1) L^(0) T^(2) A^(1)`

Solution :`M^(-1) L^(0) T^(2) A^(1)`
Mobility `mu = (v_(d))/(E)`
`[ mu] = ([V_(d)])/([E]) = ([v_(d)])/([N]) XX [C]`
` = ([M^(0) L^(1) T^(-1)] xx [A^(1) T^(1)])/([M^(1) L^(1) T^(-2) ] ) = [M^(1) L^(0) T^(2) A^(1)] `
Discussion
43419.

In Fig frame S' has velocity 0.90c relativeto frame S. An observer in frame S' measures two events as occurring at the following spacetime coordinates: event Yellowat (5.0m, 20ns) and event Green at (-2.0m, 4.5 ns). An observer in frame S wants to find the temporal separationDeltat_(GY)=t_(G)-t_(Y) between the events. (a) Which equation in Table 36-2 should be used? (b) Should +0.90c or -0.90c be substituted for v in the parentheses on the equation's right side and in the Lorentz factor gamma? What value should be substituted into the (c ) first and (d) second term in the parentheses?

Answer»


ANSWER :(B) +0.90c; (C ) 25NS; (d) -7.0m
Discussion
43420.

An engine approaches a hill with a constant speed. When it is at a distance of 0.9 km it blows a whistle, whose echo is heard by the driver after 5 sec. If the speed ofsound in air is 330 m/s, calculate the speed of the engine.

Answer»

Solution :If the speed ofthe ENGINE is V, the distance travelled by the engine in 5 SEC, will be 5V. And hence, the distance travelled by SOUND in reaching the HILL and coming back to the moving driver = 900 + (900 - 5V) =(1800 - 5V).
So time interval between the original sound and its echo
`(1800-5V)/330 = t = 5s` (given)
The above equation on solving gives V = 30 m/s.
Discussion
43421.

If two electrons are forced to come closer to each other, the P.E. of the system of 2 electrons will:

Answer»

Becomes zero
INCREASES
Decreases
Becomes `alpha`

SOLUTION :P.E increases as WORK is done against force of REPULSION.
Discussion
43422.

A very long, strainght, thin wire carries -3.60 nCm^(-1) of fixed negative charge. The wire is to be surrounded by a uniform cylinder of positive charge, radius 1.50 cm, coaxial with the wire. The volume charge density rho of the cylinder is to be selected is zero. Calculate the required positive charge density rho (in muCm^(-3)).

Answer»

6
7
5
3

Solution :We don't really need ot write an interal, we juset need the CHARGE per unit LENGTH in the cyliner to be equal to ZERO. This means that the positive charge in cylinder must be `+3.60 nC m^(-1).` This positive charge is uniformaly distributned in a circle of radius `R=1.50` cm, so
`rho =(3.60nC m^(-1))/(piR^(2))=(3.60nCm^(-1))/(pi(0.015m)^(2))=5 MUC m^(-3).`
Discussion
43423.

Which of the following devices is sometimes called an electric eye?

Answer»

LED
Photocell
Intergrated chip
Solar cell

Answer :B
Discussion
43424.

Gauss's law is true for any closed surface,no matter what its shape or size ''Say the following statements are true or false.In uniform electric field we know that the dipole experience no net force but experience a torque having a relation with p and E is given by

Answer»

SOLUTION :`barT=barpxxbarE` or `pEsintheta`
Discussion
43425.

What is the de Broglie wavelength of a. a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, b. a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and c. a dust particle of mass 1.0xx10^(-9) kg drifting with a speed of 2.2 m/s ?

Answer»

SOLUTION :`m=0.04kg, v=1xx10^(3)ms^(-1)`
a.`LAMBDA=(h)/(mv)=(6.6xx10^(-34))/(0.04xx10^(3))=1.65xx10^(-35)m`
b. `lambda=(h)/(mv)=(6.6xx10^(-34))/(0.06xx1)=1.1xx10^(-32)m`
c. `lambda=(h)/(mv)=(6.6xx10^(-34))/(1xx10^(-9)xx2.2)=3xx10^(-23)m `
Discussion
43426.

