Physics Model Question 1

 

CLASS:12 SUBJECT: PHYSICS

FM: 75 TIME: 3Hrs

GROUP-A

Attempt all questions (11x1=11)

1.       A SHM has a amplitude A and time period T. The time taken by it to travel from x= A to x= A/2 is;

(a)                  T/12 (b) T/3 (c) T/4 (d) T/6

2.       A liquid is flowing in a tube under streamline flow. If the radius of the tube is doubled , the rate flow becomes;

(a)   Four times (b) 16 times (c) 2 times (d) ¼ times

3.       15g of air is heated from 0⁰C to 5⁰C at constant volume by adding 150cal of heat. The change in internal energy per gram is;

(a)   150cal (b) 10cal (c) 750cal (d) 180cal

4.       The distance between two consecutive antinodes is 0.5m. The distance travelled by the wave half the time period is;

(a)   0.25m (b)0.5m (c)1m (d) 2m

5.       Young’s double slit experiment confirms that;

(a)   Light is source of energy (b) Particle nature of light
(c) wave nature of light (d) None of all

6.       Which of the following is the most important factor that helps to recognize a person by his voice?

(a)                  Intensity (b) pitch (c) quality (d) None of all

7.       Seebeck effect helps to determine;

(a)   Direction of current (c) magnitude of emf

(b)  Nature of material (d) all of above

8.       Magnetic field inside the solenoid is independent of;

(a)                  Length of solenoid (c) radius of solenoid

(b)     Number of turns in solenoid (d) current in solenoid

9.       An electron and proton are under the effect of same electric field. They will experience;

(a)     Same force and same acceleration

(b)     Same force and different acceleration

(c)     Different force and same acceleration

(d)     Different force and different acceleration

10.    The unit of inductance is;

(a) V/m b) J/A c) VsA^-1 d)VA

11.    Which gate is an inverter gate?

(a)NOT gate (b) OR gate c) AND gate d) none

GROUP-B

Attempt all questions[8x5=40]

1.     (a) In a physics lab, you attach a 0.200 kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring’s force constant[3]

b) At what point in the motion of a simple pendulum is the string tension greatest? Least?[2]

2.       (a)What is molecular theory of surface tension.[3]

(b).The ship rides several centimeters lower in the water in lake than it did in the ocean. Explain.[2]

3.       (a)Derive the Poisseulle’s formula[3]

(b).How can you distinguish the laminar flow of fluid with turbulent flow?[2]

4.       (a)What is molecular theory of surface tension.[3]

(b).The ship rides several centimeters lower in the water in lake than it did in the ocean. Explain.[2]

5.       What is the Wheatstone bridge? Find the balanced condition of Wheatstone’s by using Kirchoff’s laws[1+4=5]

6.       (a)How did Laplace corrected Newton’s formula for velocity of sound?[3]

(b) Calculate the speed of sound in air at 20⁰C if density of air is 1.29kg/m³, molar mass 28.8g and pressure 1.1013 x10⁴Pa (γ for air is 1.4)[2]

7.       (a)Define the coherent sources of light[1]

(b).Prove analytically that the bright and dark fringes in Young’s double slit experiment are equally spaced.{No necessary for full derivation}[2]

(C) define the closed organ pipe and hence define the resonance condition for the stationary wave formation[2]

OR

Describe the different modes of stationary wave produced in the closed organ pipe[3]

What is end correction of organ pipe[2]

8.       (a)How the electrons are emitted from metal surface in the photoelectric effect?[2]

(b).Derive the Einstein’s photoelectric equation[3]

OR

What are rectifiers? [1]

How the filter circuit added to rectifier can make uniform output voltage.[2.]

