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41851.

Twenty seven water drops of the same size are charged to the same potential .if they are combined to from a big drop, the ratio of the potential of the big drop to that of a small drop is

Answer»

3
6
9
27

Solution :`V. = N^(2//3)V ""RARR (V.)/(V) = (27)^(2//3) = 9`
Discussion
41852.

Atomic numbers of two element are 31 and 41. Find the ratio of the wavelength of the K_(alpha) are

Answer»

`4:3`
`9:16`
`16:27`
`16:9`

ANSWER :D
Discussion
41853.

sqrt 4 /sqrt 2निम्न मे से कौन सी संख्या है -

Answer»

पूर्णांक
परिमेय
अपरिमेय
इनमे से कोई नहीं

Answer :C
Discussion
41854.

When a particle is restricted to move along x-axis between x = 0 and x = 4, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends r = 0 and x= a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de-Broglie relation. The energy of the particle of mass m is related to its linear momentum as E= p^2//2m. Thus, the energy of the particle can be denoted by a quantum number 'r' taking values 1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x=0 to x=a. Take h=6.6 xx 10^(-34) Js and e=1.6 xx 10^(-19)C. The speed of the particle that can take discrete values is proportional to

Answer»

`n^(-3//2)`
`n^(-1)`
`a^(1//2)`
n

Solution :The speed of the particle that can take discrete VALUES
`p=mv=(nh)/(2A)`
`rArr v ALPHA n`
Discussion
41855.

Consider the following two statements. I. The linear momentum of a system of particles is zero. II. The kinetic energy of a system of particles is zero. Then

Answer»

I implies II and II implies I.
I does not IMPLY II and II does not imply I.
I implies II but II does not imply I
I does not imply II but II implies I.

Solution :If the kinetic energy of a system of particles is ZERO, it means all the particles are at REST. Hence, the MOMENTUM is zero. Thus II implies I.
But if the momentum of a system of particles is zero, it does not MEAN that all the particle are at rest. Hence, the particle may possess kinetic energy. Thus I does not imply II.
Discussion
41856.

When a particle is restricted to move along x-axis between x = 0 and x = 4, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends r = 0 and x= a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de-Broglie relation. The energy of the particle of mass m is related to its linear momentum as E= p^2//2m. Thus, the energy of the particle can be denoted by a quantum number 'r' taking values 1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x=0 to x=a. Take h=6.6 xx 10^(-34) Js and e=1.6 xx 10^(-19)C. If the mass of the particle is m= 1.0 xx 10^(-30) kg and a= 6.6 nm, the energy of the particle in its ground state is close to

Answer»

0.8meV
8meV
80meV
800meV

Solution :The mass of the particle in the box is `m=1.0 xx 10^(39) kg, a=6.6nm`
The energy of the particle in the ground state
`E_(n ML)=(l^(2).h^(2))/(8 ma^(2)E)` in electron volt units
`RARR` E in electron volts.
`=((6.6 xx 10^(-34))^(2))/(8 xx 1.0 xx 10^(-30) xx (6.6 xx 10^(-9))^(2) xx 1.6 xx 10^(-19))`
=7.8meV =8meV
Discussion
41857.

A ball of mass 'm' is pushed with a velocity u towards a movable wedge of mass 3m which is at rest. Height of the wedge is 60m. Assume all surfaces to be smooth. The minimum value of 'u' for which the block will reach the top of the wedge is ( There is no loss of energy when the ball tries to rise on the wedge ) ( take g = 10 m//s^(2) )

Answer»

`40 m//`SEC
`80 m//`sec
`20M //`sec
`50M//`sec

Answer :A
Discussion
41858.

When a particle is restricted to move along x-axis between x = 0 and x = 4, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends r = 0 and x= a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de-Broglie relation. The energy of the particle of mass m is related to its linear momentum as E= p^2//2m. Thus, the energy of the particle can be denoted by a quantum number 'r' taking values 1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x=0 to x=a. Take h=6.6 xx 10^(-34) Js and e=1.6 xx 10^(-19)C. The allowed energy for the particle for a particular value of n is proportional to

Answer»

<P>`a^(-2)`
`a^(-3//2)`
`a^(-1)`
`a^(2)`

Solution :
`E=p^(2)/2m`, but according to de-Broglie THEORY.
`lambda=H/p or p=h/lambda`
In this case, n LOOPS are possible
`p=(nh)/(2a) as n lambda/2=a`
`E=((n^(2)h^(2))/(4a^(2) XX 2m))=(n^(2)h^(2))/(8ma^(2))`
Hence `E alpha n^(2)" and E is also "alpha 1/a^(2)`
Discussion
41859.