A charged particle of charge 4 mC enters a uniform magnetic field of induction vecB = 3bari+6barj+6bark tesla with a velocity barv = 4bari-xbarj+y bat k. If the particles continues to move undeviated, then the magnitude of velocity of the particle is

Answer»

10 m/s
15 m/s
12 m/s
8 m/s

ANSWER :C
Discussion
43427.

One end of each of two identicalsprings, each of force-constant 0.5N are attached on the opposite sides of a wooden block of mass 0.01 Kg. The other ends of the springs are connected to separaterigid supports such that the springs are unstretchedan are collinear in a horizontal plane. To the wooden piece is fixed a pointerwhich touches a vertically moving plane paper. The wooden piece, kept on a smooth horizontal table is now displaced by 0.02 m along the line of springs and released. If the speed of paper is 0.1 m/s. The distance between two consecutive maxima on this path is 6.28 xx 10^(-y) mtrs. Find y

Answer»


ANSWER :2
Discussion
43428.

At what temperature will the average K.E. of a gas molecule be twice its value at N.T.P. ?

Answer»

`546^@C`
`213^@C`
`546^@K`
`283^@C`

Answer :B
Discussion
43429.

Mark out the correct options.

Answer»

Diamagnetism occurs an all materials.
Diamagnetism result from the partial alignment of permanent magnetic moment.
The magnetising FIELD intensity H is always zero in free space.
The magnetic field of induced magnetic moment is opposite to the applied field.

Solution :When a material is placed in magnetic field, dipole moments are induced in the atoms by the applied magnetic field. Since the direction of magnetic field due to induced dipole moment is opposite to the applied magnetic field. Therefore, resultant magnetic field is smaller than the applied magnetic field. This process is called diamagnetism. As this process takes place for all the material, therefore all the material EXHIBIT diamagnetism. Hence, option (a) and (d) are CORRECT.
Diamagnetic material do not have permanent magnetic moment on their own. When they are placed in magnetic field, dipole moments are induced by the applied magnetic field. THUS, there is no net alignment of permanent magnetic moment so these materials do not have any permanent magnetic moment of their own. Hence, option (b) is incorrect.
Magnetic field intensity is not vero in free space. Hence, option (c) is incorrect.
Discussion
43430.

Refraction index of medium-2 with respect to medium-1 is ...... .

Answer»

`n_(12)=(n_1)/(n_2)`
`n_(21)=(n_2)/(n_1)`
`n_(12)=(n_2)/(n_1)`
`n_(21)=(n_1)/(n_2)`

SOLUTION :`n_(21)=(n_2)/(n_1)`
Discussion
43431.

A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of the conductor. When the arm MN of length of 20 cm is moved towards left with a velocity of 10 ms^(-1), calculate the emf induced in the arm. Given the resistance of the arm to be 5Omega (assuming that other arms are of negligible resistance) find the value of the current in the arm.

Answer»

Solution :Here B = 0.5T, length of moving conductor l =20 cm = 0.2 m and speed `v = 10 m s^(-1)`
`therefore` INDUCED emf `|varepsilon| = Blv = 0.5 xx 0.2 xx 10 = 1.0` VOLT
As resistance of RECTANGULAR conductor `R = 5Omega`
`therefore` Current `I = varepsilon/R=(1.0)/5 = 0.2A`
Discussion
43432.

A Huygens eye-piece is made by arranging two planoconvex lenses of focal lengths 0.03m and 0.09m in the usual manner. Light falls directly on the field lens of the eye-piece. Calculate the position of the final image formed by the eye-piece and trace the path of rays showing the formation of images.

Answer»


ANSWER :`1.5` CM BEHIND the eye-lens
Discussion
43433.

A body is released from the top of a smooth inclined plane of inclination q. It reaches the bottom with velocity V. If the angle of inclination is doubled for the same lengh of the plane, what will be the velocity of the body on reaching the ground:

Answer»

`V`
`2V`
`(2 cos THETA)^(1//2) v`
`(2 sin theta)^(1//2) v`

ANSWER :C
Discussion
43434.

Nuclei having the same ................ but different ................ are called isotopes.

Answer»

SOLUTION :ATOMIC NUMBER (or number of PROTONS), mass number (or number of neutrons)
Discussion
43435.

