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3451.

(A) : Optical communication is preferred to microwave communication (R) : Microwaves provide large number of channels and bandwidth as compared to optical signals.

Answer»

Both 'A' and 'R' are true and 'R' is the correct EXPLANATION of 'A'.
Both 'A' and 'R' are true and 'R' isnot the correct explanation of 'A'.
A' is true abd 'R' FALSE
A' is false and 'R' is false

Answer :C
Discussion
3452.

Two point charges +q and - q are placed at a distance d apart. What are the points at which the resultant field is parallel to the joining of two charges?

Answer»

Solution :The RESULTANT field is PARALLEL to the line JOINING the two CHARGES at any point and also at any point on. the perpendicular bisector of the line joining the two charges.
Discussion
3453.

Give one use of each of the following y-ray

Answer»

SOLUTION :γ RAY – In TREATMENT of CANCER
Discussion
3454.

In a hydrogen atom, which of the following electronic transitions wouldinvolve the maximum energy change

Answer»

From `n = 2 " to " n = 1`
From `n = 3 " to " n = 1`
From `n = 4 " to " n = 2`
From `n = 3 " to " n = 2`

ANSWER :B
Discussion
3455.

12 identical capacitors are connected in series between two points . Out of these for .n. capacitors. the spacing between the plates is reduced to half and for the remaining it is doubled. What is the value of .n. for which the effective capacitance remains uncharged?

Answer»

4
6
8
2

Answer :C
Discussion
3456.

Fraunhoferlines give information regarding the presence of certain elements in the Sun's

Answer»

PHOTOSPHERE
CORONA
Chormosphere
All these

ANSWER :C
Discussion
3457.

A wave travelling along the x-axis is described by the equation y (x,t) = 0.005 cos (alpha x - beta t) . If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then alpha and beta in appropriate units are :

Answer»

`alpha = 25.00 PI , beta = pi`
`alpha = (0.08)/(pi) , beta = pi (2.0)/(pi)`
`alpha = (0.04)/(pi), beta = pi (1.0)/(pi)`
`alpha = 12.50 pi , beta = (pi)/(2.0)`

Solution :y(X,t) = 0.005 `(alpha x - beta t)`
COMPARE it with STANDARD form.
y(x,t) = r cos `((2 pi x)/(LAMBDA) - (2 pi)/(T) t ) `
`alpha = (2pi)/(lambda) = (2pi)/(.08) = 25 pi`
`beta = (2pi)/(T) = (2pi)/(2) = pi `
`rArr ` choice is (a)
Discussion
3458.

Two coherent sources whose intensity ratio 81:1 produce interference fringes . Calculate the ratio of intensity of maxima and minima in the fringe system .

Answer»


ANSWER :`25 : 16`
Discussion
3459.

A parallel plate capacitor with circular plates o f radius lm has a capacitance o f 1 nF. At 1=0, it is connected fo r charging in series with a resistor R = IM omegaacross a 2V battery. Calculate the magnetic field at a point P, halfway between the centre and the periphery o f the plates, after t = 10^(-3)s. (The charge on the capacitor at time t is q(t) = CV[1- exp (t//tau )], where the time constant tauis equal to CR)

Answer»

Solution :The time constant of the CR circuit is `TAU = CR = 10^(-3)` s. Then, we have
`q(t) = CV[l-exp(-t//tau)]`
`= 2 xx 10^(-9) [1-exp (-t//10^(-3)]`
The electric field in between the plates at time t is `A = PI(l)^(2) m^(2)` = area of the plates. Consider now a circular loop of radius [ 1/2) m parallel to the plates passing through P. The magnetic field B at all points on the loop is along the loop and of the same VALUE. Then flux
`E xxpixx1/2^(2)=(piE)/(4)=(q )/(4epsilon_(0))`
Thedisplacement current
Now applyingamere maxwell law to the loopwe get
or `B=0.74 xx10^(-13)` T
Discussion
3460.

If momentum (P), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula

Answer»

`[P^1 A^(-1) T^1]`
`[P^2 A^1 T^1]`
`[P^1 A^(-1//2) T^1]`
`[P^1 A^(1//2) T^(-1)]`

SOLUTION :Let energy , `E prop P^a A^b T^c`
or `E=kP^a A^b T^c`
or `[ML^2 T^(-2)]=[MLT^(-1)]^a [L^2]^b[T]^c = [M^a L^(a+2b)T^(-a+c)]`
whence , a=1 , b=`1/2` , c=-1
DIMENSIONAL FORMULA for E is `[P^1 A^(1//2) T^(-1)]`
Discussion
3461.

Which of the following displacement (X) time graphs is not possible?

Answer»




SOLUTION :is not POSSIBLE,because a ta PARTICULAR time t, DISPLACEMENT cannot have two VALUES.
Discussion
3462.