Write the forward bias characteristics of PN junction diode.[2]

GROUP-C

Attempt all questions[3x8=24]

1.       (a) What is the use of Biot and Savart’s law?[1]

(b) Describe how the Hall voltage is developed during Hall effect.[2]

(c) What is Ampere’s circuital law[2]

(d) Use Ampere’s law to determine the magnetic field due to solenoid.[3]

OR

(a)Define electromagnetic induction.[1]

(b)What is Lenz law ? How does it follow the law of conservation of energy[3]

(c) Describe about the working of potentiometer and hence write the expression of internal resistance of given cell with use of potentiometer

 

 

 

 

2. (a) A machinist is using a wrench to loosen a nut. The wrench is 25.0 cm long, and he exerts a 17.0 N force at the end of the handle at 37° with the handle . (i) What torque does the machinist exert about the center of the nut? (ii) What is the maximum torque he could exert with a force of this magnitude, and how should the force be oriented?[2+3=5]

(b)Derive the expression of acceleration of body rolling down an inclined plane[3]

3. What are Bohr’s postulates for hydrogen like atom? Hence find the expression of radius of nth orbit, velocity of electron and total energy of electron in that orbit by using the Bohr’s postulates.[3+1.5+1+2.5]

 

 

 

 

Sample Research Report

 

Sample Research proposal

 

SEE Science New Model Question 2079

 

Quantum Statistical

aQUANTUM STATISTICS PHYSICSNEED FOR QUANTUM STATISTICS In classical statistics, we assume that it is possible to simultaneously determine the position and momentum coordinates of a gaseous particle as precisely as we like. It means that these particles are distinguishable and can be labelled. But this is not true in practice. Heisenberg’s uncertainty principle forbids determination of the position (q) and the momentum (p) of a particle simultaneously with infinite precision. If the uncertainties in the measurements of q and p are Δq and Δp, respectively, the product ΔqΔp cannot be made less than ℏ/2. i.e ΔqΔp⩾ℏ

2where ℏ=h

2𝜋 and h=6.62 ⨯10-34 Js is Planck’s constant. It implies that when we study the behaviour of an assembly of identical particles statistically,we should treat it as a collection of indistinguishable particles. Further, blackbody radiation deserves a unique place in physics because it gave birth to quantum theory. BOSE EINSTEIN STATISTICSFollowing are the conditions of Bose-Einstein statistics(1) Particles are identical and indistinguishable.(2) Particles don't obey Pauli exclusion principle.(3) The sum of particles in each quantum state is the total number of particles.(4) The sum of energy of each particle in the quantum state is the total energy.Let us consider N particles in system. Let particles are in first, second, ... ith cell with average energy 𝜀1,𝜀2,…, 𝜀i for each particle. Let g is the degenency factor. Suppose that ni particles are arranged in a row and distributed among gi quantum states with (gi-1) partitions in between. The total number of possible arrangements of particles and partitions is equal to the total number of permutations of (ni+gi-1) objects in a row. Therefore the total possible ways of arranging ni particles with (gi-1) partitions =(ni+gi-1)!As the particles are identical and indistinguishable the possible number of distinct arrangements =(ni+gi-1)!

ni!(gi-1)! The total number of different and distinguishable ways of arranging N paricles in all the available energy states is given by,a

Ω	=(n1+g1-1) n1!(g1-1)×(n2+g2-1) n2!(g2-1)×…
∴Ω	= ∏i(ni+gi-1)! |ni!(gi-1)! ...............(i)

Also we have, logx!=xlogx-x, for large x∴ From (i), |logΩ= ∑ilog(ni+gi-1)!-logni!-log(gi-1)! = ∑i(ni+gi-1)log(n i+gi-1)-(ni+gi-1)-|| ||[(nilogni-ni)-(gi-1)log(gi-1)-(gi-1)]∴logΩ= ∑i(ni+gi-1)log(ni+gi-1)-nilogni-(gi-1)|log(gi-1)]For Maxima 𝛿logΩ = 0i.e 𝛿[ ∑i(ni+gi-1)log(ni+gi-1)-nilogni-(gi-1 )|log(gi-1)]=0 ∑i𝛿(ni+gi-1)log(ni+gi-1)-𝛿[nilogni]-𝛿(gi-1)|log(gi-1)=0 ∑ilog(ni+gi-1).𝛿ni+(ni+gi-1).𝛿ni

