Csir Ugc Net_physics_ Free Solved Paper

C SIR NET, GA GA TE, IIT-JA IIT-JA M, UGC UGC NET , TI FR, IISc , JE JEST , JNU, BHU BHU , ISM , IBPS IBPS , CSAT, CSAT, SLET, NIM CET, CET, CT ET

(4) 128 7.

A car ow ow ner buys buys petrol at Rs. Rs.7.50 7.50,, Rs. 8 and Rs. 8.50 per per l iter f or thr ee succ su ccess ess ive ve years. year s. What approximately is the average cost per liter of petrol if he spends Rs. 4000 4000 each year?

(1) Rs. 7.98 (2) Rs. 8 (3) Rs. 8.50 (4) Rs. 9

8.

In a certain store, the the prof it is 320% of the cost. IfIf the co c ost increase increases s by by 25% 2 5% but the selling selling price remains remains constant, approx approx ima imate ly w hat percentage per centage of the selling selling price price is the profit?

(1) 30% (2) 70% (3) 100% (4) 250%

9.

Today is Friday Friday af ter 62 days, it w ill be : (1) Thursday Thursday (2) Friday Friday (3) Wednesday (4) Tuesday

Phone: 0744-2429714 Mobile: 9001297111, 9829567114, 9001297243 Website: www.vpmclasses.com E-Mail: vpm cl classe asse [email protected] / in inff o@v pmclasses.c mclasses.co om A ddress: ddress: 1-C 1-C-8, -8, Shee Shee la Ch C howdhary Roa Road d, SFS, TA LWAND LWANDI, I, KOTA , RAJASTHA RAJASTHA N, 324 3240 0 05

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10. A car travelling travelling w ith of of its ac tual speed cov ers 42 km km in 1 hr hr 40 min min 48 se s ec. Find the actual speed of the car. 6 (1) 17 k m /h / hr 7 (2) 25 km/hr (3) 30 km/hr (4) 35 km/hr 11.

P is a w orking rking and Q is a sleeping sleepin g partne partn er. P puts in Rs. 3400 and Q puts Rs.6500. P rec eives 20% 20% of the prof its f or managing. managing. The rest rest is distributed in proportion to their capitals. capitals. Out Out of a total profit prof it of Rs.990, how how much did P get ? (1) 460

(2) 470 (3) 450 (4) 480 12.

A law law n is the f orm of of a rectangle having having its its s ide in the ratio ratio 2:3 2 :3 The ar ea of the law n is is 1/6 hectares. Find the length and breadth of the lawn. (1) 25m

(2) 50m (3) 75m (4) 100 m 13.

A n aeroplane aeroplane covers co vers a ce c ertain distance distance at a spee spe ed of 240 240 km k mph in 5 hours. T o cov er the same d istance in 1 hours , itit must trav el at a spee s peed d of :

(1) 300 kmph Phone: 0744-2429714 Mobile: 9001297111, 9829567114, 9001297243 Website: www.vpmclasses.com E-Mail: vpm cl classe asse [email protected] / in inff o@v pmclasses.c mclasses.co om A ddress: ddress: 1-C 1-C-8, -8, Shee Shee la Ch C howdhary Roa Road d, SFS, TA LWAND LWANDI, I, KOTA , RAJASTHA RAJASTHA N, 324 3240 0 05

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(2) 360 kmph (3) 600 kmph (4) 720 kmph 14.

Find Find out the missin mis sing g nu mber of o f the the given questio ques tion: n: 2

7

4

5

2

3

1

?

6

10

42

72

(1) 2 (2) 4 (3) 5 (4) 3 15.

A ll of the f ollo ollow w ing are are the th e same in a manner. anner. Find nd out the one which is different among them:

(1) BFJQ (2) RUZ G (3) GJOV (4) ILQX PART B (16 -35) 16 .

What is the solution of integral ∞

∫ [cos(3 x) + 2] δ( x – π )dx 0

(1) 0 (2) 2 (3) 3 (4) 1 ∞

17

Solve the integral equation

1 – α, 0 ≤ α ≤ 1 f ( θ) cos αθ d θ =  α >1 0, 0

∫


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(1) f(θ ) = (2) f(θ ) = (3) f(θ ) = (4) f(θ ) = 18 .

19 .

2(1 – co s θ)

πθ 2 2cos θ

πθ2 (1 – cos θ) 2πθ2 (1 – cos θ)

πθ2

Fi nd the the fun ction tion whose lapl ace ace t rans ransfo form rm is

(1)

1 [at sin at at + co s a t] t] 2a

(2)

1 ( a + sin a t − cos at ) 2a

(3)

1 [a + cos at at + si n at ] 2a

(4)

1 [at co sa t + sin at at] 2a

s2 (s2 + a2 )2

T he L Lag ag range range equa tion ion of m otion tion o f two ri gid bo dies dies of m asses ‘m’ and ‘2m ‘2m ’ are connected connected by a li ght ght fl exibl exible e spri ng of spring pring const constan t K. what i s the Lagrange Lagrange equation equation o f m otion. (1) ɺxɺ +

(2) ɺxɺ +

k x =0 m k 2m

x =0

(3) ɺxɺ +

3k x =0 2m

(4) ɺxɺ +

5k x =0 2m


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20 .

In the following indicator diagram, the net amount of work done will be (1) Positive (2) Negative (3) Zero (4) Infi Infi nity

21 .

A particle moves in a plane under the influence of a force, force, acting towards a centre of force whose magnitude is F =

1

1 – 2 

rɺ 2 – 2ɺɺ rr 

th e dista distance nce of the parti parti cl e to the centre o f force, force, th en  whe re r i s the r  c2  the Lagrangian Lagrangian for the motion in a plane plane i s

(1) L =

rɺ r ɺθ 1 1 rɺ – – 2 + 2 2 r c r

(2) L =

rɺ2 r 2θɺ 2 1 2 rɺ 2 + + + 2 2 2 2 r c r

2

(3) L =

rɺ 2 2

2

+

2

r 2 ɺθ2 2

2

–

2 r

+

1 rɺ c2 r 2

rɺ 2 1 1 rɺ2 (4) L = – + 2 2 2 r c r 22 .

Cal culate culate the Fermi energy energy in ele ctro ctron n vol t for sodium sodium assum i ng that it has has one free electron electron per –3 atom. Given Given densi densi ty of sodium = 0.97 g cm , atomic weight of sodium is 23. (1) 3.541 3.541 e V (2) 3.451 3.451 e V (3) 5.135 5.135 e V (4) 3.145 3.145 e V


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23 .

The paramagnetic contribution to the magnetic susceptibility per m3 of potassium, for which the Fermi energy is 2.1 eV is at wt. of potassium is 39.1 gm and density of potassium is 0.86 × 10 3 kgm –3. (1) 420.5 × 10 –8 (2) 450.2 ×10+8 (3) 420.5× 10 –6 (4) 425.2×10+6

24 .

