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2801.

A parallel plate capacitor has plates of area A separated by distance ‘d’ between them. It is filled with a dielectric which has a dielectric constant that varies ask(x)=K(1+ax) where ‘x’ is shown in figure. If (al)lt lt 1 , the total capacitance of the system is best given by the expression:

Answer»

`(AK in_0)/d ( 1 + alphal)_`
`(Ak in_0)/d (1 + (alpha l)/2)`
`(A in_0K)/d(1 + ((alpha l)/2)^2)`
`(A in_0K)/d (1 + (alpha^2l^2)/2)`

Solution :`DC=(K in_(0)A/ldx)/d`
`C_(eq)=INT(dc)=(K in_(0)A)/(ld) dx`

`-(in_(0)A)/(ld)int_(0)^(l)(1+alpha X)dx`
`=(K in_(0)A)/(ld)(l+(alphal^(2))/2)""(Kin_(0)A)/d(1+(alphal)/2)`
Discussion
2802.

The magnetude of a magnetic induction due to a short magnetic dipole of moment 8 Am^2 at a distance of 1 m , from its centre along a line making an angle of 30^@ with the equator is :

Answer»

`1.0584 XX 10^(-6) (WB)/m^2`
`1.0584 xx 10^6 (Wb)/m^2`
`1.5 xx 10^(-6) (Wb)/m^2`
`1.5 xx 10^6 (Wb)/m^2`

ANSWER :A
Discussion
2803.

A battery of 2V and internal resistance 2Omega is used in wheatstone bridge. Find the current through the galvanometer in the unbalanced position of the bridge when P=1Omega, Q=2Omega,R=2Omega,S=3Omega and G=4Omega.

Answer»


ANSWER :0.011 A
Discussion
2804.

A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with time at a rate (dvecB)/(dt). Loop 1 of radius R gt r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure below. Then the emf generated is

Answer»

In loop-1 `-(dvecB)/(dt)piR^2` and in loop-2 zero
In loop-1 `-(dvecB)/(dt)pir^2` and in loop-2 zero
Zero in both loop
In loop-1 `-(dvecB)/(dt) pir^2` and in loop=2 -`(dvecB)/(dt) pir^2`

SOLUTION :In loop-1
`phi=vecA.vecB=AB cos theta`
`THEREFORE` Induced EMF `EPSILON=(dphi)/(dt)=-(d(AB cos theta))/(dt)`
`epsilon=-A cos theta. (dB)/(dt)`
Here `vecA||vecB` hence `theta=0^@`
`therefore epsilon=-A . (dvecB)/(dt) "" [ cos 0^@=1]`
`therefore epsilon =-pir^2(dvecB)/(dt)`
For loop -2
Induced emf `epsilon=0` because loop-2 is not magnetic field .
Discussion
2805.

A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, then the percentage increase in the capacitance will be

Answer»

`400%`
`66.6%`
`33.3%`
`200%`

ANSWER :A
Discussion
2806.

A square plate of side length 4 cm is placed in a magnetic field of induction 20 mT in a direction making an angle 30^@ with the plane of coil then the magnetic flux linked with the coil is ?

Answer»

SOLUTION :Area `=(4 xx 10^(-2))xx(4xx10^(-2))`
`=16xx 10^(-4)m^2`
`theta=90-30=60^@`
`vecB=20XX 10^(-3)T`
`PHI =BA cos theta`
`=(20xx 10^(-3))(16 xx 10^(-4))xx1/2`
`=10xx10^(-3)xx16xx109^(-4)`
`=16xx 10^(-6)`
`=16 mu wb`
Discussion
2807.

20Omega resistor is connected with the A.C. source of V = 282sin (120 pit), the current passing through the resistor will be ……..

Answer»

14.1 A
10A
7.05 A
5A

Solution :Comparing EQUATION `V=282 SIN (120pit)`
with the equation `V=V_m sin omegat`
`V_m`=282 V, `OMEGA=120 PI` rad/s , R=20 `Omega`
`THEREFORE I_(rms)=I_m/sqrt2`
`=V_m/(sqrt2R)=282/(1.41xx20)`=10 A
Discussion
2808.

One milligram of matter converted into energy will give :

Answer»

`9xx10^5` J
`9XX10^3` J
90 J
`9XX10^10` J

Answer :D
Discussion
2809.

Mutual inductance of two coils can be increased by _____

Answer»

DECREASING the NUMBER of turn in the coils.
increasing the number of turn in the coils.
winding the coils on wooden cores
none of the above.