If the distance between the screen and the light source is reduced to 1/3 of the orginal distance, the illuminationof the screen

Answer»

INCREASES 3 TIMES
REDUCES to 1/3rd
Increases 9 times
Reduces to 1/9h

Answer :C
Discussion
41860.

Answer the following questions: (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?

Answer»

Solution :(d) Angular magnification of eye-piece is `[(25//f_(e) ) + 1] ( f_(e)" in cm")` which increases if `f_(e)` is smaller. Further, magnification of the OBJECTIVE is given by
`(v_(0))/(|u_(0)|)=(1)/((|u_(0)|//f_(0))-1)`
which is large when `|u_(0)|`is SLIGHTLY GREATER than `f_(0)` . The microscope is used for viewing very close object. So `|u_(0)|`is SMALL, and so is `f_(0)`.
Discussion
41861.

A photograph of the moon was taken with telescope. Later on, it was found that a housefly was sitting on the objective, lens of the telescope. In photograph A'

Answer»

The image of housefly will be reduced
There is a REDUCTION in the INTENSITY of the image.
There is an INCREASE in the intensity of the image
The image ofthe housefly will be enlarged

Answer :B
Discussion
41862.

Which of the graph between kinetic energy and time is correct ?

Answer»

A
B
E
F

Answer :D
Discussion
41863.

Suppose a pure Si crystal has5 xx 10^(28) atoms m^(-3) . It is droped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n_(i) = 1.5 xx 10^(16) m^(-3).

Answer»

Solution :Note that thermally GENERATED electrons`( n_(i) ~ 10^(16)m^(-3))` are negligibly SMALL as compared to those produced by doping.
Therefore, `n_(e ) = N_(D) `
Since `n_( e) n_(h ) = n_(i )^(2) `,The number of holes ,`n_(h) = ( 1.5 xx 10^(16))^(2) // 5 xx 10^(28) xx 16^(-6)`
`n_(h) =( 2.25 xx 10^(32)) //( 5xx 10^(22)) ~ 4.5xx 10^(9) m^(-3)`
Discussion
41864.

Gold has a specific heat of 130Jkg*""^(@)C and wood has a specific heat of 1,800Jkg*""^(@)C. If a piece of gold and a piece of wood, each of mass 0.1 kg, both absorbs 2.340 J of heat, by how much will their temperatures rise?

Answer»

SOLUTION :We're given c,m, and Q, and we know that `DeltaT=Q//(MC)`.
`DeltaT_("gold")=(Q)/(mc_("gold"))=(2.340J)/((0,1kg)(130J//kg*""^(@)C))=180^(@)`
`DeltaT_("wood")=(Q)/(mc_("wood"))=(2,340J)/((0,1kg)(1,800J//kg*""^(@)C))=13^(@)C`
NOTICE that the temperature of gold increased by `180^(@)C`, but the temperature of the wood increased by only `13^(@)C`.
Discussion
41865.

The coil of a tangent galvanometer is 11 cmin radius . How many turns of the wire should be wound on it if a currentof 70 mu A is to produce a deflection of 45^(@) . Given B_(H)=0.32 gauss.

Answer»


ANSWER :80 TURNS
Discussion
41866.

What is modulation ? Explain different types of modulation.

Answer»

Solution :Modulation is that proces in which some characteristics of carrier wave is varied in ACCORDANCE to intensity of audio or VIDEO signal, `therefore` Modulated wave = carrier wave + signal.
This process is done by an ELECTRONIC device known as modulator. The REVERSE process of separating the signal from the modulated wave is known as demodulation and this process is done by an electronic device called detector.
Discussion
41867.

यदि वस्तु (बिंब) उत्तल लेंस के फोकस तथा फोकस-दूरी की दूनी दूरी के बीच हो, तो प्रतिबिंब

Answer»

काल्पनिक, सीधा तथा छोटा बनेगा
काल्पनिक, उल्टा तथा बड़ा बनेगा
वास्तविक, उल्टा तथा छोटा बनेगा
वास्तविक, उल्टा तथा बड़ा बनेगा

Answer :D
Discussion
41868.