The figure shows the inverse magnetic susceptibility (1/ χ) (dimens (dimensii onl onl ess) ess) a s a funct fu nctii on of temperature for a paramagnetic material. Calculate the concentration of magnetic ions, if they are assumed to be Co 2+ with configura tion tion 3d 7. (1) 5·4 × 10 23 (2) 5·1 × 10 26 (3) 6·9 × 10 23 (4) 6·9 × 10 26

25 .

A 3D structu structu re of current current carrying carrying wire is as shown in the fi gure. The m agneti agneti c force experienced experienced by  cha rge rge parti cl e of mass mass m a nd charg cha rge e q, when when it i t is cros cross si ng origin with with velo city v along +ve Y-axis will be

 µ 0Ι qν µ 0Ι qV  + i 4πR   8R

(1) 

(2)

2 µ0 Ιqν   1 +  ˆi – kˆ    8R  π 

µ0 Ιq ν ˆ

(3) –

8

k

(4) Zero


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26 .

If E f(0) and Ef are Fermi levels of a metal at 0K and 30000 K, then what is the value of

Ef if E f(0)

E f(0) = 7 eV. (1) 0.119 (2) 0 .88 .88 (3) 1.113 (4) 1.188 27 .

–1. Find the internuclear The small (rotational) Raman displacement for HCI molecule molecule is 41·6 cm –1 –34 J s, c = 3·0 × 108 m s –1 di stance tance b etw et ween t he atoms form ing the m olecule. ecule. Given : h = 6 ·63 × 10 –34 2 3 –1 and NA = 6·023 6·023 × 10 m ol .

(1) 1·29 A° (2) 2·29 A° (3) 2·49 A° (4) 0·64 A° 28 .

For the given circuit the the open loop gain is 12000 and R1 = 120 kΩ and Rf = 600 kΩ . V i = 1.2 V. Find t he exa ct outp ut vol vol tage fo r the inverti inverti ng op erationa erationall amplifie amplifie r. (1) – 5 V (2) – 6 V (3) – 5·99 V (4) – 4·99 V

29 .

Oxygen has nuclear spin of 5/2. The NMR of oxygen gives (1) 2 lines (2) 3 lines (3) 4 lines


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(4) 6 lines 30 .

A metal strain gauge factor of two. Ιts nom inal inal resista resista nce i s 120 oh o hm s. It undergoes strai strai n 10 –5, the value of change of resistance in response to the strain is (1) 240 ohms (2) 2 × 10 5 ohm ohm (3) 2.4 × 10 –3 ohm (4) 1.2 × 10 –3 ohm

31 .

Evaluate

∫ ∫∫





Fd τ, where F = xyz 2 over the prism placed at o rigin as sho sho wn in i n the Figure.

V

1 3 2 (2) 3 1 (3) 9 2 (4) 9 (1)

32 .

A star initially has 1040 deut erons. erons. I t prod prod uces en ergy vi vi a the processes processes H2 + 1H2 → 1H3 + p and

1

H2 + 1H3 → 2He 4 + n

1

If the average power radiated by the star is 10 16 W, the deuteron supply of the star is exhausted in a tim e of the o rder of (1) 10 6 sec (2) 10 8 sec (3) 10 12 sec (4) 10 16 sec The masses of the nuclei are as follows: M (H2) = 2·014 amu; M (p) = 1·007 amu; M (n) = 1·008 am am u; M (He4) = 4·001 amu.


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33 .

A γ -ray -ray photon produces an electron-positron pair, each moving with a K.E. of 0·01 MeV. The energy of the γ -ray -ray photon i s (1) 1·02 1·02 MeV (2) 1·04 1·04 MeV (3) 2·08 2·08 MeV (4) 1·03 1·03 MeV

34 .

The temperature of the two outer surfaces of a composite slab, consisting consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T 1 (T 2 > T 1).

x

T2

4x

2K

T1

 A( T2 – T1 )K   f, with f equals to x  

The rate of heat transferred through the slab, in a steady state is  (1) 1 (2) 1/2 (3) 2/3 (4) 1/3 35 .

In quark model what is the state of η° (1) ud

(2)

(3)

(4)

1 2 1 2 1 2

(uu –dd)

(u u + d d)

(us – su)

PART C (36 -55)


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36 .

The value of the counter integral

(1)

37 .

∫

C

6

sin z

 z – 1 π    6  

3

dz , i f C i s t he he ci rc rcl e z = 1.

20 πi 17

(2)

21 πi 16

(3)

15 πi 7

(4)

12 πi 13

T he matri matri x A, def defined ined by

1 2 2   A =  2 a –b –b   –2 b –a    Is ortho ortho gonal i f

38 .

(1) a = 1,

b=1

(2) a = 1 ,

b=2

(3) a = 2,

b=1

(4) a = 2,

b =

A reversible engine works between three thermal reservoirs, A,B and C . The engine absorbs an equal amount of heat from the thermal reservoirs A and B kept at temperatures TA and TB respecti respecti vely, an d reje rejects cts heat to the the rm al reservoir reservoir C kep t at temperature TC. The efficiency efficiency of the engine is α times the efficiency of the reversible engine, which works between the two reservoirs A an d C. C. which whic h one of the follo follo wing rel rel ation statemen t i s corre co rrect ct ? (1)

TA T = ( 2α – 1) + 2 (1 – α ) A TB TC


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39 .

(2)

TA T = ( 2α – 1) A + 2 ( α – 1) TB TC

(3)

TB T = ( 2α – 1) + 2 (1 – α ) C TA TA

(4)

TA T = ( 2α – 1) A + 2 (1 – α ) TC TB

A quantum mechanical particle of mass m is confined in three-dimensional three-dimensional infinite square well 9π 2 ℏ2 po tent tentii al of side ‘a’. The ei gen-ene gen-ene rgy rgy of part pa rti cl e is given as E = . The state state i s ma 2 (1) 4 f old old deg degen en erate (2) 3-fold degenerate (3) 2-fold degenerate (4) Non-degenerate

40 .

T en gramm gramm as of wate waterr at 20° 20° C i s con verted erted i nto at – 10°C at constant constant atmos atm osph ph eric press pressure ure . Assuming the specific heat of liquid water to remain constant at 4.2 J/gK and that of ice to be half of thi th i s value, value, and and tak ta ki ng the lat l atent ent heat of fusion fusion o f ice at 0°C 0°C to to be 335 335 J/g , the to tal entropy of the system is . (1) Zero –1 (2) 16.02 JK –1 –1

(3) – 15.63 JK –1

(4) 15.63 15.63 JK 41 .

A sphere rolls do wn a rough included plane; i f x be the distance distance of the point p oint of contact contact o f th t he sph ere from a fixed point on the plane, find the acceleration. (1)

(2)

5 g sin α 7 5 14

g sin α

(3) mg sin α


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7 g sin α 5

(4) 42 .