Solution :When the number of turns of COIL increases, the flux LINKED with coil also increases and so mutual inductance increases with increasing the number of turns.
`therefore N_2phi_2 prop I_1`
`therefore N_2phi_2 = MI_1 "" therefore M prop N_2`
Discussion
2810.

Dimensional formula of magnetic dipole moment is ......

Answer»

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

Solution :We have `m=NIA`
`THEREFORE [m] = A^(1) L^(2) = M^(0) L^(2) A^(1)`
Discussion
2811.

If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer»

SOLUTION :`2 V, 4V, 6 V`
Discussion
2812.

A point object O is placed midway between on the common axis of two concave mirrors of equal focal length. If the final image is formed at the position of the object, the separation between two mirrors is ........ . Focal length of mirrors is f. HINT:- A situation is depicted n the figure [ Note : A another possible situation for which object and its image coincide is when distance between mirrors is 4f ] .

Answer»

f
2f
`3/2f`
`1/2f`

Solution :It is clear from the defination of focal POINT that when object is placed on focal point, ray incident on mirror from object becomes parallel to axis of mirror after reflection.
`therefore` In FIGURE position of object is O and position of image is O. must be on the focal point on axis of CONCAVE mirror. `therefore`DISTANCE between both mirror = 2f
Note : Similarly if object and image O. are on the centre of curvature distance between mirrors = 4f.
Discussion
2813.

Rarefield Hg gas whose atoms are practically all in the ground state was lighted by a mercury lamp emitting a resonanace line of wavelength lambda= 253.65nm. As a result, the radiation power of Hg gas at that wavelength turned out to be P=35mW. Find the number of the atom in the state of resonance exitation whose mean lifetime is tau=0.15 mu s.

Answer»

Solution :As a result of the lighting by mercury lamp a number of ATOMS are pumped to the EXCITED state. In equilibrium the number of such atoms is `N`. Since the mean life TIME of the atom is `T`, the number decaying per UNIT is `(N)/(tau)`. Since a photon of energy `(2piħc)/(lambda)` results from each decay, the total RADIATED power will be `(2piħc)/(lambda).(N)/(tau)`. This must equal `P`. Thus
`N=P tau//(2piħc)/(lambda)=(P tau lambda)/(2 piħ)=6.7xx10^(9)`
Discussion
2814.

The power of a carnot engine is

Answer»

100
maximum
very very small
50

Answer :C
Discussion
2815.

A transistor is a/an

Answer»

CHIP
insulator
semiconductor
METAL

ANSWER :C
Discussion
2816.

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be:

Answer»

`1:4`
`1:1`
`1:5`
`1:2`

SOLUTION :`hv=w+1//2mv_("max")^(2)`
`i.e. E=w+1//2v_("max")^(2)`
`:.1//2mv_("max")^(2)=E-w`
`:.(v_(1)^(2))/(v_(2)^(2))=(E_(1)-w)/(E_(2)-w)`
`(1eV-0.5eV)/(2.5eV-0.5eV)`
`=(0.5)/(2)=(1)/(4)`
`:.(v_(1))/(v_(2))=1//2`
Discussion
2817.

A cell of emf (E) and internal resistance r is connected across a variable external resistance . R. The graph of terminal potential difference V as a function of R is

Answer»




SOLUTION :`V=(ER)/((R+r))=E/((1+r/R))`
Discussion
2818.

The pass-axes of two polarisers were kept such that the incident unpolarised beam of intensity I_0 , gets completely blocked. Another polariser was introduced in between these two polarisers with its pass-axis 60^@with respect to the pass-axis of the first one. The output intensity would then become

Answer»

`0`
`3/32 l_0`
`3/16 l_0`
`3/8 l_0`

SOLUTION : Since the pass axes of two polarisers `P_1` and `P_2`were KEPT such that the incident unpolarised beam of intensity `I_0`gets completely blocked. This means that both polarisers were kept cross (perpendicular of their pass axes) to each other. When third polariser `P_3`is introduced between `P_1` and `P_2`such that the angle between the pass axes of `P_1` and `P_3`is `60^@` , i.e.`theta_1 = 60^@`
intensity of polarised light emerging from first polaroid, `I_1 = (I_0)/(2)`
Intensity of polarised light (`I_3` ) emerging from polaroid `P_3`is given by law of Malus,
i.e.,`I_3 = I_1 cos^2 THETA`
` = (I_0)/(2) cos^2 60 = (I_0)/(2) 1/4`
`I_3 = (I_0)/(8)`
Now, the angle between pass axis of `P_3`and `P_2`
` theta_2 = 90^@ - 60^@ = 30^@`
` therefore ` Intensity of polarised light emerging from lost (second) polaroid `P_2`
`I_2 = I_3 cos^2 30^@= (I_0)/(8) . 3/4 = (3I_0)/(32)`
Discussion
2819.

