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551.

Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies v_(A) gt v_(B). (i) In which case is the stopping materials A and B having threshold frequencies v_(A) gt V_(B). (i) In which case is the stopping potential more and why? (ii) Does the slope of the graph depend on the nature of the materials used ? Explain.

Answer»

SOLUTION :
The graphs showing variation of stopping potential `V_(0)` with frequency v of INCIDENT radiation for two photosensitive materials A and B, having THRESHOLD FREQUENCIES `v_(A) and v_(B)(v_(A)gtv_(B))` are shown in here. the graphs are straight line graphs.
(i) The stopping potential `V_(0)=(h)/(e)(v-v_(0))`, where `v_(0)` is the threshold frequency. As threshold frequency of A is more than that for B, it is obvious that for an incident radiation of given frequency the stopping potential for B is more than for A.
(ii) The slope of the graph is `h/e`, which is INDEPENDENT of the nature of the photosensitive material used.
Discussion
552.

At a distance 320 km above the surface of the earth , the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly ( radius of earth = 6400 km )

Answer»

`2%`
`6%`
`10%`
`14%`

Answer :C
Discussion
553.

You put three identical coins on a turntable at different distances from the centre and then turn the motor on. As the turntable speeds up, the outermost coin slides off first, followed by the one at the middle distance, and, finally, when the turntable is going the fastest, the innermost one. Why is this?

Answer»

For greater distances from the centre the CENTRIPETAL acceleration is HIGHER, and so the force of friction becomes unable to hold the coin in place.
The weight of the coin CAUSES the turntable to flex downward, so the coin nearest the edge falls off first.
Because of the way the turntable is made, the coefficient of STATIC friction decreases with DISTANCE from the centre.
For smaller distances from the centre, the centripetal acceleration is higher.

Solution :`a_(c) = omega^(2) r and f_(ext) = m omega^(2) r`
Discussion
554.

Which logic gate is represented by the following truth table ?

Answer»


ANSWER :A
Discussion
555.

There are two thermally insulated vessel.One with 0.025 moles of helium and other with n moles of hydrogen.Initially both the gases are at room temperature.Now equal amount of heat is supplied to both the vessels. It is found that in both the gases temperature rises by same amount. Find the number of moles of hydrogen in second vessel.

Answer»

Solution :As the gases ARC enclosed in closed vessels, the heating can be taken as isochoric heating and as HEAT supplied to both vessels are same, we have
`Q=n_1C_(V_1)DELTAT=n_2=C_(V_2)DeltaT`
or `0.025xx3/2RDeltaT=n5/2RDeltaT`
[As for He , `C_(V_1)=3//2` R and for `H_2,C_(V_2)=5//2R` ]
or `N=(0.025xx3)/5`=0.015 mole
Discussion
556.

The critical speed of satellite orbiting close to the earth’s surface is,

Answer»

Solution :For a satellite ORBITING very CLOSE to the surface of earth, the critical speed`V_c=SQRT((GM)/R)`but `GM=GR^2
V_c=sqrt((gR^2)/R)=sqrt(gR)`
`sqrt(10xx6.4xx10^6)=sqrt(64xx10^6)=8xx10^3 m//s`
Discussion
557.

Match column A (layers in the ionosphere for sky wave propagation) with column B (their height range) The correct answers is

Answer»

`{:(I,II,III,IV),(a,B,C,d):}`
`{:(I,II,III,IV),(d,c,a,b):}`
`{:(I,II,III,IV),(d,c,b,a):}`
`{:(I,II,III,IV),(c,d,a,b):}`

Answer :B
Discussion
558.

The K_alpha and K_(beta) lines of characteristic X-ray spectrum of molybdenum are 0.76 Å and 0.64 Å respectively, The wavelength of La line is :

Answer»

1.4Å
2.4Å
4.1Å
3.6Å

Solution :`E_(2)-E_(1)=(12375)/(0.76)EV`
`E_(3)-E_(1)=(12375)/(0.64)eV`
`E_(3)-E_(2)=(12375)/(0.64)-(12375)/((0.76)`
`or lambda K_(alpha)=(12375)/(E_(3)-E_(2))=4.1Å`
Discussion
559.

