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41251.

A box of mass 2 kg is placed on the roof of a car. The box would remain stationary untill the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and g = 10 ms^(-2). This maximum acceleration of the car, for the box to remain stationary, is

Answer»

`8ms^(-2)`
`6ms^(-2)`
`4ms^(-2)`
`2ms^(-2)`

SOLUTION :For the box to remain stationary
`ma_("MAX")=mu_(s)MG`
`a_("max")=mu_(s)g=0.2xx10ms^(-2)=2ms^(-2)`
Discussion
41252.

In a regular hexagon each corner is at a distance .r. from the centre. Identical charges of magnitude .Q. are placed at 5 corners. The field at the centre is (K = (1)/(4pi in_(0)))

Answer»

`KQ//R^(2)`
`(6KQ)/(r^(2))`
`(5KQ)/(r^(2))`
zero

Answer :A
Discussion
41253.

What do you understand by potential barrier ?

Answer»

Solution :Potential barrier . When a p-n junction is formed, there occurs migration of holes and ELECTRONS in TWO regions and the two sections of the junction diode no longer remain neutral. The p-section of the junction diode becomesa slightly NEGATIVE while the n-section is rendered positive. Due to this, there is potential graident in the depletion LAYER, negative on p-side and positive on n-side , and potential difference developed across the junction diode due to migration of majority carriers is CALLED potential barrier.
Discussion
41254.

By what percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by 10% at a constant temperature ?

Answer»

<P>8·1%
9·1%
10·1%
11-1%

Solution :`P_(1)V_(1)=P_(2)V_(2)`
`rArrP_(2)=(P_(1)V_(1))/(V_(2))=(P_(1)V_(1))/((V_(1)-10/100V_(1)))`
`rArrP_(2)=10/9P_(1)`
% Increase in PRESSURE `=(P_(2)-P_(1))/(P_(1))xx100`
`=(10/9P_(1)-P_(1))/(P_(1))xx100`
`=100/9=11.1%`
Thus correct choice is (d).
Discussion
41255.

What would be the energy of the magnetic field of the coil of the previous problem, if its core were of a nonferromagnetic material? What is the source of oxcess energy in the case of a ferromagnetic core?

Answer»


Solution :"The energy of the MAGNETIC field in the ABSENCE of the ferromagnetic core is `W_(m) 1//2mu_(0) H^2V` It is significantly less than the energy of the magnetic field in the presence of a forromagnetic core, DESPITE the fact that the current Mowing through the winding is the same in both cases. The explanation is that in the absence of a ferromagnetic core the current in the coil atlnins ils stationary VALUE very quickly and the work performed by the power SUPPLY in inducing the magnetic field is much less than in the presence of such a core.
Discussion
41256.

Intensity at A due to source is I. Without concave mirror, then find out the intensity of A after placing concave mirror.

Answer»

Solution :`P=Ixx4pi (10)^(2)`
Intensity at `P=P/("Area")`
`=P/(4pi (30)^(2))=I/9`
Now `R/60=x/20`
`x=r/3`
Power INCIDENT on mirror
`P_(p)=I/9xx PI R^(2)`
This power will be incident on `pix^(2)` area
So Intensity from mirror `=(Ixxpi R^(2))/(9pi (R//3)^(2))=I`
So TOTAL, `I = I + I = 2I`
Discussion
41257.

Decide the position of the image formed by the given combination of lenses.

Answer»

SOLUTION :V= 25 CM
Discussion
41258.

When light of a certain wavelength is incident on a plane surface of a material at a glancing angle 30^@, the reflected light is found to be completely plane polarized Determine Refractive index of given material.

Answer»

Solution :Angle of incident light with the surface is `30^(@)`. The angle of incidence `=90^(@)-30^(@)= 60^(@)`. Since reflected light is completely polarized, THEREFORE incidence takes place POLARIZING angle of incidence `theta_p`.
`theta_(p)= 60^(@)`
Using Brewster.s law `mu = TAN theta_(p)= tan 60^(@)"", mu = SQRT(3)`.
Discussion
41259.