Statement I: In seriesL−C−R AC circuit, current and voltage are in same phase at resonance. Statement II: In series L−C−R AC circuit, resonant frequency does not depend on the value of resistance. Hence current, at resonance, does not depend on resistance.

Answer»

STATEMENT I is TRUE, statement II is true, statement II is a correct EXPLANATION for statement I.
Statement I is true, statement II is true, statement II is not a correct explanation for statement I
Statement I is true, statement II is false
Statement I is false, statement II is true

ANSWER :D
Discussion
2820.

The refractiveindex of glass w.r.t. amedium is 4/3 . IfV_m-V_g=6.25 xx 10^(7) m//s , then the velocity of light in the medium will be

Answer»

` 2.5 xx 10^(8) m//s `
`2.25 xx 10^(8) m//s `
`1.875 xx10^(8) m//s `
`1.5xx10^(8) m//s `

Solution :`""_(m) n_(G)=(V_(m))/(V_g)=(4)/(3) :. V_m=(4)/(3)V_g`
`:. V_m-V_g=V_g((4)/3-1)=(V_g)/(3)`
`:. (V_g)/(3)=6.25 xx 10^(7)`
`:. V_g=1.875 xx 10^(8) m//s `
`V_m=(4)/(3)V_g=2.5 xx 10^(8) m//s `
Discussion
2821.

An 80 kg stuntman jumps out of a window that's 45m above the ground Q. How fast is he falling when he reaches ground level?

Answer»

Solution :His gravitational potential energy turns into kinetic energy: MGH=`(1)/(2)MV^(2)`, so `v=sqrt(2gh)=sqrt(2(10)(45))=30m//s`
(You COULD also have answered this question using BIG FIVE #5).
Discussion
2822.

The workdone in rotating an electric dipole of moment P in an electric field E through an angle theta from the direction of the field is

Answer»

pE `(1 - COS THETA)`
2pE
zero
pE cos `theta`

SOLUTION :W = pE `(cos theta_(0) - cos theta)"" [ theta_(0) = cos theta, cos theta = 1]`
W = pE `(1 - cos theta)`
Discussion
2823.

If epsilon_(0) and mu_(0) represent the permittivity and permeability of vaccum and epsilon and mu represent the permittivity and permeability of medium, then refractive index of the medium is given by :

Answer»

`SQRT((epsilon_(0)mu_(0))/(epsilon MU))`
`sqrt((epsilon mu)/(epsilon_(0)mu_(0)))`
`sqrt((epsilon)/(mu_(0)epsilon_(0)))`
`sqrt((mu_(0)epsilon_(0))/(epsilon))`

Solution :`mu=(c)/(V)=(1)/(sqrt(mu_(0) epsilon_(0))) XX (sqrt(mu epsilon))/(1)=sqrt((mu epsilon)/(mu_(0)epsilon_(0)))`
Discussion
2824.

Whenever a hydrogen atom emits a photon in the Balmer series, it

Answer»

NEED not emit any more photon.
may emit ANOTHER photon in the Paschen series.
must emit another photon in the Lyman series.
may emit another photon in Balmer series.

ANSWER :C
Discussion
2825.

Even carnot engine cannot give 100% efficiency because we cannot

Answer»

REACH ABSOLUTE ZERO temperature
find IDEAL sources
eliminate friction
prevent radiation

Answer :A
Discussion
2826.

What are beats ? Obtain an expression for the beat frequency ? Where and howare beatsmade us of ?

Answer»