A symmetrical top with moments of inertia Ιn = Ι y and and Ι z in the body axes frame is described by the Hamiltonian 1 2 1 2 L x + L2y + Lz 2Ix 2Ix

(

H=

)

Here moments of inertia are parameters and not operators. L x L y and L z are the angular momentum op erator erator i n the the body axe s frame. frame. (i) The eigenvalues of the Hamiltonian (1)

(2)

ℏ

2

2 Ιx ℏ

+ 1) + 

ℓ(ℓ

+ 1) + 

2

2 Ιx

 1 1  2 2 – ℏ m  2Ι z 2Ι x 

ℓ(ℓ

 1 1  2 2 + ℏ m 2 2 Ι Ι  z x 

2 2  1 1  2 ℏ m – ( 1 ) + + ℏ ℓ ℓ  2Ι x  2 Ιx 2Ι z 

(3) 

2 2  1 1  2 ℏ m ( 1 ) + ℏ ℓ ℓ + +  2Ι x  2 Ιx 2Ι z 

(4) 

(ii) Expected value for a measurement of L x + L y + Lz for a ny stat e i s (1) Zero (2) – ℏm (3) ℏm (4) 43 .

ℏ

m 2 If we take in the semi-empirical mass formula ac = 0.58 MeV and aa = 19.3 MeV. Then possible atomi atomi c number of most most stable nu cl ei of m ass nu mber 64.

(1) 26 (2) 29 (3) 32 (4) 33


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44 .

36 g of water at 30°C are converted into steam at 250°C at constant atmospheric pressure. The spe cific heat he at of water is assume assume d cons constant ta nt at 4.2 J/g J /g K and the latent heat of vapo vaporization rization at 10 0°C 0°C is Cp 22 60 J/g. For wate r vapou r, assum assum e pV = mRT mRT where R = 0.4619 kJ/ kJ/ kg K , and = a + bT + cT °, R where a = 3.634, b = 1.195 × 10 –3 K –1 and c = 0.135 × 10 –6 K –2 Calculate the entropy change of the system. (1) 0.2181 kJ/K (2) 0.0235 kJ/K (3) 273.1 J/K (4) 314.3 J/K

45 .

A perpendicularly polarized wave propagates from region 1(εr1 = 8.5, µr1=1 , σ1 = 0) to region 2, free spa ce, with wit h an angle of i ncidence of 15°. 5°. Given Given E i0 = 1 .0 µV / m , then Er0 , i s– s– (1) 1.62 µ V/m (2) 0.623 µV/m V/m (3) 4.23 µ V/m (4) 7.75 mV/m

46 .

A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on collision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B? (1) 9 m (2) 6 m (3) 3 m (4) 2 m

47 .

Cal culat culat e the amount of energy rele released ased i f all all the deuteri deuteri um at oms i n the water in the lake of area area 1 about about 1 0 5 sq. miles mile s and and o f dept depth h the mile m ile area u sed u p in fusio fusio n. 20 (1) 2.18 2.18 × 1038 M eV


Page 17


(2) 43 MeV (3) 1.56 1.56 × 1039 M eV (4) 6.9 × 10 38 MeV 48 .

T he maximum maximum wa ve length of photons photons that can can be det detected by a photo di di ode m ade of a sem iconductor iconductor of band band gap 2 eV is about (1) 620 nm (2) 700 nm (3) 740 nm (4) 860 nm

49 .

T he three three electroni electroni c ci ci rcuits m arked (i), (ii ) and (iii ii i ) i n the figu re below below can c an al al l work as logi l ogi c gates, gates, where the input signals are either 0V or 5V and the output is V 0.

Identi Identi fy the correct combina combina tion of l ogi c gates (i(i ), (i (i i ), (iii (iii ) in in the the o ptions gi ven belo w. (1) NOR, XOR, AND (2) OR, NAND, NOR (3) NAND, AND, XOR (4) AND, OR, NOR

Statement Sta tement for for Linked Answer Answer Question 50(i) 50(i) and 50 (ii) Phone: 0744-2429714 Mobile: 9001297111, 9829567114, 9001297243 Website: www.vpmclasses.com E-Mail: vpm cl classe asse [email protected] / in inff o@v pmclasses.c mclasses.co om A ddress: ddress: 1-C 1-C-8, -8, Shee Shee la Ch C howdhary Roa Road d, SFS, TA LWAND LWANDI, I, KOTA , RAJASTHA RAJASTHA N, 324 3240 0 05

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sin z Let f(z) = cos z – for non-zero z ∈ C and and f(0) = 0. Al so, let g (z) = sinh sinh z for z ∈ C. z 50(i).

T hen f(z) ha s a zero zero at z = 0 of order (1) 0 (2) 1 (3) 2 (4) Greate r than than 2

50(ii). Then

g(z) zf(z)

has a pole at z = 0 of order

(1) 1 (2) 2 (3) 3 (4) Greate r than than 3 51 .

The Lagrangian of a system is giv gi ven b y L=

1 m r 2 (θɺ 2 + sin 2 φɺ 2 ) – V (r, θ, φ) 2

The equation of motion is (1)

d ∂V 2 2 (mr sin θφɺ ) – =0 dt ∂φ

(2)

d ∂V 2 2 (mr sin θφɺ ) – =0 dt ∂φ

(3)

d ∂V 2 2 (mr sin θφɺ ) + =0 dt ∂φ

(4)

∂V d 2 2 (mr cos θφɺ ) + =0 ∂φ dt

Linked question 52(i), 52(ii), 52(iii) 1 A one-dimens one-dimensional ional harm harm onic oscillat oscillat or of a particle particle with with m ass an and potential potential energy v(x) = mω 2 x 2 2


Page 19


T his pa rti rti cl e ha s a charg charg e q and and is plac placed ed in a unifo rm ele ctric field E pa ralle l to the x – axis, E = Exˆ . 52.(i). T he Hamiltonian amiltonian of t he pa rti rti cle (1)

(2)

P2 1 + m ω2 x2 − ε x 2m 2 P2 1 + m ω2 x2 + εx 2m 2

P2 1 (3) mω2 x 2 − εx 2m 2 (4)

P2 1 mω2 x 2 + εx 2m 2

52(ii). Perform a coordinate transformation y = ax+b (where (where a and b are consta onsta nt / such uc h that i n the y coo rdi rdi nate th t he Hamiltonian i s si m ilar to that of of a o ne – dimensi dimensi ona onall harm harm oni c oscilla tor tor (with (wit h no charge) What are a and b (1) a = 1 , b = ε / mω

2

(2) a = 1 , b = – ε / mω

2

2

(3) a = ε / mω , b = 1 2

(4) a =– ε / mω , b = 1 52(iii). The energy eigenvalues of the system is

53 .

