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1451.

In the electrolytic cell, current flows from …………………. .

Answer»

SOLUTION :NEGATIVE to POSITIVE TERMINAL
Discussion
1452.

ProveBrewster's Law.

Answer»

Solution :As an unpolarised light enters the denser medium, a part of light at the point of incidence gets reflected and the rest refracted in the dense medium. It is noticed that the refracted beam of light is either partially polarised or COMPLETELY UNPOLARIZED. However, the reflected beam gets completely polarised in the plane at right ANGLES to the plane ofincidence at a particular angle of incidence called polarising angles. It is also noticed that the angle between the completely polarised reflected beam and the refracted beam are at right angles to each other. Applying Snell's law of reflection, we write absolute refractive index, (rarer medium is taken as air)

`mu=(sin i_B)/(sin r)""...(1)`
where `i_B-` polarising angle or Brewster's angle.
Also, `BOC=90^@`
i.e., `NON'=180^@=NOB+BOC+N'OC`
i.e., `180^@=i_B+90^@+r`
`therefore i_B+r=90^@""....(2)`
Angle of refraction `r=90^@-i_B""...(3)`
USING (3) in (1) we get
`mu=(sin i_B)/(sin (90^@-i_B))`
`therefore mu=(sin i_B)/(cos i_B)`
or `tan i_B=mu""...(4)`
Expression (4) is known as the Brewster's law. The tangent of polarising angle for air-medium interface is equal to the absoluate refractive index of the medium for other than air-medium interface,
`tan i_B=mu_2=(mu_2)/(mu_1)` where `mu_2=(c )/(v_2)`
and `mu_1=(c )/(v_1) and mu_2 gt mu_1, v_1 and v_2` represent speed of light in the two OPTICAL media (1) and (2).
Note: David Brewster, in the year 1811, discovered that at the polarising angle of incidence, the reflected light is completely plane polarised.
Discussion
1453.

A block of mass m sliding down an incline plane. The incline plane has fixed base length 'l' and coefficient of friction on the inclineplane is 'mu'. The plane is fixed and block slides from top to bottom. Let theta_(0) be the inclination angle for minimum sliding time and v_(0) be the block's speed when it reaches the bottom in that case. Pick the correct option (s):

Answer»

`v_(0)=sqrt(2gl(tantheta_(0)-MU))`
`v_(0)=sqrt(2gl tantheta_(0))`
`tan(2theta_(0))=(-1)/(mu)`
`tan(theta_(0))=1/(mu)`

Solution :`t^(2)=(2l)/(g[cos alpha SIN alpha -mu cos^(2) alpha])`
For min. time, `d/(dalpha)(cosalpha sin alpha-mucos^(2)alpha)=0`
`IMPLIES tan 2ALPHA =(-1)/(mu)`
`/_\K+/_\U=W_("friction")`
`impliesv_(0)=sqrt(2gl(tantheta_(0)-mu))`
Discussion
1454.

An alpha -particle of energy 5 MeV is scattered through 180^@ by a fixed uranium nucleus. The distance of closest approach is of order:

Answer»

`10^(-15)cm`
`10^(-12)cm`
`10^(-10)cm`
1Å

Solution :`r_(0)=1/(4pi epsi_(0)) .(q_(1).q_(2))/(E)=(1)/(4pi epsi_(0)) .(2E .92e)/(5 XX 1.6 xx 10^(-13))`
`=(9 xx 10^(9) xx 2 xx 92 xx 1.6 xx 10^(-19) xx 1.6 xx 10^(-19))/(5 xx 1.6 xx 10^(-13))`
`approx 10^(-12) m`
Discussion
1455.

A magnetic dipole is placed at right angles to the direction of lines of force of magnetic induction B. If it is rotated through an angle of 180^(@) then the work done is

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`2MB`
`MB`
`-2MB`
ZERO

ANSWER :D
Discussion
1456.

Scientist……proved that electric charge is quantized.

Answer»

R.A millikan
J.J Thomson
Max-Planck
Maxwell

Answer :A
Discussion
1457.

The hale telescope of mount Polamor has a diameter of 200 inches. What is its limiting angle of resolution for 600 nm light?