The moving coil of a galvanometer made of 200 turns of thin wire is suspended from an elastic thread. The area of the coil is 1 cm^2, it coincides in direction with the lines of induction of a magnetic field with induction of 15 mT. When the current of 5.0 A is passed through the coil it turns through 15^@. By what angle will the coil turn with a current of 7.5 muA? What is the torsion modulus of the thread?

Answer»


SOLUTION :A torque `M=wiSB` ACTS on the moving coil balanced by a torque M `f_(alpha)` DUE to the elasticity of the twisted thread, where is the torsion inodulus. Equating, we obtain `j alpha=W//SB` from which we see that, other conditions being equal, the coils angle of rotation is PROPORTIONAL to the CURRENT.
Discussion
43436.

In the Geiger-Marsden scattering experimenl of the number of scattered particles detected are maximum and minimum at the scattering angles respectively at

Answer»

`0^(@) and 180^(@)`
`180^(@) and 0^(@)`
`90^(@) and 180^(@)`
`45^(@) and 90^(@)`

ANSWER :A
Discussion
43437.

Find the refractive index of the material, if a prims having an angle A=60^(@) which produces a minimum deviation of 30^(@).

Answer»

`SQRT(3)`
`sqrt(2)`
`sqrt(5)`
`1/(sqrt(2))`

Solution :(b)
Discussion
43438.

The rms voltage of the generator is V_0 The frequency of the ac generator is verynear zero. What is the rms current through the circuit?

Answer»

Solution :As we just now discussed, at small frequencies, the circuit is like a dc circuit. So, the capacitor can be replaced by a open circuit and inductor by a short circuit. Figure 31-22 shows the circuits as they would appear according to these CHANGES.

It is obvious from the circuit that it is not a circuit of series elements , nor elements in parallel.

Calculation : The rms current is given by
`I_(rms)=V_(rms)/Z`.
However, the impedance Z . because the circuit is not series RCL circuit. The circuit behaves as if it contained only two identical resistors in series, with a total impedance of Z=2R. Thus circuit has
`I_(rms)=V_(rms)/(2R)`.
Learn: If the source had a very HIGH frequency (much HIGHER than the RESONANCE frequency), the capacitor would be treated as short-circuited and inductor as open-circuited. Figure 31-23 shows the circuits as they would appear according to these changes. So, at this frequency, the impedance of the circuit would be R/2. Hence
`I_(rms)=(2V_(rms))/R`.
Discussion
43439.

Making use of the Hund rules, find the numbers of electrons in the only partially filled subshell of the atom whose basic term is (a) .^(3)F_(2)(b) .^(2)P_(3//2),(c ) .^(6)S_(5//2).

Answer»

Solution :(a) `.^(3)F_(2)` The MAXIMUM value of spin is `S=1` here. This means there are `2` electrons. `L= 3` so `p` electrons are ruled out. This is simplest plssiblility is `d` electrons. This is the CORRECT choice for if we were considering `f` electrons, the maximum value of `L` allowed by Pauli priciple will be `L=5` (maximum value ofthe magnitude of megnetic quantum number will be `3+2=5`).
Thus the atom has two `d` electrons in the unfilled shell.
(b) `.^(2)P_(3//2)` Here `L=1,s=(1)/(2)` and `J=(3)/(2)`
Since `J=L+S`, Hund's rule implies the shell is more than half FULL. THSI means one electron less than a full shell. On the basis of hole picture it is easy to see that we have `p` electrons. Thus the atom has `5p` electrons.
(c )`.^(6)S_(5//2)` Here `S=(5)/(2), L=0` we either have five electrons of five holes. The angular part is antisymmertic. For five `d` electrons, the maximum value of the quantum number cosistent with Pauli exclution principle is `(2+1+0-1-2)=0` so `L=0` for `f` or `g` electrions `L gt 0` whether the shell has five electrohns or five holes. Thus the atom has five `d` electrons.
Discussion
43440.

The susceptibility (X_(m)) of bismuth is ................. .

Answer»


ANSWER :0.00017
Discussion
43441.

In the figure shown i=10 e^(-41) A.Find B_(L) "and" V_(ab)

Answer»


ANSWER :`-80 E^(-4T), -40 e^(-4t)`
Discussion
43442.

Define 'electrostatic potential".