If we increase the driving frequency in a circuit with a purely capacitive load, do (a) amplitude V_C and (b) amplitude I_C increase, decrease, or remain the same? If, instead, the circuit has a purely inductive load, do (c) amplitude V_L and (d) amplitude I_L increase, decrease, or remain the same?

Answer»


ANSWER :(a)Same ; (B) INCREASES ; (C ) same ; (d) DECREASES
Discussion
41869.

A number of tiny drops of water all of the same radius r cm combine to form a single drop of radius R cm. The rise in temperature of water is given by (if sigma = S.T. specific heat= 1 cal/g) :

Answer»

`(sigma)/J(1/r-1/R)`
`(3sigma)/J(1/r-1/R)`
`(3sigma)/J(1/(r^(2))-1/(R^(2)))`
`(sigma)/J(1/(r^(2))-1/(R^(2)))`

Solution :Here Energy RELEASED =`4pi(nr^(2)-R^(2)).sigma`
`therefore` Heat PRODUCED=`W/J=(4sigmapi)/J(nr^(2)-R^(2))`
Heat produced in drop=`4/3piR^(3)xx1xxtheta`
`therefore(4sigmapi)/J(nr^(2)-R^(2))=4/3piR^(3)theta`
or `theta=(3sigma)/J(1/r-1/R)`
`therefore` Correct CHOICE is (b).
Discussion
41870.

The area of cross section of a steel wire is 0.1cm2. What is the force required to double it's length ifY = 2 xx 10^11 N /M^2?

Answer»

Solution :strain `l/L` = `THEREFOR` F = AT
= `0.1 XX 10^(-4) xx 2 xx 10^11 = 2 xx 10^6 N`
Discussion
41871.

In a series LCR circuit, impedance Z is same at two frequencies f_1 and f_2. Therefore, the resonant frequency of this circuit is

Answer»

`(f_1+f_2)/2`
`(2f_1f_2)/(f_1+f_2)`
`SQRT(f_1^2+f_2^2)/2`
`sqrt(f_1f_2)`

ANSWER :D
Discussion
41872.

whichof thefollowingstatementis / arecorrect ?

Answer»

THEPH of `1.0xx10^(-8)` M solution ofHCl is 8.
thecnjugatebaseof `H_(2)PO_(4)^(-)` is`HPO_(4)^(2-)`
Autoprotolysis constantof waterdecreaseswithtemperature
whena solutionof a weakmonoproticacidis titrated againsta strongbase , at half - NEUTRALIZATION point`PH= (1//2)pk_(a)`

Solution :`H_(2)PO_(4)^(-)HARR H^(+)+HPO_(4)^(-2)`
Discussion
41873.

Which common name is wrong ?

Answer»




ANSWER :4
Discussion
41874.

Assertion : If the law of gravitation becomes inverse cubelaw even then a line joining the sun the planetsweeps equal areas in equal time intervals . Reason : A planet moves in an allipticalpath .

Answer»

If both the assertion and REASON are true statement andreason is correct EXPLANATION of the assertion .
If both the assertion and reason are true statement but reason is not a correct explanation of the assertion .
If the assertion is true but the reason is a false statement.
If both assertion and reason are false statements.

Solution :Sweeping equal AREAS in equal TIME intervals is a consequence of law of conservation of angular momentum. So long as the gravitational force is Radial, no torque acts on the planetand the angular momentum in CONSERVED .
It isalso true that a planet moves in an elliptical orbit. But this notthe reason of the assertion .
Discussion
41875.

A plane electromagentic wave E=E_(m) cos ( omegat - kr)propagates in vacuum. Assuming the vectors E_(m) and k to be known, find the vector H as a function of time t at the point with radius vector r=0 .

Answer»

Solution :`VEC(nabla)xxvec(E)=-(deltavec(B))/( deltat)=- mu _(0)(deltavec(H))/( deltat)`
`=nablacos ( omegat - vec(k). vec(r))xxvec(E)_(m) =vec(k) xx vec(E)_(m) sin ( omegat- vec(k). vec(r))`
At ` vec (r)=0`
`(deltavec(H))/(deltat)=-(vec(k) xxvec(E)_(m))/( mu_(0))sin omegat`
So integrating `(` ignoring a CONSTANT `)` and using `C=(1)/(sqrt(epsilon_(0)mu_(0)))`
` vec (H)= (vec(k) xx vec(E)_(m))/( mu_(0))cos c k t xx(1)/(ck)=sqrt((epsilon_(0))/( mu_(0)))(vec(k)xxvec(E)_(m))/( k ) cos c k t `
Discussion
41876.