Which-logic gate is represented by the following truth table ?

Answer»

AND
OR
NAND
NOR

ANSWER :B
Discussion
560.

State Gauss' theorem. With the help of this theorem, find out the electrical intensity at any nearby point due to a uniformly charged thin and long straight wire. Or, Define electrical dipole moment. An electrical dipole is placed within a uniform electric field (E) and is rotated to an angle angle(theta) = 180^(@). find out the work done.

Answer»

Solution :Two equal andopposite charge`pm`q, SEPARATED by a DISTANCE vector `vec(l)` directed from -q to +q, from an electric dipole. Its dipole moment is `vec(p) = q vec(l)`.
Torque on an electri dipole in an electric FIELD `vec(E)` is, `vec(tau) = vec(p) xx vec(E)`.
Its magnitude is, `tau = p E SIN theta, ` where ` theta` = angle between `vec(p) and vec(E)`.
`therefore` Work done to rotate the dipole from `0^(@)` to `180^(@)` is,
W = `int DW = int_(0)^(180) tau d theta`
= `p E int_(0)^(180) sin theta d theta = p E [ -cos theta]_(0)^(180) = 2pE`
Discussion
561.

Is the drift speed a characteristic of the material of the conductor?

Answer»

Solution :No. `V_d = i/(neA)` For a given material N and E are constants. But `V_d` DEPENDS on the AMOUNT of current i flowing through the conductor and also on the area of cross section A.
That is why we find `v_d ~ 10^(-3) ms^(-1)` etc. with DIFFERENT orders of magnitude for copper.
Discussion
562.

Two coils of self inductances 2mH and 8mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is

Answer»

10 mH
6 mH
4 mH
16 mH

Answer :C
Discussion
563.

There are 6 xx 10^(21) hydrogen molecules in a vessel of volume 200 cm^(3). Its tempera-ture is 27^(@)C and the pressure is 10^(5) Nm^(-2). If the temperature be raised to 47^(@)C, then in what ratio the following qunatities would change (i) number of molecules per unit volume (ii) pressure of the gas is the vessel and (iii) average kinetic energy of hydrogen ?

Answer»


ANSWER :(i) It remains unchanged `(3 xx 10^(19) "PER cm"^(3))` , (ii) `15 : 16` , (iii) `15 : 16`
Discussion
564.

What is relative humidity ?

Answer»

SOLUTION :It is the RATIO of mass of vapor actually PRESENT in a’ given VOLUME of air at ordinary temperature to the mass of water vapor required to saturate it at the same temperature
Discussion
565.

Equation of trajectory is given by y=xtan theta - (())/(2u^2 cos^theta)

Answer»


ANSWER :`[gx^2]`
Discussion
566.

A particle Is projected vertically up and another is let fal to meet at the same instant.If they have velocities equal in magnitude when they meet,the distance travelled by them are in the ration of

Answer»

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

ANSWER :C
Discussion
567.

The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive " "_(6)^(14)C present with the stable carbon isotope " "_(6)^(12)C. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of " "_(6)^(14)C, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of" "_(6)^(14)C dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Answer»

Solution :GIVEN, NORMAL activity `R_(0)`=15 decays per minute, Present activity R =9 decays per minute, and Half-life `T_(1/2)` = 5730 YEARS
Since `R=R_(0)e^(-lamda t) IMPLIES R_(0)/R=e^(lamdat) or t=1/(lamda)ln R_(0)/R=1/(lamda)ln(15/9)`
But `lambda=0.693/T_(1/2) = 0.693/5730 Year^(-1)`
`therefore t= 5730/0.693 In(15/9)`year =4225 years
Discussion
568.

A coil of inductance L connects the upper ends of two vertical copper bars separated by a distance l . A horizontal conducting connector of mass m starts falling with zero initial velocity along the bars without losing contact with them. The whole system is located in a uniform magnetic field B perpendicular to the plane of the bars. Find the law of motion x(t) of the connector.