(1)

ε2 1  1 ℏω  n +  − 2  2  m ω2

(2)

1  3  ε2 ℏω  n +  2  2  mω 2

(3)

1 e2  1 ℏw  n +  − 2  2  2mw 2

(4)

ε2 1  3 ℏω  n +  − 2  2  2m 2m ω2

The equation x3 – x2 + 4 x – 4 = 0 is to b e solved olved u si ng the Newton-Rep ewton-Rep hson m ethod. ethod. If x = 2 i s taken taken as the ini tial appro appro xi mati on usi usi ng this m etho d will be


Page 20


(1)

(2)

2 3 4 3

(3) 1 (4) 54 .

A mass m is released from rest at height h. Find the Hamilton characteri characteri sti c function of the sys syste m (1) (2)

55 .

3 2

∫ ∫

2m (E (E – mg z)1 /2 d z 1/2

2m (E + mg z)

dz

(3)

∫

2m(E m(E – mgz) gz) dz

(4)

∫

2m (E (E + m gz) d z

At what temperatu temperatu re will the number of Ι2 molecules in the ν = 1 l evel be one-te one-tenth nth of that i n the the ν = 0 –1 –1 le vel? Given : ω ε = 214.6 cm , –34 Js, Js, c = 3 .0 × 10 8 m s –1 and k = 1 .38 × 10 –23 J/K. ω exe = 0.6 cm –1, h = 6.63 × 10 –34

(1) 155.3 K (2) 135.5K (3) 133.5 K (4) 127.5 K

Answ Answ er key


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Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. 3 1 3 4 1 2 1 2 4 4 2 2 4 4 1

Que Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. 4 1 4 3 2 1 4 1 2 2 2 1 2 4 3

Que. 31 32 33 34 35 36 37 38 39 40 41 42(i) 42 ii 43 44

Ans. 3 3 4 2 2 2 2 1 2 3 1 1 3 2 3

Que. 45 46 47 48 49 50(i) 50(ii) 51 52 i 52(ii) 52(iii) 53 54 55

Ans. 2 4 4 1 4 3 3 3 1 2 3 2 1 3

Solutions PART A (1-15)

1.(3)

m1D1 m2 D2 = w1 w2 24 × 15 m2 × 12 = 1 80 24 m2 = 40

2.(1)

RA × SA 9×9 = PA ⇒ = PA QA 7 Diameter = PA + AQ 81 + 7 = 13 0 7 7 Rad ius =

3.(3)

Diameter ∴ Radius 2

= 65 7

R1 + R2 = a R2 + R3 = b R3 + R1 = c R1 + R2 + R2 + R3 + R3 + R1 = a + b + c


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⇒

a+b+c 2

R 1 + R2 + R3 =

4. (4) V alue alue of each tool in in 1985 =

10 × 107 18 × 10

= 5

5.(1)

7

[Since 1 crore = 10 ]

3

5 9

T housand housand

The required required perc entage

=

(since (sin ce tota totall employ mploy ed

= 360 360 – unemploy unemployed) ed) =

6.(2)

7.(1)

18 × 100 ( 360 –18 )

18 5 × 10 0 = 5 % 3 42 19

This is an alternating alternating m ultiplication ltiplication and a nd subtractin subtracting g s eries: eries: Firs t, multiply multiply b y 2 and then subtract 8.

 4000 4000 40 00  Total quantity of petrol =  litres + + 8 8.50   7.50 consumed in 3 years

2 1 2 + +  li ters 1 5 8 17  

4000 

 76700   litres  51 

=

Total amount spent = Rs. (3 x 4000) = Rs. 12000.

 12000 × 51 = Rs. 6120 = Rs.7.98  767  76700 

Average cost =  8.(2)

Let C. C. P.= Rs. Rs. 100. Then, Pr of it = Rs. 320, S.P. = Rs. Rs. 420. New C.P. .P. = 125% of Rs. 100 100 = Rs. 125 New S.P. = Rs Rs . 420. Prof it = Rs. (420 - 125) = Rs. 295.


Page 23


1 47 5  29 5 ×  Requ Req u ired percenta perce ntage ge =  1 00  = % = 70% (a (a pproxi mately ) 21  42 0 % A student multiplied a number by 9.(4)

3 5 instead of 5 3

Each day of the w eek is repeated repeated after 7 da days ys . So, af af ter 63 days, days , itit w ill ill be Friday Friday . Hence Hence after 63 days, it w ill be Thu Thurs rs day. day . Theref Theref ore the the requi r equired red day is Thursday .

4 51 12 6 hrs. 10 .(4 .( 4) 40 mi n = 1 hr s = 5 75 75 Time taken = 1 hr 40 min 48 sec = 1 hr Let the actual speed be x km/hr. 5 1 26 = 42 x× 7 75

Then,

 42 × 7 × 75  = 35km hr.  35km / hr.  5 × 126 

x =

11.(2) Given, Total prof it = Rs. 990 990 Ration of their capitals capitals = 34 : 65. Now , prof it am amoun ountt got go t by P = 20% 20% of total total profit + P’s share sha re in in balance 80% profit for his capital

 0.2 + 0. 8 × 34  = 470  3 4 + 6 5  12.(2) Now area = (1/6 × 1000)s 1000)s q m = 5000/3 sq m 2x × 3x = 5000/3 =>x × x = 2500 / 9 x = 50/3 length length = 2x = 100/3 m and breadth = 3x = 3× (50/3) = 50m 13. (4 ( 4) Distance = (240 x 5) = 1200 km. Speed = Distance/Time


Page 24


Speed Speed = 1 200/(5/3 200/(5/3)) km/hr.

[We can w rite rite 1 hours as 5/3 hou hours rs ]

Requ Req u ired speed = 12 00 x 3 km/hr km/hr = 720 km/hr. km/hr. 14.(4) A s ,

2 × 5 × 1 = 20

and

4 × 3 × 6 = 72

Sim Si mi larly ,

7 × 2 × ? = 42 ?

∴

42 =

14

=

3

15.(1) A ccording to question, question,

Therefore, B F J Q is odd.