Which of the following is used to study the structure of crystals ?

Answer»

U.V. rays
Infrared
X-rays
Microwaves

ANSWER :C
Discussion
3463.

The unit of pressure in SI system is given by

Answer»

DYNE `cm^-2`
`N/m^2`
`Nm^2`
Nm

Answer :B
Discussion
3464.

Describe with suitable block diagrams, action of pn-junction diode under forward and reversebias conditions. Also draw I-V characteristics.

Answer»

Solution :PN Junction diode under forward bias

When the diode is forward biased as shown in the figure the depletion region width decreases and the barrier height is reduced. The electrons from n-sied cross the depletion region and reach p-side also HOLES cross the junction and reach the n-side. A CONCENTRATION gradient s developed at the junction boundary. Due to this the motion of charged carriers on either side gives RISE to current. The total diode forward current is sum of hole DIFFUSION current and CONVENTIONAL current due to electron diffusion.
PN Junction diode under reverse bias

When the diode is reverse biased the depletion region width increases and the barrier height is increased. This supresses the flow of electrons from n-side to p-side and holes from p-side to n-side. Thus, diffusion current decreases. The conventional current is due to drift of the minority charge carriers which is of the order of micro amperes. The current under reverse bias is voltage independent up to a critical reverse bias voltage known as breakdown voltage.
The I-V characteristics are as shown
Discussion
3465.

False fruits are found in

Answer»

GUAVA, PEAR and sapota
Black PEPPER and beet
Apple, STRAWBERRY and cashew
Banana and apple

Answer :C
Discussion
3466.

The permeablities of paramagnetic materials are :

Answer»

zero
equal to unity
less than unity
greater than unity

Answer :D
Discussion
3467.

झारखंड इनमें से किस पठार में स्थित है?

Answer»

दक्कन पठार
मालवा पठार
कार्बी एंगलोंग पठार
छोटानागपुर पठार

Answer :D
Discussion
3468.

The radius of an electron orbit in a hydrogen atom is of the order of.

Answer»

`10^(-8) m`
`10^(-8) m`
`10^(-11) m`
2eV

Solution :Radius of an ELECTRON in 1st orbit of hydrogen ATOM `r_(1) =0.53 Å = 0.53 XX 10^(-10)` m. Hence radius is of the order of `10^(-10)` m.
Discussion
3469.

Huygen's principle state that:

Answer»

wave is transverse wave
each POINT of the wave FRONT is in different phase
each point of the WAVEFRONT is in same phase
none of the above

Answer :C
Discussion
3470.

The minimum magnifying power of a telescope is M. If the focal length of its eyelens is halved, the magnifying power will become

Answer»

M
2M
3M
4M

ANSWER :2
Discussion
3471.

What maximum power can be obtained from a battery of emf epsilonand internal resistance r connected with an external resistance R ?

Answer»

`(epsilon^(2))/(4r)`
`(epsilon^(2))/(3r)`
`(epsilon^(2))/(2R)`
`(epsilon^(2))/(r)`

Solution :`(epsilon^(2))/(4r)`
CURRENT in CIRCUIT I = `(epsilon)/(R + r) `

If we get maximwn power through battery then EXTERNAL resistance becomes R = r.
`therefore " power P " = I^(2) R = (epsilon R)/((R + r)^(2)) `
`= (epsilon^(2) r)/((2r)^(2))[ because " PUTTING R " =r]` ,
` = (epsilon^(2))/(4r)`
Discussion
3472.

An YDSE is carried out in a liquid of refractive index mu = 1.3 and a thin film of air is formed in front of the lower slit as shown in the figure. If a maxima of third order is formed at the origin O, find the thickness of the air film. The wavelength of light in air is lambda_(0) = 0.78 mu m and D/d = 1000.

Answer»

`7.8 MU m`
`3.9 mu m`
`15.6 mu m`
`0.5 mu m`

ANSWER :A
Discussion
3473.

Small identical balls with equal charges of magnitude 'q' each are fixed at the vertices of a segular 20019-gon (a polygon of 2019 sides) with side 'a'=4 mum. At a certain instant, one of the balls is released and a sufficiently long time interval later, the ball adjacent to the first released ball is freed. The kinetic energies of the released balls are found to differ by K=9xx10^(9) J at a sufficiently large distance from the polygon. determine the charge q in mC.

Answer»


ANSWER :2
Discussion
3474.

The process in which radiation is deflected by particles in the medium through which the radiation passes is called ?

Answer»

SOLUTION :SCATTERING
Discussion
3475.