Answer»

Solution :Let at any TIME `t` VELOCITY of rodis `v` and current in circuit `i` .
Induced emf in rod `e=Bvl`
`Bvl=L(di)/(dt)`
`B(dx)/(dt)l=L(di)/(dt)`
`Ldi=Bldx`
Integrating
`Li=Blx implies i=(Bl)/(L)x`
`mg-Bil=ma`
`a=g-(Bl)/(m)i=g-(B^(2)l^(2))/(mL)x`
`(dv)/(dt)=g-omega^(2)x` (where `omega=(Bl)/(sqrt(mL)))`
`(d^(2)v)/(dt^(2))=-omega^(2)(dx)/(dt)=-omega^(2)v`
`v` oscillates simple harmonically with ANGULAR frequency
`omega=(Bl)/(sqrt(mL))`.
The solution of this equation
`v=Asin(0megat+phi)`
`t=0` , `v=0` , `phi=0`
`v=Asinomegat`
`(dv)/(dt)=Aomegacosomegat`
At `t=0` ,
`((dv)/(dt)=a=g-(Bl)/(m)i=g(. :. i=0)`
`g=Aomega` `implies A=(g)/(omega)`
`v=Asinomegat`
`v=(g)/(omega)sinomegat`
`(dx)/(dt)=(g)/(omega)sinomegat`
`int_(0)^(x)dx=(g)/(omega)int_(0)^(t)sinomegadt`
`x(g)/(omega)(|-cosomegat|_(0)^(t))/(omega)`
`=(g)/(omega^(2))(1-cosomegat)`
`v=(g)/(omega)sinomegat`
Let `v_(0)((=(g)/(omega))` : velocity AMPLITUDE
`V=V_(0)sinomegat` `x=(g)/(omega^(2))(1-cosomegat)=(v_(0))/(omega)(1-cosomegat)`

Discussion
569.

A dog chases a cat 30 m ahead of it and gains 3 meters in 3 sec after the chase started. After 10 sec the distance between them is :

Answer»

6 m
14 m
18 m
24 m

Answer :C
Discussion
570.

A particle moves in a circular path of radius 5 cm in a plane perpendicular to the principal axis of a concave mirror with radius of curvature 20 cm. The centre of circle lies on the prncipal axis at distance of 15 cm in front of the mirror.The radius of the circular path of the image is

Answer»

15 cm
20 cm
10 cm
40 cm

Answer :C
Discussion
571.

The depletion region of PN junction has a thickness of the order of

Answer»

`10^(-12)`
`10^(-6)m`
`1mm`
`1CM`

ANSWER :B
Discussion
572.

An object kept on a smooth inclined plane rising with height 1 units and length I units can be kept stationary relative to the inclined plane by giving a horizontal acceleration. The value of acceleration is :

Answer»

`(g)/(sqrt(l^(2)-1))`
`gsqrt(l^(2)-1)`
`(g)/(l)`
g.l.

Solution :The HORIZONTAL acceleration should be such that its COMPONENT along the plane should be equal and opposite to the acceleration down the plane i.e.
`a cos theta=g SIN theta`or`a=g TAN theta`
But `tan theta =(1)(sqrtl^(2)-1)`
`:.a=(g)/(sqrt(l^(2)-1)`
Hence (a) is the choice.
Discussion
573.

Charges 5 mu C, -2muC, 3mu and -9mu are placed at the corners A,B,C and D of a square ABCD of side 1m. The net electric potential at the centre of th square is

Answer»

`-27 KV`
`-27 SQRT(2)KV`
`-90KV`
zero

Answer :B
Discussion
574.

Watt. Sec is same as

Answer»

Joule
Joule`//`sec
Sec`//`Joule
Joule.sec.

Answer :A
Discussion
575.

A ray of light is travelling from a denser medium to rarer medium. For these media the critical angle is C. The maximum possible deviation of the ray is ...........[Hint : The situation at total reflection is shown in the figure.]

Answer»

`pi-2`
`pi-2C`
2C
`pi/2+C`

Solution :From figure, `delta`=r-i
But if i=C`impliesr=90^@=pi/2`
`THEREFORE delta=pi/2-C`
`therefore2 delta=pi-2C` Where,`2DELTA` is maximum deviation.