PART B (16 -35) 16.(4) From the property of “di “di rac del del ta func fu nction tion ” +∞

∫ f ((xx ) δ( x – a )dx = f( a)

– ∞

+∞

so,

∫ [cos(3 x) + 2] δ( x – π )dx 0

Here He re

f(x) = cos co s(3x) (3x ) + 2

so,

δ(x – π ) = δ(0) x= π

⇒

∞

so,

∫ [cos(3 x) + 2] δ( x – π )dx = cos(3 π ) + 2 = –1 + 2 = 1 0

17.(1) We know Fourier cos transform is–


Page 25


Fc =

∫

∞ 0

2

π

f( x) cos sx dx

2

∞

π ∫0

Fc (∝ ) =

 2  ⋅ (1– ∝), 0 ≤∝< 1 f (θ ) co s ∝ θ d θ = π   0, ∝> 1

By invers inversii on form form ula of Fc (∝), we get f(θ) =

2

∞

π ∫0

Fc ( θ)x cos θ d ∝ =

2

 1 2 ∞  ∫0 ( 1− ∝) co s ∝ θd ∝ + ∫1 0 d ∝ π  π  1

2

 sin ∝ θ  − (− 1)  − co s ∝ θ  = 2 0 − co s θ − 0 + 1  f ( θ) = (1− ∝ )    π θ2 0 π  θ2 θ2   θ   f(θ ) =

2(1 − cos θ)

πθ 2

18.(4) (ii ) We We ha ve f1 (s) = f 2 =(s) =

s s2 + a2

= L{cosat}

∴ By con con volution volution theorem t

∫

−1

L {f1 (s)f 2 ( s)} = F1(y ( y)F2 (t ( t − y)dy 0

 s2  −1  s   s  t ∴L  2 = L  2 2   2 = ∫ cos ay cos a( t − y)dy 2 2  2  (s + a )    s + a   s + a  0 −1

t

t

 1 1 1 = [co s at at + co s( s( 2a 2a y − at ) ]d ]d y = y co co s at + sin( 2ay − a t )  20 2 2a 0

∫

=

 1 1 1 t cos at + sin at  = [ a t co s at at + sin at at]  2 a  2a

19.(3) Red uced uced m ass ass of two bo dy syste ystem m is

M =

m× 2m 2m = m + 2m 3


Page 26


So , ki net netii c energy energy T =

1  2m  2   xɺ 2 3 

Potential energy of system V =

1 2 kx 2

So , Lagrangi Lagrangian an L = T–V T–V L=

1  2m  2 1 2   xɺ − 2 kx 2 3 

Lagrangian equation is– d  ∂ L  ∂L =0  − d t  ∂xɺ  ∂x d  2m  xɺ − (–kx ) = 0 dt  3  2m ɺx ɺ + kx = 0 3 ɺɺ + x

3k x= 0 2m

20.(2) Cycle ‘1’ is clockwise so workdo work do ne duri duri ng cyc cycl e ‘1’ i s po si tive . Similar Simila rl y cycle ‘2’ is anti anticloc clock kwi se a nd work done during cycle ‘2’ becomes negative. But area of cycle ‘2’ is greater than area of cycle ‘1’. So resultant work is negative.

p

2

V 21.(1) Here the express expressii on for f or F represents epresen ts the th e fo rce rce between t wo charge s in Weber’s Weber’s elec el ectr trodynamics. odynamics. We have

Fr =

1 rɺ 2 – 2ɺɺ rr  1 –   2 2 r  c 

∂U d  ∂ U  + in usual notation, ∂r dt  ∂rɺ 2 

Taking U = qφ – q(A q(A ⋅ v ) and Fr = – Fr =

1 r2

 rɺ2 – 2ɺrɺ r  1  c 2 – rɺ 2 + 2ɺrɺ r  1  c2 – r 2 + 2rɺɺr  1 – = 2  = 2  c2 c2 r2   r   c  


Page 27


=

1 c

2

⋅

c

2

r

2

+

2 1  2ɺɺ rr – rɺ 

 c  2

r

2 1  2r 2ɺrɺr – rɺ 

1

 = 2 + 2 c   r

2

2

r

∂U d  ∂ U  +   =  ∂r dt  ∂rɺ 

Yields,

∂U 1 1 = – 2 whi ch gives gives an i nteg nteg ration U = = arbitrary constant say a function of r. ∂r r r Assum Assum ing U =

1 1  rɺ 2 + 2 r c  r

 , we g et 

∂U 1 rɺ2  1 = + – ∂r r 2 c2  r 2

ɺr 2 ∂U 1 2ɺr 1  = = ⋅ – – and and  2 2 2 ∂rɺ c 2 r r c r 

 d ∂U  2 d  ɺr  2  rɺɺ – ɺr 2  so that  = 2  = 2 2   d t ∂r  c dt  r  c  r  Thus

–

ɺr 2 2rɺrɺ 2rɺ 2 1 rɺ2 – 2ɺɺɺrɺr  ∂U d ∂ U 1 + = 2 + 2 2 + 2 2 – 2 2 – 2  1 –  c r c r cr r  c2 ∂r dt ∂r r 

Justifies Fr =

1

1 – r  2

rɺ – 2ɺrɺr   as given c2 

As such the generalized potential U ≡

La grangian grangian

L =T– U =

1 1 rɺ2 + 2 r c r 1 2 [ ɺr + r 2 θɺ 2 ] 2

T≡

also

∴

(on simp lification) lification)

rɺ 2 2

+

r 2 θɺ 2 2

–

1

–

r

1 rɺ2 c2 r

22.(4) We know Fermi energy of electron

 π 2N  Ef = 3  2m  V  ℏ

2

2/ 3

23

N 1 × 6. 06 × 1 0 × ρ = V M =

1 × 6. 06 × 1 0− 23 × 0 .9 7 23 22

3

28

= 2 .55×10 .55×10 /cm =2.55 × 10 –3 –31

m = 9.1 9.1 ×10

N

=

V –3 –3

m

kg

2/ 3 (1 .05 × 1 0−34 )2 2 28   ×  ×  So, E f = 3 ( 3 . 1 4 ) 2 . 5 5 1 0   2 × 9 . 1× 1 0 −31 


Page 28


–37

= 0.66× 10

= 0 .6 6 × 10

−19

–1 –19

E f = 4.97×10 Ef =

[7542]

2/3

18

(10)

(754.2) Jo ule (754.2)1/3

J

4 .9 7 × 10 −19 e V = 3 . 10 eV 1 . 6 × 10 −19

23.(1) Pa ramagnetic suscepti bility is i s gi ven by

χp =

µ0nµ B2 kBTF

where µB is Bohr magneton = 9.3 × 10 –24 J/tesla 10 –7 hnry/ m . µ0 is magnetic permeability of free space = 4 π × 10 n is no. of free electrons per unit volume. Assuming one free electron per atom the number of atoms per cubic meter of potassium is 23

