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5301.

State Huygens' principle and prove the laws of reflection on the basis of wave theory.

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Solution :Laws of reflection
(i) Ist Law. It STATES that angle of incidence is equal to angle of reflection.
(ii) 2nd Law. It states that the incident ray, the reflected ray and the normal, at the point of incidence all lie in the same plane.
First law of reflection. Let XY be a plane reflecting surface and AB be a plane wavefront incident on the surface as shown in the figure.
According to Huygens. principle, every point on wavefront AB is a SOURCE of secondary WAVELETS and the time during which wavelet from B reaches at C, the reflected wavelet from A would arrive at D.

i.e. `t=(BC)/(v)=(AD)/(v)`
or BC = AD, ...(i)
where v is the velocity of light in the medium.
In RT `/_ DeltaABC` ,
`sin i=(BC)/(AC)`
or `BC = AC sin i` ...(ii)
In rt `/_ Delta ADC, sin r=(AD)/(AC)`
or AD = AC sin r ...(iii)
Putting Eqs, (ii) and (iii) in (i), we get
AC sin i =AC sin r
or sin i = sin r
or i = r
i.e: Angle of incidence = angle of reflection
This is first law of reflection.
Second law of reflection. From the figure, we find that, the incident ray, the reflected ray and the normal all lie on the same plane XY. This proves second law of reflection.
5302.

Charged of +6 nC and -15n C,when separated by a certain distance , attract each other by a force F. If an additional charges of + 9 muCis given to each of the two charges and separation between them is kept unchanged, the force between them will still be an attractive force of magnitude F.

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SOLUTION :True - intial FORCE F = `(1)/(4pi in _0)([+6XX 10^(-9)][-15xx10^(-9)])/(r^(2)) `
and final force F. ` =(1)/(4pi in _0) = ([+15xx10^(-9)][-6xx10^(-9)])/(r^(2)) =F.`
5303.

The light rays which suffers a single TIR is

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SOLUTION :PRIMARY RAINBOW.
5304.

The circuit shows a resistance, R=0.01Omega and inductance L=3mH connected to a conducting rod PQ of length l=wm which can slide on a perfectly conducting circular arc of radius l with its center at P. Assume that friction and gravity are absent and a constant uniform magnetic field b=0.1T exists as shown in the figure. At t=0, the circuit is seitched on a simultaneously an external torque is applied on the rod so that it rotates about P with a constant angular velocity omega2 rad//sec. Find the magnitude of this torque (inN-m) at t=(0.3In 2) second.

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ANSWER :4
5305.

What one gets from velocity time graph?

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Solution :AREA under the GRAPH gives the DISTANCE COVERED.
5306.

Determine the electric field strength vector if the potential of this field depends on (x, y, z) co-ordinates as V=A(x^(2)-y^(2)).

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SOLUTION :`E_(X) =(delV)/(delx) = -2Ax and E_(y) = -(delV)/(dely)=2Ay`
`barE=E_(x) HATI +E_(y) hatj = 2A (-x hati +yhatj)`
5307.

Assertion The frequency of the electromagnetic wave naturally equals equals the frequency of oscillations of the charge. Reason The energy associated with the propagating wave comes at the expense of the energy of the source.

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If both ASSERTION and Reason are true and Reason is the correct explanation of Assertion.
If both Assertion and Reason are true but Reason is not correct explanation of Assertion.
If Assertion is true but Reason is false.
If Assertion is false but Reason is true.

Solution :The OSCILLATING ELECTRIC and magnetic FIELDS, thus REGENERATE each othe, as the wave propagates through the space.
The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. The energy associated with the propagating wave comesat the expense of the energy of the source the accelerated charge.
5308.

Three concentric conducting spherical shells of radii R, R2 and 3R carry charges Q,-2Q and 3Q, respectively compute the electric field at r = (5)/(2) R

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`(-Q)/(3piepsilon_(0)R^(2))hat(r)`
`(-Q)/(25piepsilon_(0)R^(2))hat(r)`
`(-Q)/(5piepsilon_(0)R^(2))hat(r)`
`(-Q)/(piepsilon_(0)R^(2))hat(r)`

ANSWER :B
5309.

The electric chargesare distributed in a small volume. The flux of the electric field through a surface of radius 10 cm surrounding the total charge is 20 Vm. The flux over a concentric sphere of radius 20 cm will be

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`20 Vm`
` 25 V m `
` 40 V m `
` 200 V m `

SOLUTION :Since charge enclosed REMAINS the same, hence as per Gauss. law total electric flux remains uncharged at 20 V m inspite of the INCREASE in surface area.
5310.