A satellite is orbiting in circular path near about earth surface. How much times its kinetic energy make to escape it from its orbit :

Answer»

2 times
Half
4 times
3 times

SOLUTION :`E_("es")=2K_0`
Discussion
41877.

Direction of induced emf is given by :

Answer»

MAXWELL's law
Kirchhoff's law
Lenz's law
Ampere's law

Answer :C
Discussion
41878.

A particle in the ground state is located in a unidimesional square potential well of length l with absolutely impenetrable walls (0 lt x lt l). Find the probability of the particle staying within a region (1)/(3)l le x le (2)/(3)l.

Answer»

Solution :We LOOK for the solution of Schrodinger eqn. with
`-(ħ^(2))/(2m)(d^(2)Psi)/(dx^(2))=E Psi,0lexlel`….(1)
The boundary condition of impenetrable walls means
`Psi(X)=0` for `x=0` and `x=L`
(as `Psi(x)=0` for `xlt0` and `xgtl`,)
Then solution of (1) is
`Psi(x)=A "sin"sqrt(2mE)/(ħ)x+B "cos"sqrt(2mE)/(ħ)x`
Then `Psi(0)=0implies B=0`
`Psi(l)=0implies A "sin"sqrt(mE)/(ħ)l-0`
`A~~ 0` so
`sqrt(2mE)/(ħ)l=npii`
Hence `E_(n)=(n^(2)pi^(2)ħ^(2))/(2ml^(2)),n= 1,2,3`....
Thus the ground state wave fucnction is
`Psi(x)=A "sin"(pi x)/(l)`
We evaluate `A` by nomalization
`1=A^(2)int_(0)^(i)"sin"^(2)(pi x)/(l)dx=A^(2)(l)/(pi)int_(0)^(x)sin^(2) theta d theta=A^(2)(l)/(pi).(pi)/(2)`
Thus `A+(sqrt(2)/(l))`
Finally, the probability `P` for the particle to lie in`(l)/(2) le (2L)/(3)` is
`P=P((l)/(3)lexle(2l)/(3))=(2)/(l)int_(l/3)"sin"^(2)(pix)/(l)dx`
`=(2)/(pi)int_(pi//3) sin^(2) theta d theta=(1)/(pi)int_(pi//3)^(2pi//3)(1- cos 2 theta)d theta`
`(1)/(pi)(theta-(1)/(2) sin 2 theta)^(2pi//3)=(1)/(pi)((2pi)/(3)-(pi)/(3)-(1)/(2)"sin"(4pi)/(3)+(1)/(2)`sin`(2pi)/(3))`
`=(1)/(pi)((pi)/(3)+(1)/(2)(sqrt(3))/(2)+(1)/(2)(sqrt(3))/(2))=(1)/(3)+(sqrt(3))/(2pi)= 0.609`
Discussion
41879.

A hammer of mass 1 kg strikes on the head of a nail with a velocity of 2 ms^(-1). It drives the nail 0.01 m into a wooden block. Find the force applied by the hammer and the time of 8 impact.

Answer»

`200 N , 10^(-2)` sec
`100 N , 10^(-3)` sec
`300 N , 10^(-2)` sec .
`400N , 10^(-3)` sec

Answer :A
Discussion
41880.

The energy (in W) required to excite an electron from n = 2 to n = 4 state in hydrogen atom

Answer»

`+2.55`
`-3.4`
`-0.85`
`+4.25`

SOLUTION :We know that,
`E_(n)=(13.6)/(n^(2))`
According to question, the required ENERGY is
`E=E_(2)-E_(4)`
`=(13.6)/((2)^(2))-(13.6)/(4^(2))=(13.6)/(4)-(13.6)/(16)`
`=((13.6xx4)-13.6)/(16)`
`= (40.8)/(16)=+2.55`
Discussion
41881.

A coaxial cylinder made of glass is immersed in a-liquid of surface tension S. The radius of the inner and outer surface of the cylinder are R_1 and R_2respectively . Height till which liquid will rise is (Density of liquid is rho )

Answer»

`(2S)/(R_(2)rhog)`
`(2S)/(R_(1)rhog)`
`(S)/(R_(2)-R_(1)rhog)`
`(2S)/(R_(2)-R_(1)rhog)`

ANSWER :D
Discussion
41882.