Second Method :
Deviation becomes maximum when ray of light experiences TOTAL internal refraction.
From it is clear that maximum deviation
`theta=pi-(C+C)=pi-2C`
[Note : angle of incidence for `theta_i gt C` is taken as `theta_i cong C`.]
Discussion
576.

For better resolution, a telescope must have a

Answer»

LARGE DIAMETER objective
small diameter objective
may be large
neither large nor small

Answer :A
Discussion
577.

A string breaks under a tension of 10 kg wt. if a string is used to revolve a body of mass 1.2 gm in a horizontal circle of radius 50 cm. When a body revolving at maximum speed what is its period?

Answer»

0.0155 sec
0.155 sec
0.00155 sec
1.55 sec

Answer :A
Discussion
578.

Focal lens of plane convex lens (mu=3/2) is f and radius of curvature of curved surface is R. So write the relation between f and R.

Answer»

f=R
f=`R/2`
f=2R
f=`(3R)/(2)`

SOLUTION :From lens maker.s FORMULA,
`1/f=(n-1)(1/R_1-1/R_2)`
`THEREFORE 1/f=(3/2-1)(1/R-(1)/(infty))` (Here `R_1=R` and `R_2=infty`)
`=(1/2)(1/R+0)`
`=1/2xx1/R`
`=(1)/(2R)`
`therefore f=2R`
Discussion
579.

An AC voltage is given by E = underset(o)(E)sin 2 pi t / T Then , the mean value of volatage calculated over any time interval of T / 2

Answer»

is ALWAYS zero
is never zero
is always (2`UNDERSET(o)(E) ` / `pi `)
may be zero

Solution :N//A
Discussion
580.

Components of loudspeaker : A loudspeaker consists of permanent magnets, basket, voice cails, flexible supension & rigid speaker cone as shown in figure. Loudspeaker principle A light voice coil is mounted so that it can move freely inside the magnetic field of a strong permanent magnet. the paper cone is attached to the voice coil and attached with a flexble mounting to the outer ring of the support .Because there is a definite equilibrium position for the speaker cone and there is elasticity of the mounting structure, there is inevitably a free cone resonant frequency like of a mass on a spring. The frequency can be determined by adjusting the mass and stiffness of the cone and voice coil Working : The ratio drives a rapidly changing current through the coil. The current follows that vibrations of speech and the electromagnetic force follows the current changes, pushing the paper cone. Finally the air in front of the loudspeaker is set into vibration following the cone's motion, and sound waves are transmitted to the listener The principle of loudspeaker is :

Answer»

It converts mechanical energy into electrical energy
Permanent magnet exerts force on the coil
The coil carrying current experiences torque DUE to which cone rotates and sound is produced
Paper cone attached to coil moves air to produce sound when the VOICE coil is vibrated by an AC current interacting with MAGNETIC field.

Solution :The principal can be EASILY understood from the ''working,, of loudspeaker.
Discussion
581.

Components of loudspeaker : A loudspeaker consists of permanent magnets, basket, voice cails, flexible supension & rigid speaker cone as shown in figure. Loudspeaker principle A light voice coil is mounted so that it can move freely inside the magnetic field of a strong permanent magnet. the paper cone is attached to the voice coil and attached with a flexble mounting to the outer ring of the support .Because there is a definite equilibrium position for the speaker cone and there is elasticity of the mounting structure, there is inevitably a free cone resonant frequency like of a mass on a spring. The frequency can be determined by adjusting the mass and stiffness of the cone and voice coil Working : The ratio drives a rapidly changing current through the coil. The current follows that vibrations of speech and the electromagnetic force follows the current changes, pushing the paper cone. Finally the air in front of the loudspeaker is set into vibration following the cone's motion, and sound waves are transmitted to the listener A voice coil in a loudspeaker has 40 turns of wire and loop-diameter 1 cm and the current in the coil is 1A. Assume that the magnetic field at each at the wire of the coil has constant magnitude 0.2 T and is directed at an angle 60^(@) from the normal to the plane of the coil as shown in figure. The magnitude and direction of magnetic force on the coil is

Answer»

`4 pi xx 10^(-2) N,+y`
`4 pi xx 10^(-2) N.-y`
`4 sqrt(3) pi xx 10^(-2) + y`
`4 sqrt(3) pi xx 10^(-2) N,-y`

Solution :Force on each part of ring is shown in FIGURE

`= 20.12 pi (0.5 xx10^(-3))(0.2)((sqrt(3))/(2))`
`=4 sqrt(3)pi xx10^(-2)N`
CLEARLY the force is ALONG -y direction.
Discussion
582.