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`7.2 xx 10^(-8) RAD`
`7.2 xx 10^(-4) rad`
`1.44 xx 10^(-7) rad`
`14.4 xx 10^(-10) rad`

Answer :C
Discussion
1458.

A force vec(F ) = b( - y hat(i) + x hat( j ))/( x^(2) + y^(2)) N( b is a constant ) acts on a particle as it undergoes counterclockwise circular motion in a circle : x^(2) +y^(2) = 16. The work done by the force when the particle undergoes one complete revolution is ( x,y are in m )

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0 Joule
`2 pi B` Joule
2b Joule
`4 pi b ` Joule

Answer :B
Discussion
1459.

The speed of electromagnetic wave in a medium of dielectric constant 2.25 and relative permeability 4 is

Answer»

`1 xx 10^8 m//s`
`2*5 xx 10^8 m//s`
`2 xx 10^8 m//s`
`3 xx 10^8 m//s`

Answer :`1.5 xx 10^8 m//s`.
Discussion
1460.

Glycerine is filled is 25 mm wide space between two large plane horizontal surfaces. Calculate the force required to drag a very thin plate 0.75 m^(2) in area between the surface at a constant speed of 0.5 m//s if it is at a distance of 10 mm from of the surfaces in horizontal position? Take coefficient of viscosity eta=0.5Ns//m^(2). Fill the value of X where X= force required to "drag"//6.25

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SOLUTION :`F=sum ETA A (DV)/(DT)=0.5xx0.75[0.5/(10xx10^(-3))+0.5/(15xx10^(-3))]=31.25`
Discussion
1461.

In the given circuit P.D. between A and B in the steady state is

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`25 V`
`50 V`
`75 V`
`45 V`

Answer :A
Discussion
1462.

A spherical surface of radius of curvature R, separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point and PO = OQ. The distanace PO is equal to

Answer»

5R
3R
2R
1.5R

Answer :A
Discussion
1463.

What are analog ?

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Solution :The COMMUNICATION system, which MAKES use of analong signals (the signal which VARIES CONTINUOUSLY with time) is called analog communicaion system.
Discussion
1464.

What is meant by a surface film ?

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Solution :The layer of the LIQUID surface of THICKNESS EQUAL to the range of molecular ATTRACTION is called a surface FILM.
Discussion
1465.

In question number 91, maximum value ofmagnetisation of the given domain is (Dipolemoment of an iron atom 9.27 xx 10^(-24)A m^(2))

Answer»

`8.0xx10^(5)Am^(-1)`
`6.0xx10^(4)Am^(-1)`
`8.0xx10^(3)Am^(-1)`
`6.0xx10^(3)Am^(-1)`

Solution :NET dipole moment, `M_(N)=Nxxm`
`=6.9xx10^(11)xx9.27xx10^(-24)=6.4xx10^(-12)Am^(2)`
`"Net MAGNETISATION"=(6.4xx10^(-12))/(8xx10^(-18))=8xx10^(5)Am^(-1)`
Discussion
1466.

A steady current flows in a metallic conductor of non-uniform cross-section. Which of thesequantities is constant along the conductor: current, current density, electric field, drift speed ?

Answer»

SOLUTION :Out of the physical quantities given only the electric CURRENT is constant ALONG the conductor.All other quantities DEPEND on cross-section AREA.
Discussion
1467.

Answer the following questions: (i) Why is the thin ozone layer on top of the stratosphere crucial for human survival ? Identify to which part of electromagnetic spectrum does this radiation belong and write one important application of the radiation.

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SOLUTION :(i) The thin ozone layer on top of the stratosphere is crucial for human survival, as it absorbs a major PART of highly energetic radiation, harmful for human being, coming from the SUN. These radiations lie in ultraviolet region of e.m. spectrum. UV rays are used to kill germs in WATER PURIFIERS. These are also used in LASIK eye surgery.
Discussion
1468.

How does a charge q oscillating at certain frequency produce electromagnetic waves ? Sketch a schematic diagram depicting electric and magnetic fields for an electromagnetic wave propagating along the Z-direction.