Solution :Beats `:` Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing and waning in the intensity of the resultant sound waves at regular intervals of time are called Beats.
If `v_(1)` and `v_(2)` are the frequencies of two sound notessuperimposed ALONG the same direction, no of beats heard per second `= Delta v = v_(1)- v_(2)` .
Maximumno. of beats heard per heard persec is 10 due to persistence of hearing .
Expression for the beat frequency `:`
(1) Consider the two wave trains of equal amplitudebut of nearlyequal frequencies .
(2) Let the frequencies of the waves be `v_(1)` and `v_(2)`. SAY `v_(1)` is slightly greater than `v_(2)`.
(3) Let the beatperiod be T seconds.
(4)No.of vibrations , made by the firstwave train in T seconds `= v_(1)T `
`[ :'` no. of oscillations in 1 sec = v]
`[ :'` no. of oscillationsin T sec = vt ]
(5) No. ofvibrations ,made by the second wave train inT secons =`v_(2) T `
(6)During the timeinterval T, the first wave trainwould have completed one vibration more than the second wave train.
(7) Hence, `v_(1)T -v_(2)T = 1 ` or`v_(1) - v_(2)= (1)/(T )`
(8) Since, T is the beat period, no.of beats per seconds`= (1)/(T )`
(9) Hence the beatfrequency `= (1)/( T ) = v_(1) - v_(2) = Deltav `
(10)That is the beat frequency is the difference between the frequencies of the two wave trains.
PRACTICAL applications of beats `:`
(i) Determinationof an unknown frequency `:` Out of two tunning forks, one is loaded with wax and the other is filled. Theexcited tunning forks are close together and no. of beats can be heard.Then after unknownfrequencies of them will be found practically.
(ii) For tuning musical instruments `:` Musicians use the beatphenomenon in tuning their musicalinstruments.
(III) For producing COLOURFUL effects in music`:` Sometimes , a rapid succession of beats is knowingly introduced in music . This produces an effect similarto that of human voice and is appreciated by the audience.
(iv) For detection of Marsh gas ( dangerous gases) in mines.
Discussion
2827.

Slope of V_(0)toV graph shows…..

Answer»

H
`(h)/(e)`
EH
`e^(h)`

Solution :Equation of photo-electric EFFECT,
`eV_(0)=HV-hv_(0)`
`thereforeV_(0)=(h)/(e )V-(hv_(0))/(e )`
COMPARING with y=mx+c
Slope `m=(h)/(e )`and`-(hv_(0))/(e )` is constant.
Discussion
2828.

In the explanation of laws of refreaction by Newton.s corpuscular theory, identify the true statement

Answer»

he assumed that light travels faster in a denser medium than in a rarer medium
he assumed HYPOTHETICAL medium CALLED ether, near a refracting SURFACE
density of ether near the refracting surface is variable within certain RANGE
light corpuscles within this range of variable density EXPERIENCES no force

Answer :D
Discussion
2829.

Three similar cells, each of emf 2 V and internal resistance rOmega send the same current through an external resistance of 2Omega, when connected in series or in parallel. The strength of the current flowing through the external resistance is

Answer»

`1 A`
`1.5 A`
`2 A`
`0.75 A`

Solution :
In series, the EFFECTIVE emf of cells = `3epsilon`, effective internal RESISTANCE = 3R where `epsilon` is the emf and r is the internal resistance of each cell. In parallel, the effective emf of cells = `epsilon`, effective internl resistance = `(r )/(3)`
As per QUESTION, `I=(3epsilon)/(2+3r)=(epsilon)/(2+(r)/(3))`
or `6+r=2+3rorr=2Omega`
`therefore I=(3xx2)/(2+3xx2)=(6)/(8)=0.75A`
Discussion
2830.

In an a.c circuit with phase voltage V and current I, the power dissipated is

Answer»

V.I
Depends on PHASE angle between VAND I
1/2 * V.I
`[ 1/(sqrt 2) * V.I ]`

Answer :D
Discussion
2831.

Velocity of emw in free space is given by C = _____.

Answer»

SOLUTION :`(1)/SQRT(mu_0epsilon_0)`
Discussion
2832.

The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25cm in front of it to 50cm, the magnification of its image changes from m_(25)to m_(50). The ration m_(25)//m_(50) is

Answer»


SOLUTION :m=(F)/(f+u)`
Discussion
2833.

Photodiode used for detecting optical signal is invariably used in ___ arrangement.

Answer»

SOLUTION :REVERSE BIAS
Discussion
2834.

Two parallel plate of area A are separated by two different dielectrics as shown in figure. The net capacitance is

Answer»

`(4 epsilon_(0)A)/(3d)`
`(3 epsilon_(0)A)/(R)`
`(2 epsilon_(0)A)/(d)`
`(epsilon_(0)A)/(d)`

Answer :A
Discussion
2835.

What we call the ability of the eye to change the focal length of the eye lens by adjusting the curvature of a lens ,for viewing near and far objects?

Answer»

SOLUTION :POWER ACCOMMODATION
Discussion
2836.

How many time constansts will elapse before the current in a charging RC circuit drops to half of its initial value? Answer the same question for a discharging RC circuit?

Answer»

1.9 1.23
1.23, 0.69
0.69, 1.23
0.69, 0.69

ANSWER :D
Discussion
2837.