(ni+gi-1)- logni𝛿ni-ni.𝛿ni

ni=0 [ indipendent on gi] ∑ilog(ni+gi-1).𝛿ni+𝛿ni-logni𝛿ni-𝛿ni=0∴𝛿logΩ= ∑i[log(ni+gi-1)𝛿ni-logni𝛿ni]=0 ∴ ∑ i[-log(ni+gi-1)+logni]𝛿ni=0.............(ii)Also, ∑i𝛿ni=0 ..............(iii)and ∑i𝜀i𝛿n=0 ................(iv)Multiplying (iii) by 𝛼,(iv) by 𝛽 and adding to (ii) we get, ∑i[-log(ni+gi-1)+logni+𝛼+𝛽𝜀i]𝛿n i=0 or, -loge(ni+gi-1)+logni+𝛼+𝛽𝜀i=0 ∴-log(ni+gi)+logni+𝛼+𝛽𝜀i=0 {1 can be neglected compared with |(ni+gi)}∴log(ni

ni+gi)=-𝛼-𝛽𝜀ior.-log(ni +gi

ni)=-(𝛼+𝛽𝜀i)ni+gi

ni=e𝛼+𝛽𝜖ior.1+gi/ni=e𝛼+𝛽𝜖i or.a

gi ni	=(ea+𝛽𝜖i-1)
	[ni=gi (e𝛼+𝛽𝜖i-1)]

This is Bose Einstein distribution law. Fermi Dirac Statistics The condition for Fermi Dirac statistics are(1) The sub levels consists only ni= 0 or 1. That is particle obey Pauli exclusion principle.(2) The particles are identical and indistinguishable.(3) The sum of energy possessed by a particle in each sub-levels of different energy levels is the total energy. Let g be the degeneracy factor. Let ni be the number of particles for energy level 𝜀i. The first particle can be placed in any one of the available gi states i.e., this particle can be assigned to any of gi sets of quantum numbers. Thus first particle can be distributed in gi different ways and the process continues. Thus the total number of different ways of arranging ni particles among the gi states with energy level 𝜀i is=gi(gi-1)…[gi-(ni-1)]=gi!

(g i-ni)!Further, if the particles are taken to be indistinguishable, it will not possible to detect any difference when ni particles are reshuffled in to different states occupied by them in the energy level 𝜀i.∴ The total number of different and distinguishable ways isgi!

ni!(gi-ni)!∴ The total number of different and distinguishable ways of getting the distribution n1,n2,n3 etc. among the various energy levels 𝜀1,𝜀2,…𝜀i ete can be calculated by multiplying the various factors.a

∴Ω	=g1! n1!(g1-n1)!⋅g2! n2!(g2-n2)!⋯= ∏igi! ni!(gi-ni)!
∴ logΩ	= ∑i[loggi!-logni!-log(gi-ni)!]
∴logΩ	=∑(( giloggi-gi)-(nilogni-ni)-(gi-ni)log(gi-ni)-(gi-ni)]

a

logΩ	= ∑i[gi-loggi-nilogni-(gi-ni)log(gi-ni) ]

a

𝛿logΩ

= ∑i[𝛿gi-𝛿loggi-𝛿(nilogni)-𝛿(gi-ni)log(gi-ni)]

= ∑i-ni.1

ni𝛿ni- logni𝛿ni-[(gi-n i).1

gi-ni)𝛿(-ni)+log(gi-ni)𝛿(-ni)}a

∴𝛿logΩ

= ∑i[logni-log(gi-ni)]𝛿ni

For maximum 𝛿(logΩ)=0|⇒ ∑i[logn i-log(gi-ni)]𝛿ni=0....................(i)But ∑i𝛿ni=0 ...........................(2)and ∑i𝜖i𝛿ni=0............................(3) Multiply (2) by 𝛼 and (3) by 𝛽 and add to (1)[∑logni-log(gi-ni)+𝛼+𝛽𝜖i] 𝛿ni=0|here, 𝛿ni≠0so, or logni-log(gi-ni)+𝛼+𝛽𝜀i=0 logni-log(gi-ni)=-𝛼-𝛽𝜀i log(gi-ni)-logni=𝛼+𝛽𝜀ior, logg i-ni

ni=𝛼+𝛽𝜖i gi-ni

ni=e𝛼+𝛽𝜖ior, gi

ni-1=e𝛼+𝛽𝜖ior, gi

ni=1+e(𝛼+𝛽𝜖i)∴ni=gi

e𝛼+𝛽𝜀i+1 This is Fermi Dirac distribution law. Comparision of THREE STATISTICSThe expressions for the most probable distributions in the thres statistics are:(1) Maxwell's Boltzmann statistics:n1=gi