Intensity at A due to source is I. Without concave mirror, then find out the intensity of A after placing concave mirror.

Answer»

Solution :
`P=IxxP`
`=Ixx4pi (60)^(2)`
Intensity at `P=P/("AREA")`
`=(4pi (60)^(2) I)/(4pi (30)^(2))`
`=4I`
Now `Delta PAB ~ Delta ACD`
`rArr (AP)/(PB)=(AC)/(CD)`
`rArr CD=30/60xxR=R/2`
Energy at area of R radius `=4 I=pi R^(2)`
Now energy will fall on the screen but at an Area of radius `R/2`.
So intensity from mirror `=(4IxxpiR^(2))/(pi(R//2)^(2))=16 I`
Total Intensity `=16 I+I=17 I`
Discussion
41260.

Intensity at A due to source is I. Without concave mirror, then find out the intensity of A after placing concave mirror.

Answer»

SOLUTION :
`I=P/(4PI (10)^(2))`
`1/f=1/v+1/u`
`RARR (-1)/10=1/v-1/10`
`rArr v=oo`
Intensity at A due to reflection `=I`.
TOTAL `=I+I=2I`
Discussion
41261.

In a n-p-n transistor circuit, the collector current is 10 mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?

Answer»

The EMITTER current will be 8 mA.
The emitter current will be 10.53 mA.
The base current will be 0.53 mA.
The base current will be 2 mA.

Solution :Here, `I_(C )=95% I_(E)=0.95I_(E )`
`therefore 10=0.95I_(E ) (because (I_(C )=10mA)`
`therefore I_(E )=(10)/(0.95)=(1000)/(95)=10.526`
`therefore I_(E )~~10.53mA`
Here, `I_(B)=5% I_(E )=10.53xx(5)/(100)=0.5265mA`
`therefore I_(B) ~~ 0.53mA`
Discussion
41262.

An electric box contains three e.m.f sources as shown in the figure

Answer»

EMF of the ELECTRIC box is `1/3` V
point B is at higher potential than point A
internal resistance of the box is `5/3Omega`
terminal voltage for `2V` SOURCE is `4/3`v

Answer :A::B::C::D
Discussion
41263.

A double convex lens made of glass (refractive index n=1.5) has the radii of curvature of both the surfaces as 20cm. Incident light rays parallel to the axis of the lens will converge at a distance L such that

Answer»

`L=20cm`
`L=10cm`
`L=40cm`
`L=20/3cm`

Solution :(a) Here `n=1.5` as per sign convention followed
`R_(1)=+20cm` and `R_(2)=-20cm`
`:.1/f=(n-1)(1/(R_(1))-1/(R_(2)))=(1.3-1)[1/((+20))-1/((-20))]`
`=0.5xx2/20=1/20impliesf=20cm`
Incident rays travelling parallel to the axis of lens will converge at its SECOND principal focus. Hence `L=+20cm`.
Discussion
41264.

What is the justification in appIyin principIe of Iinear superposition of wave dispIacement in expIaining the distributions in interference and diffraction patterns.

Answer»

Solution :The Iiear COMBINATION of wave EQUATION is AISO a wave equation. This is the very basis of SUPERPOSITION of waves .
Discussion
41265.

Draw the variation of magnetic field(B) with magnetic intensity(H) when ferromagnetic material is subjected to a cycle of magnetisation.

Answer»

SOLUTION :
Discussion
41266.

Magnetic susceptibility of ferromagnetic substance is ________ .

Answer»


ANSWER :`(chigtgt1)`
Discussion
41267.

A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature?

Answer»

Solution :Power of convex lens `= 1// 0.25 = 4 D`
Power of concave lens `= - 1 / 0.20 = - 5D`
Power of the combination, `P = P_(1) + p_(2) = -1D`
NATURE `:` Diverging
Discussion
41268.