Components of loudspeaker : A loudspeaker consists of permanent magnets, basket, voice cails, flexible supension & rigid speaker cone as shown in figure. Loudspeaker principle A light voice coil is mounted so that it can move freely inside the magnetic field of a strong permanent magnet. the paper cone is attached to the voice coil and attached with a flexble mounting to the outer ring of the support .Because there is a definite equilibrium position for the speaker cone and there is elasticity of the mounting structure, there is inevitably a free cone resonant frequency like of a mass on a spring. The frequency can be determined by adjusting the mass and stiffness of the cone and voice coil Working : The ratio drives a rapidly changing current through the coil. The current follows that vibrations of speech and the electromagnetic force follows the current changes, pushing the paper cone. Finally the air in front of the loudspeaker is set into vibration following the cone's motion, and sound waves are transmitted to the listener With reference to the figure of Q.& when the current in the coil is given by I=I_(c) cos (2000 pi t) where t is in seconds. the coil will experience magnetic force in the positive y-direction in the time intervals of : (Takethe initial direction of cirrent shown in figure to be positive)

Answer»

0 to `5 xx 10^(-4) SEC`
`2.5 xx 10^(-2)` to `5 xx 10^(-4) sec`
0 to `2.5 xx 10^(-4) sec`
`2.5 xx 10^(-4)` to `7.5 xx10^(-4) sec`

Solution :Force will be in `+y` direction when current is in `-ve` direction
i.e., when `cos (2000 pi t) LT 0`
i.e. when `2.5 xx 10^(-4) lt t lt 7.5 xx 10^(-1)`.
Discussion
583.

Components of loudspeaker : A loudspeaker consists of permanent magnets, basket, voice cails, flexible supension & rigid speaker cone as shown in figure. Loudspeaker principle A light voice coil is mounted so that it can move freely inside the magnetic field of a strong permanent magnet. the paper cone is attached to the voice coil and attached with a flexble mounting to the outer ring of the support .Because there is a definite equilibrium position for the speaker cone and there is elasticity of the mounting structure, there is inevitably a free cone resonant frequency like of a mass on a spring. The frequency can be determined by adjusting the mass and stiffness of the cone and voice coil Working : The ratio drives a rapidly changing current through the coil. The current follows that vibrations of speech and the electromagnetic force follows the current changes, pushing the paper cone. Finally the air in front of the loudspeaker is set into vibration following the cone's motion, and sound waves are transmitted to the listener If the diameter of the cylindrical magnet, number of turns of the coil and cross-section area of the wire of the coil are all doubled, then the magentic force on the coil assuming the same potential difference , is (assume that value of magnetic field also gets doubled)

Answer»

same
doubled
becomes 8 times
becomes FOUR times

Solution :`F=NI B 2 pi r SIN 60^(@)`
Initial resistance
`R = (rho l)/(A) = (rho N(2pi t))/(A)`
New resistance
`R=(rho l')/(A')=(rho(2N)2pi(2r))/(2A)=2((rho N(2pil))/(A))=2R`
`rArr I'=(V)/(R) =(1)/(2)`
`rArr F'=2N(1)/(2)2B,2pi(2r)sin 60^(@) =4R`
HENCE becomes 4 times.
Discussion
584.

Components of loudspeaker : A loudspeaker consists of permanent magnets, basket, voice cails, flexible supension & rigid speaker cone as shown in figure. Loudspeaker principle A light voice coil is mounted so that it can move freely inside the magnetic field of a strong permanent magnet. the paper cone is attached to the voice coil and attached with a flexble mounting to the outer ring of the support .Because there is a definite equilibrium position for the speaker cone and there is elasticity of the mounting structure, there is inevitably a free cone resonant frequency like of a mass on a spring. The frequency can be determined by adjusting the mass and stiffness of the cone and voice coil Working : The ratio drives a rapidly changing current through the coil. The current follows that vibrations of speech and the electromagnetic force follows the current changes, pushing the paper cone. Finally the air in front of the loudspeaker is set into vibration following the cone's motion, and sound waves are transmitted to the listener If the voice coil is wound loosely and there is an approachable gap between two consecutive turns. When the current is passed through coil :

Answer»

it tries to contract
It tries to expand
Current has no effect on coil
Coil will GET contracted WITHOUT current DUE to FIXED magnetic is loudspeaker.