Two capacitors of capacitance 600 pF and 900 pF are connected in series across a 200 V supply .Calculate the effective capacitance of the combination.

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Solution :Given: `C_1`=600 PF , `C_2` =900 pF
V=200V
`1/C_S=1/C_1+1/C_2=(C_1+C_2)/(C_1+C_2)`
`C_S=(C_1C_2)/(C_1+C_2)=(600xx900)/(600+900)=(54xx10^4)/(15xx10^2)=5400/15`=360 pF
5311.

The gravitational force exerted by the sun on the moon is greater than that exerted by the earth on the moon. Why then does not the moon escape from the earth.

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SOLUTION :For the MOON `V_e=sqrt(2) V_0V_e-V_o=sqrt(2)V_0-V_0=V_0(sqrt2-1)=V_0(1.414-1)=0.414V_0` To escape from the gravitational FIELD of the EARTH, the moon should faster by41.4%
5312.

Two boys are standing at the ends A and B of a ground where AB= a. The boy at B starts running in a direction perpendicular to AB with velocity v. The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is

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`(a)/(SQRT(NU^(2)+nu_(1)^(2)))`
`(a)/(nu+nu_(1))`
`(a)/(nu-nu_(1))`
`sqrt(a^(2)/(nu^(2)-nu_(1)^(2)))`

SOLUTION :
`t=(a)/(nu)=(a)/(sqrt(nu^(2)-nu_(1)^(2)))`
5313.

Two capacitors of capacitance 600 pF and 900 pF are connected in series across a 200 V supply .Calculatethe pd across each capacitor .

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Solution :P.d across `C_1` is `V_1=Q/C_1=(720cancel(00)xxcancel(10^(-12)))/(CANCEL(600)xxcancel(10^(-12)))`
`V_1=120`V
P.d across `C_2` is `V_2=Q/C_2=(720cancel(00)xxcancel(10^(-12)))/(cancel(900)xxcancel(10^(-12)))`
`V_2` = 80 V
5314.

Two capacitors of capacitance 600 pF and 900 pF are connected in series across a 200 V supply .Calculatethe total charge stored in the system.

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SOLUTION :TOTAL CHARGE `Q=VC_S=200xx360xx10^(-12)`
`Q=72000xx10^(-12)C`
5315.

The difference in mass of nucleus and the combined mass of electrons is called______.

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SOLUTION :MASS DEFECT
5316.

In an accelerator experiment on high energy collisions of electrons with positron, a certain event is interprected as annihilation of an electron-positron pair of total energy 10.2 BeV into two gamma-rays of equal energy. What is the wavelength associated with each gamma-ray ? (1 BeV=10^(9)eV)

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Solution :ENERGY of electron - positron pair `=10.2 BeV=10.2 xx 10^(9)eV`
`therefore` Energy of ONE `gamma`-ray PHOTON
`E=(10.2xx10^(9))/(2)eV=5.1xx10^(9)xx1.6xx10^(-19)J=5.1xx1.6xx10^(-10)J`
But `E=h upsilon="h"(c )/(lambda)`
`lambda=(hc)/(E )=(6.63xx10^(-34)xx3xx10^(8))/(5.1xx1.6xx10^(-10))=2.44 xx 10^(-14)m`
5317.

In Fig. 6-52, two blocks, in contact, slide down an inclined plane AC of inclination 30^(@) . The coefficient of kinetic friction between the 2.0 kg block and the incline is u_(1)=0.20 and that between the 4.0 kg block and the incline is mu_(2)=0.30. Find the magnitude of the acceleration.

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SOLUTION :`a=2.6 m//s^(2)`
5318.

Find the distance of object from a concave mirror of focal length 10 cm so that image size is four time the size of the object.

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Solution :Concave mirror can form real as well as VIRTUAL IMAGE. Here nature of image is not given in the question. So, we will consider two POSSIBLE cases.
CaseI (When image is real) Real image is formed on the same side of the object, i.e.u,v and f all are negative. So let,
u=-x
then, v=-4X as `abs(v/u)=abs(m)=4` and f=-10 cm
Substituting in, `1/v+1/u=1/f`
We have `1/(-4x)-1/x=1/(-10)OR5/(4x)=1/10`
x=12.5 cm
Note `absxgtabsf` and we know that in case of a concave mirror, image is real when object lies beyond F
case II (When image is virtual) In case of a mirror image is virtual when it is formed behined the mirror, i.e., u and f are negative while v is positive. So let,
u=-y
then, v=+4y and f=-10cm
Substituting in, `1/v+1/u=1/f`
we have, `1/(4y)-1/y=1/(-10)`
or `3/(4y)=1/10ory=7.5 cm` Note Here `absyltabsf` as we know that image is virtual when the object lies between F and P.
5319.