3

6 . 02 × 10 × 0. 86 × 1 0 −3 3 .9 1× 1 0

23

Hence no. of free electrons (per m 3) n = or

n = 1.2 × 1028 m –3 E F = kBT F or TF =

∴

3

6 .02 × 10 × 0. 86 × 1 0 −3 3 9. 9.1× 1 0

EF kB

2. 1 × 1. 6 × 10 −19 = 2 .4 3 × 10 4 K. TF = −23 1. 38 × 1 0

Su bsti bsti tuting the the valu es we have have

χp =

4 π × 10

−7

× 1. 3 × 10 28 × (9. 3 × 1 0− 24 )2 = 4 20 .5 × 1 0 −8. −19 × × 21. 1 .6 1 0

kno w ma m agne gnetic tic suscep suscep ti bilit y 24.(2) We kno 2

χ= For co

+2

Nµ 0µ B 3KT 3KT

2

g2J J((J + 1) =

Nµ 0 µ θ 4S (s + 1) 3kT

7

ha ving 3d configuration ,

m s=


Page 29


∑m

s

1 2

1 2

1 2

3 2

=s= + + =

From grap Slope

dy 1 4 00 0 − 3 00 0 = = dx χ T 20

=

1000 = 50 20

1 = 50 χT

µ0 = 4π × 1 0−7 N / A 2 µB = 9.27 × 1 0 −24 Am2 –23

K = 1.38 ×10

J/K

J = S = 3/2 g =2 So , N =

3K (χ T)  3 5 µ 0µ B2 4      2 2 −2 3

=

3 × 1 . 38 × 1 0 −7 −2 4 2 4 π × 10 ( 9. 27 × 1 0 ) × 1 5 × 5 0 N = 5. 1× 1 0 26

25.(2) T he magn magn etic fi eld at O is 

B =

µ0 Ι ˆ µ 0 Ι ˆ µ0Ι ˆ µ 0 Ι ˆ  2  ˆ  k + k+ i=  i + 1 + π k 4π R 8R 8R 8R    

   µ Ι   ˆ 2     F = q (v × B) = q  νˆj ×  0 ˆi +  1 +  kˆ   k π       8R   =

µ0Ι qv  ˆ  – k + 8R 

26.(2) We know E f = E f(0)

 2  ˆ 1 + π  i   

 π 2  K T 2  1 –  B    12  E f(0)    

Given Ef(0) = 7 eV = 7 × 1.6 × 10 –19 J


Page 30


T = 3 × 10 4 K Ef π2  1 .4 × 1 0 –23 × 3 × 104  So , = 1 –   E f(0) 12  7 × 1. 6 × 10 –19 

π2  0. 2 × 3  = 1 –  12  1 .6 

2

2

2

π ×9 = 1 – 1 2 × 64 2

= 1 –

(3 .1 4) × 3 = 1 – .115 4 × 64

Ef = 0.88 E f(0)

27.(1) We know in Raman effect

 3 ∆υ =4B =4B  J +   2 –1

So giv given 4 B = 41.6 c.m

So , rotation con con stant ta nt B =

41.6 = 10.4cm – 1 4 –1

B = 1040 m h

We kno w

B=

So,

I=

I=

6. 6 × 10 −34 8 × ( 3. 14 )2 × (1 04 0) × ( 3 × 10 8 )

8π 2Ι C h 8 π2BC

I = 2.7 × 10 –47 kgm 2 So,

r=

Ι µ

µ → reduce mass Of HCl molecule So ,

µ=

MHMCl MH + MCl

=

(1× 3 5) /( /( 6. 0 23 × 10 23 )2 (1 + 35 ) / ( 6. 02 3 × 10 23 )


Page 31


24

= 1.61 × 10 gm –27 µ = 1.6× 1.6× 1 0 kg.

So,

r=

2. 7 × 10 −47 −27 1 .6 × 1 0

–10 = 1.29× 10 –10 m

r = 1 .2 9 A °

28 . (2) Given Rin = R1 = 120 kΩ Rf = 600 kΩ V i = 1.2 1.2 V V0 = ? For inverting inverting O.P A mp voltage gain gain

∵

AV =

–R f –R f –600 –6 00 = = Rin R1 120

A V = –5 AV =

∵

V0 Vi

So, Vout = AV × Vi = – 5 × 1.2 Vout = – 6V

29.(4) Given Ι P =

5 2

We know number of spectral levels (lines) in NMR is = (2Ι p + 1) 5 +1 2

⇒

2×

⇒

6 lines

30.(3) g = 2

∆ℓ ℓ

= 10 –5


Page 32


R = 120 ohm So, New resistance R’ = Rg

∆ℓ ℓ –5

= 120 × 2 × 10 –5

= 240 × 10 R ' = 2 . 4 × 10 –3 oh m

31.(3) x runs from 0 to (1– y) d x dy d z ∫ ∫ ∫ Fd τ = ∫ ∫ ∫ xyz dx 2

2  (1 − y)2 dy  dz ∫ 2   0

1

=

∫

z2  y

0

 2  = ∫ z ∫ y(1 − y) 2dy dy  dz 20   0 1

1

2

 1 2 z  ( y + y3 − 2 y2 )d y  dz 2 0 0  1

=

2

∫ ∫

2

1 2 4 3 1 2 y y 2y  = z  – +  dz 20 2 4 3 0

∫

1 2  4 16 2 .8  = z dz + − 2 0  2 4 3  1

∫

3 1

1

1 2 2 12 Z = z   dz = 2 0 3 23 3

∫

= 0

11 1 = 33 9

32.(3)

→ 1H3 + P H2 + 1H2 

1

2

2

4

→ 2He + n H + 1H 

1

2

4

31 H → 2He + P + n

∆m = (3M(1H2) – M(2He 4) – MP – Mn) Phone: 0744-2429714 Mobile: 9001297111, 9829567114, 9001297243 Website: www.vpmclasses.com E-Mail: vpm cl classe asse [email protected] / in inff o@v pmclasses.c mclasses.co om A ddress: ddress: 1-C 1-C-8, -8, Shee Shee la Ch C howdhary Roa Road d, SFS, TA LWAND LWANDI, I, KOTA , RAJASTHA RAJASTHA N, 324 3240 0 05

Page 33


∆m = 3 × 2.014 – 4.001 4.001 – 1.0 1.007 07 – 1 .008 008 ∆m = 0.026 amu 2 –13 So , energy energy E = ∆m c = 0.026 × 931× 1.6 × 10 J –13

E = 38.72 × 10

J

So, energy of total deuteron W = ET = 10 P=

∵

t=

40

–1 –13

× 38.72 × 10

J

W t

W 3 8.72 × 10 40 × 10 –1 3 11 = 38.72 38.72 × 10 = 16 P 10 12

t = 3 .87 × 1 0

sec .

33.(4) In pair production process, electron-positron pair is produce d. So, 2

h ν = E+ + E – + 2m 0c

E + → kine tic energ y of of positron E – → kinetic energy of electron h ν → i ncide ncide nt pho pho ton energy Given E+ + E – = 0.01 MeV 2

–31 –31

Rest mass energy 2m0c = 2 × 9.1 × 10

82

× (3 × 10 ) J.