A ball of mass 150 g moving with an acceleration 20 m//s^(2) is hit by a force, which acts on it for 0.1 s. The impulsive force is

Answer»

0.5 N-s
0.1 N-s
0.3 N-s
1.2 N-s

Answer :C
Discussion
41269.

Inside a parallel plate capacitor the electric field E varies with time as t^2. The variation of induced magnetic field with time is given by ......

Answer»

`t^2`
no variation
`t^3`
t

Solution :`E PROP t^2`
`B=(DE)/(dt)`
`=d/(dt)(kt^2)` k is proportionally CONSTANT
=2kt
`THEREFORE B prop t`
Discussion
41270.

If n(A) = m and n(B) = n thenTotal number of relations from A to B is-

Answer»

`2^m`
`2^n`
`2^(MN)`
NONE of these

Answer :C
Discussion
41271.

Radioactive ._27^60Co is transformed into stable ._28^60Ni by emitting two gamma-rays of energies

Answer»

1.33 MeV and 1.17 MeV in succession
1.17 MeV and 1.33 MeV in succession
1.37 MeV and 1.13 MeV in succession
1.13 MeV and 1.37 MeV in succession

Solution :The successive EMISSION of gamma RAYS of energies 1.17 MeV and 1.33 MeV from the de-excitation of `._28^60Ni` NUCLEI formed from `beta`-decay of `._27^60Co`. This process is shown in the figure through an energy LEVEL diagram .
Discussion
41272.

What is decay law ?

Answer»

SOLUTION :It states thatnuclei disintegrating PER unit time is PROPORTIONAL to the number of undisentegrated NUCLEI present at that instant
`dN/dt=-lambdaN`
Discussion
41273.

An LCR series circuit consists L = 80 mH, C = 60 muFand R = 15Omega. The combination is connected to a 230 V, 50 Hz AC supply. Calculate the impedance in the circuit.

Answer»

Solution :Z = 31.77`Omega ` , Average power ACROSS R = 782 W, Average power across L and C = ZERO
Discussion
41274.

In Young.s double slit experiment the band width is minimum for the colour

Answer»

red
yellow
green
blue

Answer :D
Discussion
41275.

Explain the parinciple and working of a moving coil galvanometer .

Answer»

Solution :Moving coil galvanometer is a device which is USED to indicate the FLOW of current in an electrical circuit.
Principle When a current carrying loop is placed in a uniform magnetic field it experiences a torque .
Construction: A moving coil galvanometer consists of a rectangular coil PQRS of insulated thin copper wire. The coil contains a large number of turns wound over a light metallic frame . A cylindrical soft - iron core is placed symmetrically inside the coil as shown in Figure.
The rectangular coil is SUSPENDED freely between two pole pieces of a horse - shoe magnet .
The upper end of the rectangular coil is attached to one end of fine strip of PHOSPHOR broze W and the lower end of the coil is connected to a hair spring S which is also made up of phosphor bronze . In a fine suspension strip W, a small plane mirror is sttached in order to measure the deflection of the coil with the help of lamp and scale arrangement . The other end of the mirror is connected to a torsion head T . In order to pass electric current through the galvanometer, the suspension strip W and the spring S are connected to terminals .

Working : Consider a single trun of the rectangular coil PQRSwhose length be l and breadth b. `PQ = RS = l and QR = SP = b ` . Let I be the electric current flowing through therectangular coil PQRS as shown in Figure.
The horse - shoe magnet has hemi - spherical magnetic poles which produces a RADIAL magnetic field . Due to this radialfield , the sides QR and SPare always parallel to the B - field ( magnetic field) and experience no force. The sides PQ andRSare always parallel to the B - field and experience force and due to this, torque is produced.
For single turn, the deflection couple as shown in Figure is
` tau = bF = bBIl= (lb) BI = ABI `
since, area of the coil `A = lb ` For coil with N turns, we get
`tau = ` NABI ......(1)
Due to this deflecting torque, the coil gets twisted and restoring torque ( also known as restoring couple ) is developed . Hence the magnitude of restoring couple is proportional to the amount of twist `theta ` . Thus
`tau = K theta`.....(2)
where K is the restoring couple per unit twist or torsional constnat of the spring .
At equilibrium, the deflection couple is equal to the restoring couple. Therefore by comparing equation (1) and (2) , we get
`NABI - K theta `
`rArrI = K/(NAB) theta`.....(3)
(or) `I = G theta`
where , `G = K/(NAB) `is called galvanometer constant or current reduction factor of the galvanometer .
Discussion
41276.