The modulation index in Amplitude modulating, value is

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`MU LT 1`
`mu gt1`
`mu GE1`
`mule1`

ANSWER :D
5320.

Assertion:There occurs a chain reaction when uranium is bombardedwith slow neutrons. Reason:When uranium is bombarded with slow neutrons more neutrons are produced

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If both assertion and REASON are TRUE and reason is the correct explanation of assertion .
If both assertion and reason are true but reason is notthe correct explanation of assertion .
If assertion is true but reason is false .
If both assertion and reason are false

Solution :When uranium is bombarded by SLOW neutrons the reaction is represented as `._92^235U + ._0^1n to ._56^144Ba + ._36^89Kr + 3_0^1n`+Energy
As more neutrons are produced , the reason is correct.
There are additional neutrons strike uranium NUCLEI to PRODUCE even more neutrons. Thus a chain reaction is established.
5321.

The power of a transmitter 19 kW. The power of the Carrier wave is , if the amplitude of modulated wave is 10 V and that of Carrier is 30V,

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18 KW
1 kW
4.75 kW
9.5 kW

Answer :A
5322.

An explosion of atomic bomb release an energy of 7.6 xx 10^13J. If 200 Mev energy is released on fission of one .^235U atom calculate (i) the number of uranium atoms undergoing fission, (ii) the mass of uranium used in the bomb.

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`4.375xx10^24`, 926.66 G
`2.375xx10^24` , 826.66 g
`3.375xx10^24` , 926.66 g
`2.375xx10^24` , 926.66 g

Answer :D
5323.

For which position of the object, magnification of convex lens is -1. (minus one)?

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SOLUTION :For position of an OBJECT at 2F on the PRINCIPAL AXIS, the magnification in a convex lens will be -1.
5324.

Two parallel plate capacitors are arranged perpendicular to the common axis. The separation d between the capacitors is much larger than the separation / between their plates and than their size. The capacitors are charged to q_1 and q_2respectively (figure). Find the force of interaction between the capacitors.

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`1/2(q_1q_2)/(piepsilon_0d^4)l^2`
`2/3(q_1q_2)/(piepsilon_0d^4)l^2`
`3/2(q_1q_2)/(piepsilon_0d^4)l^2`
`1/3(q_1q_2)/(piepsilon_0d^4)l^2`

ANSWER :C
5325.

A piece of copper and another of germanium are cooled to 77K , the resistance of :

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each of them DECREASES
each of them increases
of Cu decreases and GE increases
of Cu increases and Ge increases

ANSWER :C
5326.

L = 8.1 mH, C = 12.5 muF and R = 100 Omega are connected in series with A.C. source of 230 V and frequency 500 Hz. Calculate voltage across two ends of resistance.

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Solution :Impedance for a L-C-R series circuit,
`|Z|=sqrt(R^2+(X_L-X_C))^2`
where `X_L=omegaL=2pifL`
`=2xx3.14xx500xx8.1xx10^(-3)=25.4 Omega`
`X_C=1/(OMEGAC)=1/(2pifC)`
`=1/(2xx3.14xx500xx12.5xx10^(-6))=25.4Omega`
`therefore |Z|=sqrt((10)^2+(25.4-25.4)^2)=10Omega`
Now `I_"rms"=V_"rms"/"|Z|"=230/100`=2.3 A
`therefore` The voltage across the ends of RESISTANCE ,
`V_R=I_(rms)xxR`=2.3 x 100
`therefore V_R`= 230 V
5327.

there in no current part of this circuit for time t lt o. Switch S is closed at t = 0. Current through the 6 Omega resistor

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increases linearly with time
increase non-linearly with time
decreases non-linearly with time
remains constant