2

2m 0c = 1.02 MeV So, h ν = 0.01 MeV + 1.02 MeV h ν = 1.03 MeV 34.(2) Ph otons are pa rti rti cl es ha vi ng spin spin 1 (intege (intege r) and pio ns are spin le ss parti parti cl e so, the t he y are Bosons B osons.. 35.(2) η° i s m eson hav ha vi ng zero ero charge. charge. So, quar quark k stru structure cture is uu or dd So, normalized quark structure is

ψ=

1 2

(u u – dd )

PART C (36 -55) 36.(2) We know f(n) (1) =

n! 2 πi

∫

f (z )dz , C ( z – a )n + 1


Page 34


6

Put f(z) = sin z, which is analytic within and on the circle |z| = 1, whose centre is z = 0 and radius 1, 1 al so put n = 2 and a = π, which lies within the given circle C, in the above formula then we get 6

 1  2! f”  π  =  6  2 πi

∫

6

sin z C

 z – 1 π    6  

3

dz.

….(1)

6

5

Now f(z) = si si n z whi ch gi ves f’(z) = 6 si si n z cos z 4

2

6

4

2

2

an d

f”(z) = 6[5 si n z cos z – sin z] = 6 sin z (5 cos z – si si n z)

∴

1  f”( π /6) = 6 sin4  π  6  4

2

  21 2  1  5 co s  6 π  – sin  6 π       2

= 6(1/2) [5(√3/2) – (1/2) ] = (3/32) (3/32) [5(3) – 1]

∴

From (1) we ha ve

∫

C

6

sin z dz 3

 1   z – 6 π   

=

21 π i. 16

37.(2) (i) If we take a = 1, b = 1 then

 1 2 2  1  A =  2 1 –1 3  –2 1 –1  1 2

∴

T

A =

1 3

2

1

–2   1

  2 – 1 –1 

A square finite inite matr m atrii x A is said to orthogonal if T

AA = Ι

 1 2 2   1 2 ––22   1+ 4 + 4 2+ 2 – 2 –2 + 2 – 2  1 1 1    A A =  2 1 –1   2 1 1  =  2 + 2 – 2 4 + 1+ 1 – 4 + 1+ 1  3  –2 1 –1  3  2 –1 –1  9  –2 + 2 – 2 –4 + 1 + 1 4+ 1+1  T

 9 2 –2  1 =  2 6 – 2  ≠ Ι 9  –2 –2 6  So, this is not correct value of a, b. (ii) If we take a = 1, b = 2


Page 35


 1 2 2  1  T hen A =  2 1 – 2  3  –2 2 –1 

∴

 1 2 –2  1  A = 2 1 2   3  2 – 2 –1  T

 1 2

2   AA =  2 1 –2  9  –2 2 –1  T

1

 1 2 –2  9 0 0 2 1 2  = 1  0 9 0      2 –2 –1  9  0 0 9 

1 0 0  1 0 0   = ×9 0 1 0 = 0 1 0 = Ι     9  0 0 1   0 0 1  1

So, the matrix is orthogonal if a = 1, b = 2. 38.(1) η of H.E. between A and C T   ηA = 1 – C  TA  

 T  η of engine = α 1 – C  TA   Here Q 2 =

∴

Q1 Q × TC, Q 3 = 1 × TC TA TB

Total Heat rejection

 1

(Q2 + Q3) = Q1TC 

 TA

+

1  TB 

Total Heat input = 2Q1

  1 1  Q1TC  +    TA TB   engin e =  1 – η of engin   2Q1    


Page 36


∴

αTC

α –

=1 –

TA

TC T – C 2 TA 2 TB

M ultiply ultiply bo th side by T A and divide by T C TA T 1 1 TA – α = A – – TC TC 2 2 TB

or

α

or

TA T = ( 2α – 1) + 2(1 – α ) A TB TC

39.(2) Energy eigen value of three dimension well is ℏ

E=

π2

2

2

2

2

(n x + n y + nz )

2

2ma 2 2 9π ℏ gi ven E = 2 ma 18 π2 ℏ2 E= 2ma2 compere equation (1) and equation (2). n x2 + n2y + n2z = 18

. . . (1)

...(2)

If n x = 4, ny = 1, n z = 1 then nx2 + n 2y + n 2z = 16 + 1 + 1 = 18 18 So, possible values (combinations) of nx n y n z are (n x, ny, nz) = (4, 1,1) (1, 4, 1), (1, 1, 4) So, the state is 3-fold degenerate.

273

40.(3)

S2 − S1 =

∫

mcp dT

293

T

= 0.01 × 4. 2 × I n

273 kJ / K 293

S 2 – S 1 = – 0.00297 kJ/K kJ/K = 2 .97 J/ K S3 − S2 =

=

−m L T

− 0.01 × 33 5 × 10 00 273

S 3 – S 2 = – 12.271 J/K 268

S4 − S13 =

∫

2 73T

mcp dT T

2 68  4. 2  kJ / K = 0. 0 1×  × In  2 73  2 


Page 37


= – 0.38 82 J/K J/K

∴

–1 –1

So, S4 – S1 = S2 – S1 + S3 – S2 + S4 – S 3 = –15.63 JK

41.(1) For one-dimensional free electron gas, energy level separation

∆E =

n 2 π2 ℏ n '2 So , E ∆ ∝ 2mL2 L2

42.(i)(1) We begin by writing the Hamiltonian as H=

1

2

2 Ιx

2

1  2 1 2  1 1  2  1 − − Lz = L +  L z 2 Ιx  2Ι z 2 Ι x   2Ι z 2Ι x 

2

(L (L x + L y + L z ) + 

where L is the total angular momentum. 2

2

We know eigen values of L = ℓ(ℓ + 1) ℏ and, L 2 = m ℏ E ℓm =

ℏ

2

2Ι x

ℓ(ℓ

 1 1  2 2 + 1) +  − ℏ m Ι Ι 2 2  z x 

So the eigenstates of the Hamiltonian are those of L2 and Lz, i.e., the s [hetrical harmonic with the eigen energies E λm.. L + L− L+ − L−  〈 Yℓm (θ, φ) | (L x + L y + L z ) | Yℓm( θ, φ) > < Yℓm( θ, φ) |  + + + L z  | Yℓm( θ, φ )〉 2i  2 

(ii)(3)

m

m

= < Yℓ ( θ, φ) | L z | Yℓ ( θ, φ ) > = m ℏ 43.(2) Binding energy according to semimpirical mass formula E b =avA = asA2/3 – a cz (z–1 (z–1 ) A –1/3 –aa(A-2z)2A –1 – apA –3/4 A nuclie will be most stable isobar which has maximum binding energy.