The Fig 12.17 shows energylevel diagram of hydrogen atom. (a) Find thetransition whichresultsin the emission of aphoton of wavelenght496 nm . (b) Whichtransition correspondsto theemissionof radiationof maximumwavelenght? Justify your answer .

Answer»

SOLUTION :(a) transition n = 4to n = 2
(B) transition n =4to n = 3
Discussion
41277.

Assertion:A sail boat cannot be propelled by air blown at the sail from a big fan attached to the boat Reason:Action of the air from the fan and reaction of the sail, both act on the boat

Answer»

Both ASSERTION and REASON are true and Reason is the CORRECT EXPLANATION of Assertion
Both Assertion and Reason are true but Reason is not a correct explanation of Assertion
Assertion is true but Reason is false
Assertion is false but Reason is true

Answer :A
Discussion
41278.

In a Young's double slit experiment set up, source S of wavelength 500 nm illuminates two slit S_(1) and S_(2) which act as two coherent sources. The source S oscillates about its won position according to the equation y = 0.5 sin pi t where y is in mm and t in seconds. The minimum value of time t for which the intensity at point P on the screen exactly infront of the upper slit becomes minimum is

Answer»

1s
2s
3s
`1.5 s`

ANSWER :A
Discussion
41279.

A dynamo acts as a

Answer»

CONVERTER of energy
Source of ELECTRIC charge
Source of MAGNETIC charge
Source of energy

Answer :B
Discussion
41280.

An Ac source is rated 222 V, 60 Hz. The average voltage is calculated in a time interval of 16.67 ms.IL

Answer»

MUST be zero
May be zero
is NEVER zero
Is `(111sqrt(2))` V

Answer :A
Discussion
41281.

Light of wavelength 589 nm is used to view an object under a microscope. The aperture of the objective has a diameter of 0.900cm.What happens to the limit of resolution if the objective is immersed in oil? Explain.

Answer»

SOLUTION :Decreases.Refractive INDEX of oil is more than that of air.Hence WAVELENGTH of the LIGHT decreases.
Discussion
41282.

On the basis of electrical conductivity, how we classify solids ?

Answer»

Solution :They are classified into THREE.(i) CONDUCTORS, (II) INSULATORS,(iii) Semiconductors.
Discussion
41283.

In figure a rectangular loop is being pulled to the right, away from the long straight wire carrying current I in the upward direction. The induced current in the loop is ____

Answer»

zero
in clockwise direction
in anticlockwise direction
None of the above

Solution :
The magnetic FIELD DUE to wire carrying current go into the plane of the loop shown by `ox` .
When the loop moves away, on right, from the wire the FLUX linked with it-decreases. Hence, EMF will be so induced as the increase the flux. This happen only when the induced emf is in clockwise direction.
Discussion
41284.

A long narrow slit is illuminated by blue light and the diffraction pattern is obtained on a white screen:-What do you mean by limit of resolution of an optical instrument?

Answer»

Solution :The minimum DISTANCE between two objects at which they can be observed as separated by an OPTICAL instrument is CALLED the limit of resolution of the instrument.
Discussion
41285.