Solution :Initally there is no current in the inductor.
So in initially, `V_(A) - V_(B) = ((6)/(1 + 5)) XX 1 = 1 V` (i)
`V_(A) - V_(C) = ((6)/(2 + 4)) xx 2 = 2 V` (ii)
from (i) and (ii), `V_(B) - V_(C) = 2 - 1 = 1 V`
`rArr V_(B) - V_(C) = L(di)/(dt)`
`rArr 1 = 0.1 (di)/(dt) rArr (di)/(dt) = 10 As^(-1)`
Current through `6 W` resistor will remain constant because it is independently connected to `6 V`. After a long time, inductor will behave LIKE a simple wire.
`1 Omega` and `2 Omega` are in parallel, their equivalent is `(2)/(3) Omega`
`5 Omega` and `4 Omega` are in parallel, their equivalent is `(20)/(9) Omega`.
`V_(1) = ((2)/(3) xx6)/((2)/(3) + (20)/(9)) = (18)/(13) V,V_(2) = V - V_(1) = 6 - (18)/(13) = (60)/(13)V`
`I_(1) = (V_(1))/(2) = (18)/(13 xx 2) = (9)/(13)A, I_(2) = (V_(2))/(4) = (60)/(13 xx 4) = (15)/(13) A`
Current through inductor: `I = I_(2) - I_(1) = (15)/(13) - (9)/(13) = (6)/(13)A`
5328.

A digital signal possesses:

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Coninuously possesses:Only TWO DISCRETE values
Only four discree values
None of the above

Answer :B
5329.

An electric kettle has two windings. When one of them is switched on, the water in the kettle begins to boil in 15 minutes, and when the other is switched on it takes 30 minutes for water to boil. If the two windings are joined in series and switched on, water in the kettle begin to boil after time: (in minutes) [Assume that the kettle loses no heat to the surroundings]

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45
60
75
90

Solution :Let the resistace of winding 1 and 2 are `R_1 and R_2` RESPECTIVELY
`therefore` When winding 1 is PRESENT `p=v^2/R_1 therefore m_(v ATER)L_f=V^2/R_1 times 15`
For `2^(nd)` winding `P=V^2/R_2 therefore m_(VATER)L_f=V^2/R_1 times 15`………(ii)
When the winding are joined in series,
`R_(eq)=(R_1+R_2) therefore P=V^2/((R_1+R_2)) therefore m_(vater) L_f=(V^2/(R_1+R_2))t`.......(iii)
from eqs (i), (ii) and (iii), t=45 min
5330.

A flexible chain of weight W hangs between two fixed points A & B which are at the same horizontal level. The inclination of the chain with the horizontal at both the points of support is theta . What is the tension of the chain at the mid point ?

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`W/2cosectheta`
`W/2tantheta`
`W/2tantheta`
none

Answer :C
5331.

A coil of 100 turns having an average area of 100cm^2 for each turn is held in a uniform field of 50 gauss, the direction of field being at right angles to the plane of coil. The field is removed in 1/100 sec. Average e.m.f. (in volt) induced in coil is:

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0.5V
5V
25V
50V

Answer :A
5332.

The fraction of world's human population using the services of internet is

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`1//2`
`1//3`
`1//4`
`1//5`

ANSWER :B
5333.

A small solid sphere of radius r rolls down an incline without slipping which ends into a vertical loop of radius R. Find the height above the base so that it just loops the loop

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2.1 R
2.3 R
2.7 R
2.5 R

Answer :C
5334.

Half the surface of atransparent sphere of refractive index 2 is silvered. A narrow, paralel beam of light is incident on the unsilvered surface, symmetrically with respect to the silvered part. The light finally emerging from theh sphere will be a

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PARALLEL beam
converging beam
slightly divrgent beam
widely DIVERGENT beam

Answer :A
5335.

A spherical mirror froms an erect image three times the size of the object . if the distance between the object and the image is 100cm/sec . What is the velocity of image at that instant ?

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15 cm
25 cm
37.5 cm
50 cm

Answer :C
5336.

Dispersive power ofprism is the ratio between the angular dispersion to_____.

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SOLUTION :MEAN DEVIATION
5337.

A 500 g block rests on a frictionless horizontal table at a distance of 400 nm from a fixed pin O. Then block is attached to pin O by an elastic cord of constant k = 100 N//m & of underformed length 900 mm. If the block is set in motion perpendiculariy as shown

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the speed V to be set initially for which the distance from O to the block P will REACH the maximum value of 1.2 m is 4.5 m/s
the speed u of the block at the moment when `OP = 1.2 m` is 1.5 m/s
the radius of curvature of the path of the block when `OP = 1.2 m` is 3.75 cm
A force of 30 N acts on the block along the length of the cord when `OP = 1.2 m`

SOLUTION :By conservation of angular momentum `0.5 xx V xx 0.4 = 0.5 xx u xx 1.2`
`v = 3u`
also by energy conservation