Solution :As current in the loops is FLOWING in the same direction. They will attract each other.
Discussion
585.

Excess pressure inside a liquid drop of radius r and surface tension T is

Answer»

T/r
2T/r
3T/r
4T/r

Answer :B
Discussion
586.

Magnetic Intensity is defined by H = B/(mu_0) - I hence B = _____

Answer»

SOLUTION :`[mu_0 (H + I)]`
Discussion
587.

Two wires of equal length and equal diameter and having resisticities rho_(1) and rho_(2) are connected in series. The equivalent resistivity of the combination is .... .

Answer»

`(rho_(1) + rho_(2)) `
`(rho_(1) rho_(2))/(rho_(1) + rho_(2))`
`(rho_(1) + rho_(2))/(2) `
`SQRT(rho_(1) rho_(2))`

SOLUTION :`(rho_(1) + rho_(2))/(2) `
R = `R_(1) + R_(2) `
`(RHO (l + l))/(A) = (rho_(1)l)/(A) = (rho_(2)l)/(A)`
`2 rho = rho_(1) + rho_(2)`
`THEREFORE rho = (rho_(1) + rho_(2))/(2) `
Discussion
588.

suppose the ends of the coil in the previousproblem are connected to a resistane of 100 Omega. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.

Answer»

Solution :Since`H=int_(0)^(1 MIN) i^2 R.dt`
`=int_(0)^(1min) (B^2A^2omega^2)/(R^2) sin omega t.R dt`
`=(B^2 A^2 omega^2)/(2R). Int_(0)^(1 min) (1-cos 2 omega t).dt`
`=(B^2A^2omega^2)/(2R)(1-(sin 2 omega t)/(2 omega))_(0)^(1 min)`
`(B^2 A^2 omega^2)/(2R)(60-(sin.2xx80xx2 pi//60xx60)/(2xx80xx2 pi//60))`
`=(60)/(2R)xxpi^2r^4xxB^2xx(80xx(2 pi)/(60))^2`
``
`= 60/200xx10xx64/9xx10xx625xx10^(-8)xx10^(-4)`
`=(625xx6xx64)/(9xx2)xx10^(-11)=1.33xx10^(-7)J.`
Discussion
589.

For a transistor, beta=50. To change the collector current by 350muA, the base current should be changed by

Answer»

`(50/350)MUA`
`(350-50)muA`
`(350//50)muA`
`(350/50)muA`

ANSWER :C
Discussion
590.

An air bubble in glass (mu = 1.5) is situated at a distance 3 cm from a convex surface of diameter 10 cm as shown. The distance from surface at which the image of bubble appears is

Answer»

2.5cm
5cm
4cm
1.5cm

Solution :`mu_(2)/V - mu_(1)/U=(mu_(2)-mu_(1))/R`
`1/v - (1.5)/((-3))=(1-1.5)/((-5))`
Discussion
591.

What will happen if the field were not uniform?

Answer»

Solution :If the DIPOLE is PLACED in non-uniform electric FIELD the NET FORCE on the electric dipole is not zero.
Discussion
592.

Waves from two different sources overlap near a particular point. The amplitude and the frequency of the two waves are same. The ratio of the intensity when the two waves arrive in phase to that when they arrive 90^(@) out phase is

Answer»

`1:1`
`SQRT2` : 1
`2:1`
`4:1`

ANSWER :C
Discussion
593.

Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are

Answer»

4I and I
5I and 3I
9I and I
9I and 3I

Answer :C
Discussion
594.

If mass density of earth varies with distance 'r' from centre of earth as rho = kr and 'R' isradius of earth, then find the orbital velocity of an object revolving around earth at adistance '2R' from its centre.