A spherical conducting shell of inner radius r_(1) and outer radius r_(2) has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Answer»

Solution :(a) `-q(4PI r_(1)^(2)), (Q+q)//(4pi r_(2)^(2))`
(b) By Gauss’s law, the NET charge on the inner surface enclosing the cavity (not having any charge) must be zero. For a cavity of arbitrary shape, this is not enough to claim that the electric FIELD inside must be zero. The cavity may have POSITIVE and negative charges with total charge zero. To dispose of this possibility, take a closed loop, PART of which is inside the cavityalong a field line and the rest inside the conductor. Since field inside the conductor is zero, this gives a net work done by the field in carrying a test charge over a closed loop. We know this is impossible for an electrostatic field. Hence, there are no fieldlines inside the cavity (i.e., no field), and no charge on the inner surface of the conductor, whatever be its shape.
Discussion
41286.

Band width of an optical fiber is

Answer»

more than 100 GHz
few kHz
less than 1MHz
less than 1GHz

Answer :A
Discussion
41287.

There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is

Answer»

`+(1)/(2)`
`-(1)/(2)`
`+(1)/(4)`
`-(1)/(4)`

ANSWER :B
Discussion
41288.

An elastic ciruclar wire of length l carries a current I. It is placed in a uniform magnetic field vecB (out of paper) such that is plane is perpendicular to the direction of vecB . The wire will experience

Answer»

No force
A STRETCHING force
A COMPRESSIVE force
A torque

Answer :N//A
Discussion
41289.

Bullwinkle in reference frame S' passes you in reference frame S along the common direction of the x' and x axes, as in Fig. 36-9. He carries three meter sticks: meter stick 1 is parallel to the x' axis, meter stick 2 is parallel to the y' axis, and meter stick 3 is parallel to the z' axis. On his wristwatchhe counts off 10.0 s, which takes 30.0 s according to you. Two events occur during his passage. According to you, event 1 occurs at x_(1)=33.0m and t_(1)=22.0 ns, and event 2 occurs at x_(2)=53.0 and t_(2)=62.0 ns. According to your measurements, what is the length of (a) meter stick 1, (b) meter stick 2, and (c ) meter stick 3? According to Bullwinkle, what are (d) the spatial separation and (e ) the temporal separation between events 1 and 2, and (f) which event occurs first?

Answer»


ANSWER :(a) 0.333m; (B) 1.00m; (C ) 1.00m; (d) 26.1m; (E ) -68.7 ns; (f) EVENT 2 occurrsbefore event 1
Discussion
41290.

A coherent portion with a width of d = 0.5mm is cut out of a convergent lens having a focal length of 20 cm. Both halves are tightly fitted against each other and a point source of monochromatic light (A = 2500 Å)is placed in front of the lens at a distance of 10 cm. Find the maximum possible number of interference bands that can be observed on the screen.

Answer»
Discussion
41291.

Why do the electrostatic field lines not form closed loop?

Answer»

Solution :ELECTROSTATIC fieldlines cannot form CLOSED LOOPS, because POTENTIAL decreased in the direction of the field and itsfield energy is consumed by the BODY.
Discussion
41292.

If a liquid is placed in a vertical cylinerical vessel and the vessel is rotated about its axis, the liquid will take the shape of figure.

Answer»




Solution :For the given ANGULAR VELOCITY of rotation, the centrifugal force `F PROP r`. Therefore, more liquid will be ACCUMULATED NEAR the wall of tube and the liquid meniscus will become concave upwards.
Discussion
41293.

If the momentum of a body is increased by 20%, find the increase in kinetic energy.

Answer»

SOLUTION :`K=P^2/(2M),K^1=(((120p)/(100))/(2m))^2=1.44 p^2/(2m)=1.44 k`
THEREFORE DeltaK=K^1-K=0.44K=44%`
Discussion
41294.

STATEMENT-1: A current flows in a conductor only when there is an electric field within the conductor. Because STATEMENT-2: The drift velocity of electron in presence of electric field decreases.

Answer»

Statement-1 is TRUE , Statement-2 is True , Statement-2 is a CORRECT explanation for Statement-1.
Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-1.
Statement-1 is True , Statement-2 is False
Statement-1 is False , Statement-2 is True

ANSWER :C
Discussion
41295.