`(1)/(2) xx 0.5 xx v^(2) = (1)/(2) xx 100 xx (0.3)^(2) + (1)/(2) xx 0.5 xx u^(2)`
`rArr (v^(2))/(4) = (9)/(2) + (u^(2))/(4)`
`rArr (9u^(2))/(4) = (9)/(2) + (u^(2))/(4) rArr (8u^(2))/(4) = (9)/(2)`
`rArr u = sqrt(9)/(4) = (3)/(2)`
So `v = 3u = 3 xx 1.5 = 45`
(C ) `a_(n) = (u^(2))/(r) "" rArr r = (u^(2))/(a_(n))`
`(a_(n) = (K_(h))/(m) = (100 xx 0.3)/(0.5) = 60m//s^(2))`
`rArr r = ((1.5)^(2))/((60)) = (2.25)/(60) = 0.0375m`
`= 3.75cm`
5338.

What we call the phenomenon of change in path of light without change in medium ?

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SOLUTION :REFLECTION
5339.

The value of electric potential at the surface of a charged conductor is 10V. Find the value of intensity of electric field and potential at a point interior to it.

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SOLUTION :The SURFACE of CHARGED conductor is an EQUI potential surface. In equipotential surface, potential is same and hence electric field is ALWAYS normal to surface of a charged conductor.
5340.

A wave progagates on a string in the positive x-direction at a velocity v. The shape of the string at t = t, is give by f(x ,t_0) = A sin(x//a). Write the wave equation or a general time t

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SOLUTION :`F(X,t) = A SIN (x - V(t-t_0))/a`
5341.

Drawings I and II show two samples of electric field lines

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The electric fields in both I and II are produced by negative charge located SOMEWHERE on the left and positive charges located somewhere on the RIGHT 
In both I and II the electric field is the same every where 
In both cases the field becomes STRONGER on moving from left to right 
The electric field in I is the same everywhere, but in II the electric field becomes stronger on moving from left to right

Answer :D
5342.

A conducting sphere of radius R, and carrying a charge Q , is joined to an uncharged conducting sphere of radius 2 R . The charge flowing between them will be :

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Q/4
Q/3
Q/2
2Q/3

Answer :D
5343.

The mass number "A" of a nucleus is related to surface area "X" of nucleus as -

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`X ALPHA A^2`
`X alpha A^(2//3)`
`X alpha A^(1//3)`
`X alpha A^(3)`

ANSWER :B
5344.

Three resistance 2 Omega , 3Omegaand 5 Omega are connected in parallel and a p.d. of 30V is applied across the terminals ofcombination. The current flowing through 6 Omegaresistance is

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`10 A`
`5 A`
`15 A`
`20A`

Answer :B
5345.

The half life of a radioactive substance is 20 minutes, The time taken between 50% decay and 87.5% decay of the substance will be

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40 mintues
10 mintues
30 mintues
25 mintues

Solution :50% decay is HALF LIFE =20 min
NEXT 25% decays again in 20 min
Further 12.5 decays in 20 min
Total time of decay 50%+25%+12.5%=60min
So difference in time of 87.5% decay and 50% decay is =60-20=40min.
5346.

The magnetic dipole moment of a current carrying coil is independent of the_____ of the magnetic field.

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SOLUTION :STRENGTH
5347.

Two ions having masses in the ratio 1 : 2 and charges 1 : 2 are projected into uniform magnetic field perpendicular to the field with speeds in the ratio 2 : 3. The ratio of the radii of circular paths along which the two particles move is

Answer»

`4:3`
`2:3`
`3:1`
`1:2`

ANSWER :B
5348.

A double convex lens forms a real image of an objectt on a screen which is fixed. Now lens is given a constant velocity v=1m/s along its axis and away from the screen for the purpose of forming image always on the screen the object is also required to be given appropriate velocity . Find the velocity (in m/s) of the object at the instant its size is doubled the size of the image.

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ANSWER :3
5349.

In Foucoult's method of determining the velocity of light, the distance between the rotating mirror and the concave mirror is made to pass along a tube of water (mu = 1.33) instead of air, the displacement is

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`t/1.33`
1.331
`(1.33)^(2)t`
`1.33t^(2)`

Solution :The DISPLACEMENT with WATER is GIVEN by, `mu t=1.33 t`
5350.

A proton of charge 1.6 xx 10^(-19)C" and mass "1.67 xx 10^(-27)kg is accelerated from rest in air through a p.d. of 1000 volts. Calculate its final speed.

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ANSWER :`4.37 XX 10^(5)m//s`