In example 8, if rectangular loop is replaced by circular loop then will the induced emf be of constant magnitude or not while entering the magnetic field?

Answer»

SOLUTION :VARIABLE
Discussion
2838.

Assertion : Two points in adjacent loops of a string wave in the case of standing wave formation attain their maximum kinetic energy simultaneously. Reason: The two points in the adjacent loops of standing wave are out of phase exactly by pi.

Answer»

ASSERTION is true and Reason is CORRECT EXPLANATION of Assertion.
Assertion is true and Reason is FALSE
Assertion and Reason both are false
Assertion is false but Reason is true.

Answer :A
Discussion
2839.

The electric field at a point, distant r from a charge q is E. The potential at that point is

Answer»

`ER^(2)`
`(E)/(R^(2))`
Er
`(E)/(r )`

ANSWER :C
Discussion
2840.

In the circuit, shown in fig v_(s)=0.2V, v_(0)=-10V. Find v_(i) and gain A_(v)=v_(0)//v_(i) and A_(v)^(')=v_(0)/v_(s)?

Answer»


ANSWER :NA
Discussion
2841.

While converting a galvanometer into ammeter and voltmeter, how it is connected ?

Answer»

SOLUTION :SERIES, PARALLEL
Discussion
2842.

Cellular Mobile works in the frequency range of

Answer»

`840` to `935 MHz`
`3.7` to `4.2 GHz`
`420` to `890 MHz`
`30` to `300 GHz`

Answer :A
Discussion
2843.

The number of turns in the coil of an A.C. generator are 100 and its cross-sectional area is 2.5 m^2. The coil is revolving in a uniform magnetic field of strength 0.3 T with the uniform angular velocity of 60 rad s^(-1). The value of maximum value produced is ...... kV.

Answer»

4.5
2.25
6.75
1.25

Solution :MAXIMUM INDUCED ENERGY ,
`V_m=NABW`
`=100xx2.5xx0.3 xx60`
=4500 V=4.5 KV
Discussion
2844.

If pressure of 8 xx 10^8 N/m^2 is applied to a lead block so that its volume reduces by 20% . The Bulk modulus of lead block is

Answer»

a)`4 XX 10^7 N/m^2`
B)`4 xx 10^8 N/m^2`
C)`4 xx 10^9 N/m^2`
d)`4 xx 10^10 N/m^2`

ANSWER :C
Discussion
2845.

A parallel plate capacitor has square plate of side 5 cm and separated by a distance of 1 mm . (a) Calculate the capacitance of this capacitor . (b) If a 10 V battery is connected to the capacitor what is the charge stored in any one of the plates? ( The value of epsilon_(0)= 8.85 xx10^(-12)Nm^(2)C^(-2)).

Answer»

Solution :(A) The capacitance of the CAPACITOR is
`C=(epsilon_(0)A)/(d) =(8.85xx10^(-12)xx25xx10^(-4))/(1XX10^(-3))=221.2xx10^(-13) F `
`C=22.2 xx10^(-12)F = 22.12 pF`
(b) The charge STORED in any on of the plates is Q = CV Then
`Q=22.12 xx10^(-12)xx10=221.2xx10^(-12)C = 221.2 PC`
Discussion
2846.

"Heat cannot by it self flow from a body at lower temperature a body at higher temperature” is a statement of consequence of :

Answer»

FIRST LAW of thermodynamics
conservation of mass
conservation of momentum
second law of thermodynamics

Answer :D
Discussion
2847.

The rms value of a sinusoidal ac current is equal to its instantaneous value at an angle of _____ degree.

Answer»

`30^@`
`60^@`
`90^@`
`45^@`

ANSWER :D
Discussion
2848.

(A) : The magnetic field at the ends of a very long current carrying solenoid is half of that at the centre. (R) : If the solenoid is sufficiently long, the field within it is uniform.

Answer»

Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'.
Both 'A' and 'R' are true and 'R' is not the correct explanation of 'A'
'A' is true and 'R' is FALSE
'A' is false and 'R' is false

Answer :B
Discussion
2849.

An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed ?

Answer»

Solution : No, the POSITION of the IMAGE does not CHANGE on REVERSING the LENS.
Discussion
2850.

A man wants to draw a bucket full of water in two different ways. As shown in figure 'a' he draws the bucket directly and as shown in figure (b) he uses a pulley. The weight of the man is 50kg and the bucket with full water weighs 25kg. Find the action on the floor by the man in the two cases. (Take g=10ms^(-2)).

Answer»


ANSWER :(a) 750 N; (B) 250N
Discussion