A.T.V. tower is 150 m tall.If the area around the tower has a population density of 750 km^(2) ,find the population covered by the broadcasting tower (R_(e)=6400 km)

Answer»

SOLUTION :`4.5xx10^(6)`
Discussion
41883.

A potentiometer wire of length 100 cm has a resistance of 10Omega. It is connected in series with a resistance and an accumulator of emf 2V and negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. what is the value of the external resistance?

Answer»


ANSWER :`790Omega`
Discussion
41884.

Electric current density is ..... .

Answer»

VECTOR QUANTITY
scalar quantity
fundamental quantity
having unit` Am^(-1)`

Solution :vector quantity
We have J = `sigma` E where `sigma` scalar and E is vector. Hence j MUST be vector.
Discussion
41885.

According to Bohr only those orbit are permissible for which the angular momentum of an electron is integral multiple of:

Answer»

H
`2PI h`
`h/(2pi)`
`h/(PI)`

ANSWER :C
Discussion
41886.

The following table has 3 columns and 4 rows. Based on table there are THREE questions. Each question has FOUR options (A) , (B) , ( C) and (D) . ONLY ONE of these four options is correct Pick the correct combination .

Answer»

(II ) (iv) (R )
(I) (ii) (S)
(II) (iii) (S)
(III) (iii) (P)

Solution :de = Bvdr
`epsilon = int_(0)^( R) (mu_(0)i)/(2PI(R+rcos theta)) xx( omegar ) dr`
Find `epsilon ` for different `theta`
ELECTRIC field at MID point = `(depsilon )/(dr) = B ( omega (R )/(2))`
Here `B =(mu_(0)i)/(2pi(R+rcos theta))+(mu_(0)r)/(2pi(2R-r cos theta))`
Discussion
41887.

In a Fraunhoffer diffraction experiment at a single slit using light of wavelength 400 nm , the first minimum is formed at an angle of 30^(@) . Then the direction theta of the first secondary maximum is

Answer»

`tan^(-1)(4/3)`
`60^(@)`
`SIN^(-1) (3/4)`
`tan^(-1) (3/4)`

Answer :C
Discussion
41888.

For the velocity time graph shown in the figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds ?

Answer»

`1/2`
`1/4`
`1/3`
`2/3`

SOLUTION :`(S_(2))/(S_(1))=("AREA of "/_\" CEF")/("area of Fig.ABCF")`
`=(10)/(10+20+10)=(1)/(4)`
Discussion
41889.

Radon has 3.8 days as its half-life . How much radon will be left out of 15 mg mass after 38 days ?

Answer»

1.05mg
0.015 mg
0.231 mg
0.50 mg

Solution :Here, `N=1/T_(1//2)=38/3.8=10`
`N=N_0(1/2)^n=m=m_0(1/2)^n =15 mg XX (1/2)^10`
`=15/1024`=0.015 mg
Discussion
41890.

A capacitor of capacitance C is connected in series with a resistance R and a DC source of emf E through a key. The capacitor starts charging when the key is closed. By the time the capacitor has been fully charged, what amount of energy is dissipated in the resistance R?

Answer»

`1/2 CE^2`
0
`CE^2`
`(E^2)/( R)`

SOLUTION :A dcsourceusesitschemicalenergyin charginga capacitorwhichis sumof
(I )itspotentialenergystoredin thecapacitor
(ii )heatenergydissipatedacrossresistance(R )
Nowwhenthecapacitorisfullychargedthenenergysuppliedby dcsource
` Q_(NET) . V=1/2CV^2+H `
`implies(CE )/(E )= 1/2CE^2+ H[ :.V = E ]`
`impliesCE^2= 1/2CE^2+H impliesH= CE^2-1/2CE^2= 1/2 CE^2`
Discussion
41891.

A substance absorbs an amount of heat Q_1 in going from one state to another and releases an amount of heat Q_2 in coming back from the second state to the first state. How much work is done by the substance and what is the change in the internal energy of the substance ?

Answer»

SOLUTION :`W=Q_1-Q_2,thetaU=0`
Discussion
41892.

What would be the duration of the year if the distance between the earth and the sun gets doubled ?

Answer»

a)229 days
b)329 days
C)129 days
d)None

Answer :C
Discussion
41893.