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Solution :An electric charge at REST has an electric field in the region AROUND it, but no magnetic field. A MOVING charge, however, produces both electric and magnetic fields. If the charge is moving with constant velocity (that is, the currentis not changing with time), the fields will not change with time and no electromagnetic wave can be produced. If, however, the motion of the charge is accelerated, the electric and the magnetic fields will change with space and time, then it produces electromagnetic waves. Hence we conclude that an accelerated charge emits electromagnetic waves.
In an oscillatory LC circuit, charge oscillates across the CAPACITOR plates. An oscillating charge has a nonzero acceleration, hence it emits electromagnetic waves of frequency same as that of the oscillating charge.

Fig. shows the GRAPHICAL representation of an electromagnetic wave in which the electric field vector `vecE` and the magnetic field vector `vecB` are vibrating along Y and X-directions respectively , and the wave is propagating along Z-direction. Both `vecE and vecB` vary with time and space and have the same frequency
Discussion
1469.

When 10 resistors, each with resistance (1)/(10) Omega are connected in parallel , equivalent resistance will be ......

Answer»

<P>`1 OMEGA`
`100 Omega `
`(1)/(100) Omega`
`10 Omega`

SOLUTION :`(1)/(100) Omega`
`(1)/(R_(p))= (1)/((1)/(10))+ (1)/((1)/(10)) + (1)/((1)/(10))+ .... + 10 ` times
` = (10)/((1)/(10)) `
`(1)/(R_(p)) = (100)/(1) "" THEREFORE R_(p) = (1)/(100) Omega`
Discussion
1470.

How does Nelson Mandela define the meaning of courage?

Answer»

over fear
Triumph over love
Triumph over selfish
Triumph over poverty

Answer :A
Discussion
1471.

A galvanometer of resistance 40Omega and current passing through it is 100muA per divison. The full scale has 50 divisions. If it is converted into an ammeter of range 2A by using a shunt, then the resistance of ammeter is

Answer»

`(40)/(399)OMEGA`
`(4)/(399)Omega`
`0.01Omega`
`0.4Omega`

ANSWER :A
Discussion
1472.

A circular coil of radius 0.08m consisting of 100 turns is carrying a current of0.4A. Calculate the magnitude of the magnetic field i) at the center of the coiland ii) at a point 0.2m from the center of the coil on its axis.

Answer»

Solution : GIVEN : r=0.08m, n = 100 turns, I = 0.4A
i) Magnetic FIELD at the centre of the coil
`B = (mu_0 nI)/(2r)`
`B = (4pi xx 10^(-7) xx 100 xx 0.4)/(2 xx 0.08)`
`B = (502.4 xx 10^(-7))/(0.16)`
`B = 3.140 xx 10^(-7)`
`B = 3.14 xx 10^(-4) T`
ii) At a POINT 0.2m from the centre of the coil on its axis
`B = (mu_0)/(4pi) (2PI n I r^2)/((x^2+ r^2)^(3//2))`
`B = 10^(-7) xx (2 xx 3.14 xx 100 xx 0.4 xx (0.08)^2)/([(0.2)^2 + (0.08)^2]^(3//2))`
`B= (1.607 xx 10^(-7))/((0.04 + 0.0064)^(3//2))`
`B= (1.607 xx 10^(-7))/(0.3593)`
`B = 4.47 xx 10^(-7) T`
Discussion
1473.

A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the direction of induced current in each coil.

Answer»

<P>

SOLUTION :CLOCKWISE in both coils/Direction of current is from
`P RARR Q ` and `C rarr D`.
Discussion
1474.

The electric field of an electric dipole at a point on its axis , at a distance d from the center of the dipole, varies as

Answer»

`1/d`
`1/d^2`
`1/d^3`
`1/d^(3/2)`

ANSWER :C
Discussion
1475.

What is the direction of the magnetic field ?

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TOWARD the RIGHT
Into the page
Toward the bottom
Out of the page

Answer :B
Discussion
1476.

Sand drops fall vertically at the rate of 2 kg/sec on to a conveyor belt moving horizontal with a velocity 0.2 m/s. Then the extra force required to keep the belt moving is

Answer»

`0.4 N`
`0.08 N`
`0.04N`
`0.2 N`

ANSWER :A
Discussion
1477.