Answer»

Solution :The electric potential at a point P is equal to the work DONE by an external force to bring a unit positive charge with constant VELOCITY from INFINITY to the point in the region of the external electric field `vecĒ`
Discussion
43443.

A ray of light is incident on the upper surface of a glass plate of thickness t (mu = "refractive index of glass"). If the angle of incidence I is very small, then lateral displacement of the emergent ray will be

Answer»

`(timu)/(MU + 1)`
`(TI(mu-1))/(mu)`
`(timu)/(mu - 1)`
`(ti(mu+1))/(mu)`

ANSWER :B
Discussion
43444.

A small block is projected with a speed v_( 0) on a horizontal track placed on a sufficientlyrough surface which turns into a semi circle ( vertical) of radius R. Find the min value of V_(o) so that the body will hit the point A after leaving the track at its highest point. The arrangement is shown in the figure, given that the straight part is rough & the curved path is smooth. The coefficient of friction is mu.

Answer»



Solution :Let the block escape the point at C with a velocity V HORIZONTALLY. Since it hits the initial spot A after falling through a height 2R we can write
`(2R)=1/2g t^(2)`, whrer `t=` time of its fall.
`rArr t=2 sqrt(R//g)`
`:.` The distance `AB=2vsqrt(R//g)`
`rArr d=2v_(x) sqrt(R//g)`...(a)
Work energy theorem is applied to the motion of the body from A to B leads `Delta KE=W_(f)`
`rArr 1/2 mv_(o)^(2)-1/2 mv_(1)^(2)=mu mgd`
`rArr v_(o)=sqrt(v_(1)^(2)+2mu gd)` ...(b)
Energyconservation between B & C yields
`1/2 mv_(1)^(2)-1/2 mv^(2)=mg(2R)`
`rArr v_(1)=sqrt(v^(2)+4gR)` ...(c)
When the block escapes C, its minimum speed v can be given as `(mv^(2))/(R)=mg`
( `:.` the normal CONTACT force `=0`)
`rArr v=sqrt(gR)`...(d)
(NOTE that if the particle has speed LESS than `sqrt(gR)` then it will not reach to the highest point of the curvature.
By using (c) and(d) we obtain
`v_(1)=sqrt(5gR)` ...(e)
using (a) and (d) we obtain
`d=(sqrt(gR))2(sqrt(R/g))=2R`...(f)
`v_(o)=sqrt(5gR+2 mu g (2R))=sqrt((5+4 mu )gR)`.
Discussion
43445.

How can you distinguish between hard boiled egg and a raw egg by spinning them on table top ?

Answer»

Solution :When the RAW and hard boiled eggs are SPUN together the raw egg will have a greater moment of INERTIA as the FLUID in it will CONCENTRATE along it.s surface.
Discussion
43446.

The internal energy U is a unique function of any state because change in U

Answer»

Does not DEPEND upon PATH
Corresponds to an adiabatic process
DEPENDS upon path
Corresponds to an ISOTHERMAL process

Answer :A
Discussion
43447.

The basic instrument employed to detect current is

Answer»

GALVANOMETER
WATTMETER
AMMETER
Voltmeter

Answer :C
Discussion
43448.

Amplitude of a wave .

Answer»

SOLUTION :AMPLITUDE of a wave is DEFINED as the magnitude of the maximum diplacement of a particle of the medium from its equilibrium position .
Discussion
43449.

When monochromatic, light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why?

Answer»

Solution :Reflaction and refraction arise through INTERACTION of incident light with the ATOMIC constituents of matter. Atoms may be viewed as OSCILLATORS, which take up the frequency of the external agency (light) and undergo FORCED oscillations. The frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.
Discussion
43450.

Sphere 1 with radius R_1 has positive charge q. Sphere 2 with radius 2.00 R_1 is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to retain only negligible charge, (a) is potential V_(1) of sphere 1 greater than, less than, or equal to potential V_(2) of sphere 2? What fraction of q ends up on (b) sphere 1 and ( c) sphere 2? ( d) What is the ratio sigma_(1)// sigma_(2) of the surface charge densities of the spheres ?

Answer»

SOLUTION :(a) `EQUAL`, (B) `0.333`,( C) `0.667`, (d) `2.00`
Discussion