The following table has 3 columns and 4 rows. Based on table there are THREE questions. Each question has FOUR options (A) , (B) , ( C) and (D) . ONLY ONE of these four options is correct

Answer»

(III) (iv) ( R)
(I) (ii) (Q)
(IV) (iii) (R )
(I) (iv) (Q)

Solution :de = Bvdr
`epsilon = int_(0)^( R) (mu_(0)i)/(2PI(R+rcos theta)) xx( omegar ) dr`
Find `epsilon ` for DIFFERENT `theta`
Electric field at mid point = `(depsilon )/(dr) = B ( omega (R )/(2))`
Here `B =(mu_(0)i)/(2pi(R+rcos theta))+(mu_(0)r)/(2pi(2R-r cos theta))`
Discussion
41894.

Evaluate how many lines there are in a true ratational spectrum of CO milecules whose natural vibration frequency is omga= 4.09.10^(14)s^(-1) and moment of inertial I= 1.44.10^(39)g,cm^(2).

Answer»

Solution :We PROCEED as above. CALCULATING `(2Iomega(0))/( ħ)` we get
`(2I omega_(0))/( ħ)~= 1118`
Now this must equal `J(J+1)~~(J+(1)/(2))^(2)`
Taking the square root we get `J ~~33`
Discussion
41895.

The following table has 3 columns and 4 rows. Based on table there are THREE questions. Each question has FOUR options (A) , (B) , ( C) and (D) . ONLY ONE of these four options is correct The combination that corresponds to minimum induced emf in rod is

Answer»

(III) (ii) (S)
(I) (IV) ( R)
(II) ( i) (Q)
(IV) (ii ) ( S)

Solution :de = Bvdr
`epsilon = int_(0)^( R) (mu_(0)i)/(2pi(R+rcos THETA)) XX( omegar ) dr`
Find `epsilon ` for different `theta`
Electric field at mid point = `(depsilon )/(dr) = B ( omega (R )/(2))`
Here `B =(mu_(0)i)/(2pi(R+rcos theta))+(mu_(0)r)/(2pi(2R-r cos theta))`
Discussion
41896.

rदुरी से विलग (separated) दो इलेक्ट्रॉनों के बीच लगने वाला बल समानुपाती होता है

Answer»

`R^2` के
`r` के
`r^-1` के
`r^(-2)` के

Answer :D
Discussion
41897.

The stress applied to the wire having mass per unit length as 0.1g/(cm) and density 7X10^(-3)g/(cm)^3 when 10g.wt is attached to its free end is

Answer»

a)`668 "dyne"/(cm^2)`
b)`68.6 "dyne"/(cm^2)`
C)`66.8 "dyne"/(cm^2)`
d)`686 "dyne"/(cm^2)`

Answer :D
Discussion
41898.

Which of the following is quantised according to Bohr’s theory of hydrogen atom

Answer»

Linear MOMENTUM of electron
Angular momentum of electron
Linear velocity of ELECTRO
Angular velocity of electron

Answer :B
Discussion
41899.

Gamma rays and radio waves travel with the same velocity in free space. Distinguish between them in tems of their origin and the main application.

Answer»

SOLUTION :Gamma rays and radio waves both TRAVEL with the same velocity in free space because both of them are electromagnetic waves. Differences between them are:
(i) Origin: Gamma rays are produced in nuclear reactions and during RADIOACTIVE decay but radio waves are produced by the accelerated motion of charges in conducting wires
(II) Application: Gamma rays are used in MEDICINE to destroy cancer cells but radio waves are used in radio and TV communication systems.
Discussion
41900.

Give reasons for (a) Ligther elements are better moderators for a nuclear reactor than heavier elements. (b) In a natural uranium reactor, heavy water is prefered moderator to ordinary water. (c) Cadmium rods are provided in a reactor. (d) Very high temperatures as those obtained in the interior of the sun are required for fusion reaction to take place.

Answer»

Solution :(a) A good moderator has two properties. It slows down neutrons by elastic collision and it dose not remove them form the core by absorbing them. That is why ligther elements are better moderators.
(b) Heavy water is used in reactors using natural uranium as FUEL. This is becuase it has lesser absorption probability of neutrons than ordinary water.
(c) Cadmium RODS have a high cross section for neutron ansorption. They are used for controlling the nuclear chanin reaction responsible for producing nuclear energy
(d) for fusion reactions, very high TEMPERATURES are required. It enables protons to have enough kinetic energy and overcome their mutual electrostatic repulsion and come CLOSER than the range of nuclear forces of attraction.
Discussion