(a) A point object is placed in front of a double convex lens( of refractive index n = n_(2) // n_(1) with respect to air ) with its spherical faces of radii ofcurvature R_(1)and R_(2). Show the path of rays due to refraction at first and subsequently at the second surface to obtain the formation of the real image of the object. Hence obtain the lens-maker's formula for a thin lens. A double convex lens havingboth faces of the same radius of curvature has refractive index 1.55. Find out the radius of curvature of the lens required to get the focal length of 20cm.

Answer»

Solution :(a)
The first refracting surface ABC forms the IMAGE `I_(1)` of the object O. The image `I_(1)` acts as a VIRTUAL object for the second refracting surface ADC, which forms the real image I as shown in the diagram.
For refraction at ABC.
`(n_(2))/( v_(1)) - ( N _(1))/( u ) = ( n _(2) -n_(1))/( R_(1)) `...(i)
For refraction at ADC
`( n_(1))/( v )- ( n_(2))/( v_(1)) = ( n _(1) - n_(2))/ (R_(2))`....(ii)
Adding equations (i) and (ii)
`(n_(1)) /( v) - ( n_(1))/( u ) = ( n _(2) - n_(1)) ((1)/( R_(1)) - ( 1)/( R_(2)))`
`(1)/( v) - ( 1)/( u )= (( n_(2) )/( n_(1)) - 1) ((1)/( R_(1))- ( 1)/( R_(2)))`
We know, If `u = oo, v = f `
`(1)/( v) - (1)/( u ) = (1)/( f)`
`(1)/(f) = (( n_(2))/( n_(1)) - 1) ((1)/( R_(1)) - ( 1)/( R_(2)))`
`(1)/( f) = ( MU _(21) - 1) ((1)/( R_(1)) - ( 1)/( R_(2)))`
(B) `(1)/(f) = ( mu _(21)-1) ((1)/(R_(1)) - ( 1)/( R_(2)))`
`(1)/(20) = ( 1.55-1) ((1)/( R) - (1)/( -R))`
`= 0.55 xx (2)/( R )`
`R = 0.55 xx 2 xx 20 = 22 cm`
Discussion
1478.

नाइट्रिक अम्ल का आणविक द्रव्यमान क्या है (u)

Answer»

60
62
61
63

Answer :D
Discussion
1479.

The length of a spaceship is measured to be 25% of its rest length. (a) To three significant figures, what is the speed parameter beta of the spaceship relative to the observer's frame? (b) By what factor do the spaceship's clocks run slow relative to clocks in the observer's frame?

Answer»


ANSWER :(a) 0.968; (B) 4.00
Discussion
1480.

Monochromatic light of wavelength 5000 A is incident on narrow slit of width 0.001 mm. Now this light is focused on screen placed on focal plane of convex lens. For this interference first minima will be obtained at ...

Answer»

`0^(@)`
`15^(@)`
`30^(@)`
`50^@)`

Solution :For first minima `d sin theta= N lambda=1lambda`
`sin theta=(lambda)/(d)`
`=(5000xx10^(-10))/(0.001xx10^(-3))=0.5`
`:. Theta=30^(@)`
Discussion
1481.

Consider a system of two particles having masses 2kg and 5 kg the particle of mass 2 kg is pushed towards the centre of mass of particles through a distance 5 m,by what distance would particle of mass 5kg move so as to keep the centre of mass of particles at the original position?

Answer»

2 m
12.5m
8m
5m

Answer :A
Discussion
1482.

A circular loop of radius r, carrying a current I lines in y - z plane with its centre at the origin. The net magnetic flux through the loop is

Answer»

DIRECTLY PROPORTIONAL to R.
ZERO
INVERSELY proportional to r.
directly proportional to I.

Answer :d
Discussion
1483.

An iron rod 0.2 m long, 10 mm in diameter and of permeability 1000 is placed inside a long soleniod wound with 300 turns per meter. If a current of 0.5 ampere is passed through the rod, find the magnetic moment of the rod.

Answer»


ANSWER :`0.2325 Am^(-1)`
Discussion
1484.

The current density varies with radial distance r as J=ar^(2) in a cylindrical wire of radius R. The current passing through the wire between raidal distance R//3 and R//2 is

Answer»

`(65piaR^(4))/(2592)`
`(25piaR^(4))/(72)`
`(65pia^(2)R^(3))/(2938)`
`(81pia^(2)R^(4))/(144)`

ANSWER :A
Discussion
1485.