A cell of emfepsiand internal resistance 'r' is connected across a variable load resistor 'R'. Draw the plots of the terminal voltage 'V' versus (i) R and (ii) the current I. It is found that when R = 4 Omega , the current is 1 A and when R is increased to 9 Omega , the current reduces to 0.5 A. Find the values of the emf epsiand internal resistance r.

Answer»

Solution :We know that current`I = (epsi)/(R + r)`
When `R = 4 Omega , I = 1A , ` HENCE`1 = (epsi)/( 4+ r)`.....(i)
Again if `R. = 9Omega` then `I. = 0.5A` , hence ` 0.5 = (epsi)/(9 + r)`..(II)
From (i) and (ii) , we get , r= 1 `Omega` and ` epsi` = 5V
Discussion
41296.

Sensitive electrical instruments in the vicinity of an electromagnet may be damaged when the electromagnet is turned on or off.

Answer»

Solution :True - When the ELECTROMAGNET is turned on or off, the magnetic FLUX in the neighbourhood changes. As a result of which an induced current MAY be set up in the electrical INSTRUMENT and may DAMAGE it.
Discussion
41297.

A beam of unpolarised light passesthrough a tourmaline crystal A andthen it passes through a second tourmaline crystalB oriented so that isprincipalpalne is parallel to that of A . The intnsity of emergent light is I_(0).Now B is rotatedby 45^(0) aboutthe ray . The emergent lightwill have intensity

Answer»

`(I_(0))/2`
`(I_(0))/(SQRT(2))`
`I_(0) sqrt(2)`
`2I_(0)`

ANSWER :A
Discussion
41298.

What will happen when we put kerosene oil on the waterbody?

Answer»

It will help mosquitoes FIND their breeding ground
It will help mosquitoes to breed
It will take away AIR from the larva
None of the above

Answer :C
Discussion
41299.

What is simple microscope ? Obtain the equation of magnification for the image formed at normal vision distance.

Answer»

SOLUTION :A simple magnifier or microscope is a converging lens of SMALL focal length.
In order to use such a lens as a microscope, the lens is held near the object one focal length away or less and the eye is positioned close to the lens on the other side. An erect, magnified and VIRTUAL image of the object at a distance so that is can be viewed comfortably i.e., at 25 cm or more. If the object is at a distance f the image is at infinity.

If the object is at a distance SLIGHTLY less than the focal length of the lens, the image is virtual and closer than infinity. ALTHOUGH the closest comfortable distance for viewing the image is when it is at the near point (distance D = 25 cm), it causes some strain on the eye.
The linear magnification m for the image formed at the near point D, by a simple microscope.
`m=v/u` ... (1)
But, lens formula,
`1/f=1/v-1/u therefore 1/u=1/v-1/f`
Multiplying by .v. on both sides,
`therefore v/u=v/v-v/f` ... (2)
From equation (1) and (2),
`m=1-v/f` [`because v/u`=magnification m]
But image distance v=-D
`therefore m= 1-(-D)/(f)`
`therefore m= 1+D/f`
Discussion
41300.

Ice at 0^(@)C enclosed in an adiabatic shell and is compressed to a pressure of 600 atm. It is known that an increase in the pressure of 138 atm causes the melting point of ice to drop by 1 K. Assuming the phase diagram in this part to be linear, find the fraction of the ice that is going to melt.

Answer»

SOLUTION :First find the melting points `t=-Deltap//k=-4.35^(@)C`. As the ICE is cooled to this temperature, the heat liberated is `Q=mcDeltat= mc|t|` This heat will be SPENT to melt the ice `Q_(m)=m_(1) lambda`. Hence the fraction of the ice that will melt will be `x=(m_(1))/(m)=(c|t|)/(lambda)`.
In the calculation assume that the specific heat of ice and the heat of melting remain CONSTANT.
Discussion