A microscope is focused on a coin lying at the bottom of a beaker. The microscope is now raised up by 1 cm. To what depth should the water be poured in to the beaker so that coin is again in focus

Answer»

1cm
4/3cm
3cm
4cm

Answer :D
Discussion
3476.

A point P, is 40 m away from charge 2 mu C and 20 m away from charge 4 muC. Find the electric potential at point P. How much work to be done for a charge ot - 0.4 C brought from infinity to point P ?

Answer»

SOLUTION :2250 V , - 900 J
Discussion
3477.

A charge on electron is 1.6 xx 10^(-19) C. How many electrons will be passing in 2 seconds through a cross-section of a conducting wire carrying 0.7 A electric current ?

Answer»

4.4 `xx 10^(18)`
`4.4 xx 10^(-18)`
`8.8 xx 10^(18)`
`8.8 xx 10^(-18)`

Solution :`8.8 xx 10^(18)`
`I = (Q)/(t) = ("NE")/(t)`
` therefore n = (I t)/(E) = (0.7 xx 2)/(1.6 xx 10^(-19))= 0.875 xx 10^(19)`
`therefore n APPROX 8.8 xx 10^(18)`
Discussion
3478.

A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field vecB in the core for a magnetising current of 1.2 A?

Answer»

Solution :Here R = 15 cm = 0.15 m , N = 3500 , `mu_r =800 , and I = 1.2 A`
`therefore ` Magnetic FIELD in the core of Rowland RING
`B = (mu_0 mu_r NI)/(I) = (mu_0 mu_r N I)/(2pi R)=(4pixx 10^(-7) xx 800xx3500 xx 1.2)/(2pi xx 0.15) = 4.48 T `.
Discussion
3479.

What is the value of fringe width beta in case of two consecutive dark or bright fringes ?

Answer»

SOLUTION :`BETA = (LAMBDA D )/d`
Discussion
3480.

For an electron miroscope , which of the following is false ?

Answer»

It uses MAGNETIC lens to CONVERGE electron BEAM
Its resolving power is DIRECTLY proportional to accelerating potential of electrons
Its resolving power is inversely proportion to wavelength of electrons
Magnification attained with the HELPOF it of the order of ` 10^(6)`

Answer :B
Discussion
3481.

The wave speed is :

Answer»

62.9 m/s
96.8 m/s
7.28 m/s
9.25 m/s

Solution :WAVE speed `v = SQRT((T)/(mu)) = sqrt((45)/((0.525))) = 925 MS^(-1)`
Discussion
3482.

A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true:

Answer»

PQ is horizonta
QR is horizontal
RS is horizontal
Either PQ or RS is horizontal

Answer :2
Discussion
3483.

In a biprims experiment, the slite is illuminatedby red light of wavelenght 6400. A.U. and the crosswire in the eypeice is adjusted to be at the centre of the 3^(rd) bright band . By using blue light it is found that the 4^(th) bright band is at the centre of the croswiere, find teh wavelength of blue light.

Answer»

4800 A.U.
4500 A.U.
5800 A.U.
4600 A.U.

Solution :For interence by thin films, the order of THICKNESS of film is approximately equal to WAVELENGHTOF light ,
`i.e.,~~1(t=10000Å)`
Discussion
3484.

Nuclear force result from the exchange of what particles ?

Answer»

SOLUTION :MESON (`PI`-MESOS)
Discussion
3485.

5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. Taking the initial temperature to be T_(1), the work done in the process is

Answer»

`(9)/(8)RT_(1)`
`(3)/(2) RT_(1)`
`(15)/(8)RT_(1)`
`(9)/(2) RT_(1)`

Solution :`TV^(gamma-1)=C`
`T_(1)(5.6)^(2//3)=T_(2)(0.7)^(2//3) rArr T_(2)=T_(1)(8)^(2//3)=4T_(1)`
`THEREFORE Delta W` (WORK done on the SYSTEM) `=(NR Delta T)/(gamma-1) =(9)/(8) RT_(1)`
Discussion
3486.

Three concentric conducting spherical shells of radii R,2R carry charges Q, -2Q and 3Q, respectively. Find the electric potential at r = R and at r = 3 R, where r is the radii distance from the centre.

Answer»

`(Q)/(4piepsilon_(0)R),(Q)/(6piepsilon_(0)R)`
`(Q)/(6piepsilon_(0)R),(Q)/(piepsilon_(0)R)`
`(Q)/(piepsilon_(0)R),(Q)/(4piepsilon_(0)R)`
`(Q)/(2piepsilon_(0)R),(Q)/(4piepsilon_(0)R)`

ANSWER :A
Discussion
3487.

In meter bridge when P is kept in left gap, Q is kept in right gap, the balancing length is 40cm. If Q is shunted by 10Omega, the balance point shifts by 10cm. Resistance Q is

Answer»


ANSWER :5
Discussion
3488.