One end of a uniform bar of weight W_(1) is clamped to the ceiling and a weight W_(2) is suspended to the other end. If the area of cross-section is A, what is the stress at the mid-point of the rod ?

Answer»

`(W_(1)+W_(2))/A`
`(W_(1)-W_(2))/A`
`((W_(1))/2+W_(2))/A`
`((W_(2))/2+W_(1))/A`

Solution :Stress=`("WEIGHT" W_(2) + "Half the WT. of rod")/("area")`
`=(W_(2)+(W_(1))/2)/A`
Correct CHOICE is (c).
Discussion
1486.

Given the electric field in the region vec(E ) = 2xi, find the net electric flux through the cube and enclosed by it.

Answer»

Solution :`lambda=500 A = 5000 XX 10^(-10)m=5xx10^(-7)m`
Reflected ray : No CHANGE in wavelength and frequency.
REFRACTED ray : Frequency remains same, wavelength decreases
Wavelength `lambda = (lambda)/(mu)`
Discussion
1487.

A charge is placed at the centre of the line joining two charges q and q. For the system of three charges to be in equilibrium, what should be the value of Q?

Answer»

Solution :Here Q is in equilibrium as it is at the centre and NET FORCE on it is ZERO
For equilibrium of q, net force on it is zero
`1/(4 pi epsilon_0) (Qq)/((r//2)^2) + 1/(4 pi epsilon_0) (q^2)/(r^2) = 0`
where r is separation between q and q `IMPLIES Q = (-q)/(4)` .
Discussion
1488.

A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes in Young's double-slit experiment. What is the least distance (in m) from a central maximum where the bright fringes due to both the wavelengths coincide ? The distance between the slits is 3 mm and the distance between the plane of the slits and the screen is 150 cm.

Answer»


ANSWER :1.3
Discussion
1489.

An open switch in an RL circuit is closed at time t = 0, as shown in the following figure. The curve that best illustrates the variation of potential difference across the resistor as a function of time is

Answer»

1
2
3
4

Answer :C
Discussion
1490.

The advantage of BJT (Bipolar junction transistor) over vacuum tube triode is :

Answer»

BJT is cheap
It is LESS bulky
there is no heating ELEMENT in it
all of the above

Answer :D
Discussion
1491.

In a semiconducting material the mobilities of electrons and holes are mu_(e ) and mu_(h) respectively.Which of the following is true?

Answer»

`mu_(e ) gt mu_(h)`
`mu_(e ) lt mu_(h)`
`mu_( e)=mu_(h)`
`mu_( e)lt 0, mu_(h) gt 0`

SOLUTION :`mu_(e ) gt mu_(h)`
In SEMICONDUCTOR as electrons are free and as well as its mass is less so electrons is mobile where as HOLES are in covalentbond so do not move, THEREFORE `mu_( e) gt mu_(h)`.
Discussion
1492.

Unpolarized light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the transmitted light is one third of intensity of the inciddent beam?

Answer»

Solution :Intensity of the light transmitted through the first polarizer `I_(1)= I_(0)"/"2`, where `I_0` is the intensity of the incident unpolarized light.
Intensity of the light transmitted through the second polarizer is `I_(2)= I_(1) cos^(2) theta" where "theta` is the angle between the characteristic directions of the polarizer SHEETS.
But `I_(2)= I_(0)"/"3` (given) `""THEREFORE I_(2)= I_(1)cos^(2)theta= (I_0)/(2)cos^(2) theta= (I_0)/(3)`
`cos^(2) theta = 2"/"3 implies theta = cos^(-1) sqrt((2)/(3))`.
Discussion
1493.