e𝛼+𝛽𝜀i⇒gi

ni=e𝛼+𝛽𝜖iwhere 𝛼,𝛽 are constant, gi is the degeneracy of ith level and ni are the number of particles for energy 𝜀i.(2) Bose Einstein statistics:a

n i	=gi e𝛼+𝛽𝜖i-1
or, gi ni+1	=e𝛼+𝛽𝜖i

(3) Fermi Dirac statistics:ni=gi

e𝛼+𝛽𝜀i+1or. gi

ni-1=e𝛼+𝛽iIt is to be noted that if g1

n i is very large in comparison to unity we may write, gi

ni≈gi

ni+1≈gi

ni-1i.e., For large values of gi

ni Bose Einstein and Fermi Dirac distributions approach the Maxwell-Boltzmann distribution. This is the case for normat existence of gases when the temperature is not too law and pressure is not too high.ELECTRON GASElectrons in a metal belong to a most characteristics system of fermions because electrons obey the exclusion principle. For electrons in a metal, the energy levels are grouped in bands. Practically at all temperature, the lower level energy bands are filled with electrons. The upper level energy bands are partially filled with electrons. The distribution of electrons is to be considered only in the upper bands called conduction bands. The zero energy level is taken at the lowest level of conduction band. It is assumed that the electrons have frce movements within the conductor, provided the energy associated with the electrons is of the order of upper energy bands.We know from Fermi-Diarc distribution law,ni=gi

e𝛼+𝛽𝜀i+1where n i is the number of particles in ith cell and gi is the degeneracy factor for ith cell. As the energy of electron in the conduction band is continuous the degeneracy factor gi is replaced by g(E)dE and ni by n(E)dE i.e., n(E)dE=g(E)dE

eaeE/KT+1Here n(E) refers to number of electrons, g(E) refers number of phase cells.But g(E)dE in terms of momentum is,g(p)dp=8𝜋v

h3 P2dP=8𝜋v

h3P.PdPBut E=1

2mv2 and P=mvaP=(2mE)1/2∴P2=2mE⇒2PdP=2mdE⇒PdP=mdEa∴g(E)dE=8𝜋v

h3(2mE)1/2mdE =82𝜋v

h3m3/2 E1/2dE∴n(E)dE=82𝜋v

h3⋅m3/2E1/2dE

e𝛼eE/KT+1But for clectron gas 𝛼=-EF

KT,EF is Fermi energy∴n(E)dE=82𝜋v

h3m3/2E1/2dE

e(E-EF) /KT+1This is exact form of Fermi Dirac law of energy distribution electrons. At the temperature T=0K, number of electrons is equal to the total number of energy states occupied by the electrons from zero to EF.a

∴n	=EF∫0g(E)dE=82𝜋v h3m3/2EF∫0E1/2dE
	=82𝜋Vm3/2 h3[2 3E3/2]EF0
	=162𝜋Vm3/2 3h3⋅E3/2F
or, E3/2F	=3nh3 162𝜋Vm3/2
or, EF=	h2 2m(3n 8𝜋V)2/3

FERMI LEVEL AND FERMI ENERGYThe topmost filled energy level at absolute zero temperature is called Fermi level and corresponding energy at topmost level (Fermi level) is called Fermi energy. i.e., the energy values upto which all the states are full at 0K and above which all the energy states are empty is known as Fermi energy.

Numerical Suppose that we have two state system, in which the first state has energy 𝜀 and second state 3𝜀. Give the ratio of probability of occupancy of first level to the probability of occupancy of second level, and simplify it.[TU 2075]Solution:There are two energy states 𝜀1 and 𝜀2 are 𝜀 and 3𝜀 for n1 amd n2 level with probability of occurrence P1 for state 1st and P2 for that of 2nd . Then probability P1 and P2 for two staes are,P1=Cexp(-𝜀1

kBT)And P2=Cexp (-𝜀2

kBT)Dividing (ii) by (i)Now, P1

P2=exp(𝜀2-𝜀1

kBT)[Taking room temperature]or, P1

P2=exp( 3𝜀-𝜀

kBT)or, P1

P2=exp(2𝜀

kBT)∴ The ratio of probability of occupancy of first level to the probability of occupancy of second level is exp(2𝜀

kBT).