Answer»

`sqrt((pikR ^(3)G)/(4))`
`sqrt((pi k R ^(3)G)/(2))`
`sqrt((pi k R ^(3)G)/(8))`
`sqrt(pi k R ^(3)G)`

Solution :
Let 'M' be total mass of EARTH. Consider a shell of thickness 'DR' and mass 'DM' at a distance 'r' from centre inside earth,
`implies dm = rho 4 pi x ^(2) dr `
`M int dm`
`= underset(0) overset(R ) int 4 pi kr^(3) dr`
`=(4 pi k R ^(4))/(4) = pi kR ^(4)`
Let field due to earth's gravity at a distance '2R' from centre be `T.IxxA=4 pi G m _("inside).`
`impliesIxx 4pi (2R) ^(2) =4 pi G (pi kR^(4))`

`impliesI =(pikR ^(4)G)/(4R^(2))`
`implies I =(pi k R ^(4) G )/(4R^(2)) `
For a satellite of mass 'm' MOVING in orbit of '2R' RADIUS.
`mI =(mv^(2))/((2R))`
`implies I =(V^(2))/(2R)`
`implies (pi kR ^(2)G)/(4) =(V^(2))/(2R)`
`V = sqrt((pi kR ^(5)G)/(2))`
Discussion
595.

Figure shows a point object placed between two parallel mirrors. Its distance from M_(1) is 2 cm and that from M_(2) is 8 cm. Find the distance of images from the two mirrors considering reflection on mirror M_(1) first.

Answer»

Solution :To UNDERSTAND how images are formed see the following figure and table.
You will require to know what SYMBOLS like I121 stands for. See the following diagram.


Similarly images will be formed by the rays STRIKING MIRROR `M_(2)` first. Total number of images `=oo`
Discussion
596.

Define polarizing angle for a material.

Answer»

SOLUTION :The angle of incidence for which the REFLECTED LIGHT is completely plane polarised iscalled POLARISING angle or Brewster.s angle.
Discussion
597.

Assertion:- Net work done by all the internal force of a system is independent of choice of reference frame.

Answer»

If both Assertion `&` REASON are TRUE `&` the Reason is a CORRECT explanation of the Assertion.
If both Assertion `&` Reason are True but Reason is not a correct explanation of the Assertion.
If Assertion is True but the Reason is False.
If both Assertion `&` Reason are False.

ANSWER :B
Discussion
598.

Two insulated copper sphere of charges +10 mu C and -20 mu Chave their centres seprated by a distance of 10 cm . What is the force of attraction if they placed. a in free space b. in water of dielectric constant 81.

Answer»

Solution :Data supplied
` r=10 xx 10 ^(-2) m.`
` q_1 =10 mu C =10 xx 10^(-6) C , q_2 = 20 MUC =20 xx 10 ^(-6 ) C `
a. ` "" F_("AIR") =(1)/( 4pi in _0) (q_1q_2)/( r^(2)) `
` F_(air) =( 9xx 10^(9) xx10 xx 10 ^(-6)xx-20xx10 ^(-6))/( (10xx 10 ^(-2))^(2)) =-180 N`
b. ` F_("water")=(1)/( 4piin _0) =(q_1q_2)/( r^(2)) =(air )/(in_r) =(-180)/(81)=-2.22N`
NEGATIVE sign indicates that the force is attractive.
Discussion
599.

A certain length of a uniform wire of resistance 12Omega is bent into a circle and two points, a quarter of circumference apart, are connected to a battery of emf 4 V and internal resistance 1Omega. Find the current in the different parts of the circuit.

Answer»


ANSWER :`(12)/(13)A,(4)/(13)A`
Discussion
600.

(i) Define the term drift velocity. (ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms ofnumber density of free electron and relaxation time. On what factors does resistivity of aconductor depend ?(iii) Why alloys like constantan and manganin are used for making standard resistors ?

Answer»

SOLUTION :(iii) Alloys like CONSTANTAN and manganin are USED for making standard resistors because (a) theirresistivity is comparatively HIGHER, and (b) their temperature coefficient of resistivity is extremelysmall so that the resistance does not CHANGE with change in temperature.
Discussion