Answer questions on the basis of your understanding of the followingparagraph and the related studied concept. Electric flux, in general , through any surface is defined as per relation: phi_E= int oversetto E. oversetto (ds), whereintegration has to be performed over the entire surface through which flux is required. The surface under consideration may be a closed one or an open surface. Whenflux through a closed surface is required we use a small circular sign on the integrationsymbol. Thus flux over a closed surface oint E= oint oversetto E. oversetto (ds) .it is customary to take the outward normal as positive in this case. A German physicist Gauss established a fundamental law to find electric flux over a closed surface. As per Gauss' law , the flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by in_0 . Mathematically oint oversetto (E) . oversetto (ds) =(1)/( in_0)[q_(en) ], where q_(cn)is the net charge enclosedwithin the surface. It is possible to derive Gauss' lawfrom Coulomb's laws. Gauss' lawcan be applied to obtain electric field at a point due to continuous charge distribution for a number of symmetric charge configurations. Consider a closed surface having certain charges both within and outside as shown. The total electric flux of the given closed surface is :(##U_LIK_SP_PHY_XII_C01_E03_011_Q01.png" width="80%">

Answer»

` + 2.26 xx 10^(5)V m `
` +7.91 xx 10 ^(5)V m`
` -2.26 xx 10 ^(5)V m `
` +1.36 xx 10 ^(-6) V m`

Solution :`phi_E=(1)/(in_0) sum q_(en) =(-2MU C)/(8.85xx10^(-12) )= )(-2XX10^(-6))/(8.85 xx 10 ^(-12)) =-2.26 xx 10 ^(5) V m ]`
Discussion
1494.

Rate of evaporation (volume evaporated per unit time) of water kept in an earthen pot is proportional to the volume of water present in the pot. It is observed that it takes 48 hrs for 75% of the water kept in the pot to evaporate. The empty pot was placed below a tap from which water leaks in small drops each of volume v_(0). The drops fall at a uniform rate of n drops per hour. Calculate the volume of water in the pot 24hrs after it was kept below the tap.

Answer»


ANSWER :`(12m_(0))/(LN2)`
Discussion
1495.

A body starts climbing up an inclined plane having inclination 30^@ and stops at distance of 17.3 m. If the angle of inclination is 60^@, how long will it be able to climb up starting with same speed ?

Answer»

8.6 m
10.0 m
12.9 m
17.3 m

Answer :B
Discussion
1496.

A mild steel wire of length 2L and cross-sectional areaA is stretched, well within elastic limit, horizontally between two pillars, A mass m is suspended from the midpoint of the wire. Strain in the wire is

Answer»

`x^2/(2L^2)`
`x/L`
`x^2/L`
`x^2/(2L)`

Solution :
CHANGE in LENGTH , `triangleL=AC-AO` (From `triangleACO` )
`=[L^2+x^2]^(1//2) -L=L[1+1/2 x^2/L^2]-L=x^2/(2L)`
`therefore` Longitudinal STRAIN = `(DELTAL)/L = (x^2//2L)/L=x^2/(2L^2)`
Discussion
1497.

Monochrmoatic light of wavelength 500nm from a narrow slit is incident on the double slit. If the separation of 10 fringes on the screen 1m away is 1cms. Find the slit separation.

Answer»

SOLUTION :`nbeta=1xx10^(-2)`
`:.beta=(10^(-2))/(n)=(10^(-2))/(10)/(10)=10^(-3)m`
`beta=(lambdaD)/(d)`
`impliesd=(lambdaD)/(beta)=2.5xx10^(-4)m`
Discussion
1498.

Who was Ram Narayan?

Answer»

OLYMPIC MEDAL winner
Swimmer
Commonwealth GAMES Winner
None of the above

Answer :A
Discussion
1499.

निम्न मे से किस आवेश का अस्तित्वसंभव नहीं है

Answer»

`3.2xx10^(-19) C`
`6.4xx10^(-19) C`
`2.4xx10^(-19) C`
`1.6xx10^(-19) C`

ANSWER :C
Discussion
1500.

An electron having mass 9.1xx10^(-31)kg, charge 1.6xx10^(-19)C and moving with the velocity of 10^(6)m//s enters a region where magnetic field exists. If it describes a circle of radius 0.2 m then the intensity of magnetic field must be _______ xx10^(-5)T.

Answer»

14.4
5.65
2.84
1.13

Solution :`(MV^(2))/r=Bqv`
`thereforeB=(mv)/(qr)=(9.1xx10^(-31)xx10^(6))/(1.6xx10^(-19)xx0.2)`
`thereforeB=2.84xx10^(-5)T`
Discussion