Derive the law of radioactive decay on the basis of the fact that the decay probability for a nucleus is independent of the number of nuclei and is proportional to the period of observation.

Answer»


Solution :The decay probability is the ratio of the number of the NUCLEI that have EXPERIENCED decay to the TOTAL number of nuclei: `omega=-(dN)/N`.
It is proportional to the TIME of observation, `omega=lamdadt,i.e.(dN)/N=-lamdadt`. We have
`int(dN)/N=-lamdaintdt,"so "lnN+C=-lamdat`
At the initial point of time t = 0, N = `N_(0)`, so `lnN_(0)+C=0andC=-lnN_(0)`. Substituting this into the above formula, we obtain
`lnN-lnN_(0)=-lamdat,orN=N_(0)e^(-lamdat)`
Discussion
3489.

If the electric flux entering and leaving a closed surface in air are phi_(1) and phi_(2) respectively, the net electric charge enclosed within the surface is ____________ .

Answer»


Solution :As PER Gauss.s law net flux `(phi_(2)-phi_(1))=(1)/(in_(0))` (net charge ENCLOSED).
Discussion
3490.

An alternator produces voltage wave equal to the number of ……… used.

Answer»

SLIP rings
windings
magnets
brushes

Answer :D
Discussion
3491.

किसी एकपरमाणविक पदार्थ की एक फलक केन्द्रित घनीय इकाई सैल में परमाणुओं की संख्या होती है -

Answer»

1
2
3
4

Answer :D
Discussion
3492.

n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance?

Answer»

N
`1/n^2`
`n^2`
1/n

Answer :C
Discussion
3493.

In the figure AB and BK represents incident and reflected rays. If angle BCF=theta then /_BFP will equal to

Answer»

`THETA`
`2THETA`
`3theta`
`3.5 theta`

Solution :(b) Here, incident RAY is PRALLEL to principal axis, so
`/_AMN=/_BCF=theta`
also `/_CBF=/_KMN=/_AMN`
`:. /_BFP=theta+/_CBF=theta+theta=2theta`
Discussion
3494.

In a photoelectric cell, the product of the stopping potential and electronic charge is equal to

Answer»

the MOMENTUM of every EMITTED ELECTRON
the kinetic ENERGY of every emitted electron
the photoelectric work function of the emitter material
the maximum kinetic energy that an emitted electron can have.

Answer :D
Discussion
3495.

One ferromagnetic substance has magnetic susceptibility chi_(m) at 27^(@) C temperature. At which temperature, it would become half ?

Answer»

`600^(@)C`
`300^(@) C`
`54^(@) C`
`327^(@) C`

SOLUTION :`chi_(m) prop (1)/( T)`
`therefore (chi_(m_(1) ) )/( chi_(m_(2)) ) = (T_(2) )/( T_(1))`
`(2 chi_(m_(1) ) )/(chi_(m_(1) ) ) = (T_2)/( 273+ 27) (because chi_(m_(2) ) = (chi_(m_(1 )))/( 2))`
`therefore T_(2) = 600 K`
`therefore t_(2) = T_(2) = 273 = 600 - 273 = 327 ^(@) C`
Discussion
3496.

The ratio of magnetic potentials due to magnetic dipole in the end on position to that in broad on position for the same distance from it is :

Answer»

0
`OO`
1
2

Answer :B
Discussion
3497.

A thin film (n = 1.25) coats a thick glass plate (n = 1.50). White light is incident normal to the film. In the reflections, fully destructive interference nccurs at 600 mm and fully constructive interference at 700 nm. Calculate the thickness of the film.

Answer»

SOLUTION :840 nm
Discussion
3498.

Explain the use of transformer for distribution of power over long distances.

Answer»

Solution :TRANSFORMER can be used for the large scale TRANSMISSION and distribution of electric energy over LONG distances.
The voltage output of the generator is stepped up so that CURRENT is reduced and the `I^(2) R` loss is cut down.
It is then transmitted over long distances to an area sub-station near the consumers.
There the voltage is stepped down. It is further stepped down at distributing sub-station and utility poles beofre a power supplyof 240V reaches our HOMES.
Discussion
3499.

Determine the resistance R_(AB) between points A abd B of the frame made of thin homogeneous wire (as shown in figure), assuming number of successively embedded equilateral tiangles ( with sides decreasing by half ) tends to infinity. SideAB is eual to a, and the resistanceof unit length of the wire is rho.

Answer»


ANSWER :`R_(AB)=(arho(sqrt7-1))/3`
Discussion
3500.

Give the uses ofelectromagnets.

Answer»

Solution :ln ELECTRIC BELL, LOUDSPEAKER, diaphragms of TELEPHONES, CRANES.
Discussion