∴ For maximum binding energy E b =Ebmax when

∴

dEb =0 dz

For maxim maxim um bindi ng en ergy ergy E b = Ebmax when

∴

dEb =0 dz

–1/3 –a c(2z–1)A –1/3 + 4aa (A– 2z) A –1 =0

−1/ 3

∴

z=

+ 4a a aA −1/ 3 + 8 aa A − 1 2 ac A


Page 38


∴

=

0.58 (6 4)− 1/3 + 4 × 1 9. 3 8 × 19 .3 −1/3 + 2 × 0. 58 58(6 4) 4) 64 0 .1 45 + 77 . 2 = 28.62 0 .2 .29 + 2. 41 41 25 25 z ≈ 29

44.(3) m = 36 36 g = 0.03 0.03 6 kg kg T 1 = 30° 30 °C = 30 30 3 K T 2 = 523 K (∆S) water

373   373  kJ / K  303 

= m cP ℓn 

= 0.03143 kJ/K mL

(∆S) Vaporization = =

T2

0 .0 36 × 2 26 0 373 373

= 0.21812 kJ/K 523

(∆S) Vapour =

∫ mc

p

373

523

= mR

dT T

a



∫  T + b + CT d T

373


Page 39


5 23

2  CT  = m R a ℓn T + b T +  2 3 73  5 23

2  CT  = m R a ℓn T + b T +  2 3 73 

52 3 C   + b × (5 23 – 3 73 ) + (5 23 2 – 3 73 2 )  = m R a ℓn 37 3 2  

= 0.023556 kJ/K (∆S) System= (∆S)water + (∆S)vaporizat ion + (∆S)vapour = 273.1 J/K. 45.(2) T he intri intri nsic impedances a re

η1 =

η0 120 = = 129Ω an d η2 = η0 = 120 π Ω εr1 8.5

and the angle of transmission is given by sin15° sin θt Then

=

ε 8 . 5ε 0

or

θt = 48.99°

E r0 η0 co s θi − η1 cos θ t = = 0.623 0.623 or E i0 η2 co sθ i + η1 cos θ t

E r0 = 0.62 0.623 3 µV/m

46.(4) Sin ce the ki ki netic energy of A after af ter coll collii sion i s one one-n -nii nths nths of i ts i nitial kine tic energ energ y, the momentum o f A after collision is one-third of its initial momentum. Since the momentum is to be conserved, we have p = p’ – p/3 p/ 3 whe re p is ini tial momentum momentum of A and p ’ is the mo momen menttum of B af ter the colli colli sion. [Th e final final m om entum of part pa rtii cl e A is neg neg ati ve si si nce its di rection ec tion is reversed reversed ]. T heref herefore, p ’ = 4p/3 T he kin kin etic ene rgy gained gained by particl particl e B due to the coll oll i sion is p’2 /2M where wh ere M i s the m ass of p arti cl e B. 2

The kinetic energy lost by particle A due to the collision is (8/9)×p /2m. /2m. 2

2

[Note [Not e that the i nitial nitial ki ne ti c energy energy o f parti parti cle A i s p /2m /2m and its final ki ki ne tic energ energ y is (1/ (1/9) 9) p /2m /2m ]. Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have ’2

2

p /2M = (8/9) p /2m Substituting for p’ = 4p/3, we have 2

2

(16/9) (p /2M) = (8/9) (8/9 ) p /2m from which M = 2 m .


Page 40


47.(4) The volume of water = 10 5 ×

∴

1 = 5000 5000 cub c ubic ic miles mile s. 20

Mass of water = 5000 × (1.609 × 10 3)3 × 103 kg = 2.08 2. 08 × 10 16 kg.

or No. of molecules of water = 2.08 × 1016 × 6.0221 6.0221 4 × 10 26 /18 /18 = 6.97 × 1041 molecules. molecules. As the abundance of deuterium is 0.0156% so that the total number of deuterium atoms = 6.97 × 10 41 × 2 × 0.0156 × 10 –2 = 2.18 2.18 × 1038. As the fusion of 6 deu terium atom s give s an an en ergy ergy rel rel ease of 43 43 MeV , hen hence ce the total energy rel eas eased = 2.18 2.18 × 1038 × (43/6). = 1.56 × 10 39 MeV. 48.(1) The wave length λ (i n An gstrom unit) uni t) of a photon photon o f energ energ y E (in (in ele ctron volt) is i s gi gi ven by nearl y. λE = 12400, very nearl Therefore, λ = 12400/E [The abo ve express expression ion can be easily obtained obtained by rememb eme mber ering ing tha thatt a photon photon of energy energy 1 eV ha s wave leng th 12400 Ǻ and th e energ y i s inversel inversel y proportio proportio nal to the th e wave wave le ngth]. gt h]. Since E = 2 eV we have have λ = 12400/2 12400/2 = 6200 Ǻ = 620 nanom nanom etre. etre. Ph otons with wave leng length th g rea ea ter than than 6 40 nm nm will il l have en e nerg y le ss tha n 2 eV so that th ey will be un abl e to produ produ ce electron electron hole pai rs i n the semi condu condu ctor of band gap gap 2 eV. So the correct option i s (1). 49.(4) Circuit (i) is shown the logic circuit of AND GATE and here output Y =A⋅B Circuit (ii) is sho sho wn the logic ci ci rcui rcui t of OR G AT E and output Y = A +B Similarly circuit (iii) is shown the logic circuit of NOR GATE and output is Y= A+B sinz 50(i).(3)Since 50(i).(3)Since f(z) = cosz – z to find zeros of f(z) put f(z) = 0

⇒

sinz cos z – =0 z


Page 41


⇒

 z3 z 5   z z   z – 3 ! + 5 ! – . . ...  =0 1 – + – . .. .. .. . . –  2 ! 4 ! z      

⇒

 z2 z4   z2 z4   1– + . . ... . ..  –  1– + . . .. .. ..  = 0 4! 5!  2 ! 4!   3 ! 5! 

⇒

1 1 z2  –  + ........= 0 3 ! 2 ! 

2

4

so z = 0 is a zero of f(z) of order 2

50(ii).(3)

g( z) sin h z = zf (z (z ) z co cos z – s in in z To find poles z cos z – sin z = 0

⇒

 z3 z5   z3  + z – – . . . . . . .   – z – + .. . ... . = 0  2 ! 4 !   3! 

⇒

1 5 3 1 z  – + z  3 ! 2 !

⇒

g(z) have pole at z = 0 of order 3 zf(z)

1 1 5 ! + 4 !  = 0  

51.(3) T he Lag Lag rang rangian ian of a i s given by L = T– V =

1 m r 2 (θɺ 2 + sin 2 θφɺ 2 ) – V (r, θ, φ). 2

..(1)

In this case the only two generalized co-ordinates are θ and φ, therefore there will be only two La grangian grangian equations, equations, one i n θ and the other in φ. The Lagrangian equation in coordinate θ i s given given by by d  ∂L  ∂ L = 0.   – dt  ∂θɺ  ∂θ

..(2)

From equation (1), we have

∂L ∂L ∂V 2 2 2 mr θɺ an d mr si n θ co s θɺφ – = = ∂θ ∂θ ∂θɺ With these substitutions, equation (2) becomes

∂V d (m r2 θɺ ) – mr 2 sin θ cos θφɺ 2 + =0 ∂θ dt

…(3)


Page 42


The Lagrangian equation for conservative system in the variable φ i s given b y d  ∂L  ∂ L =0   – dt  ∂φɺ  ∂φ Again equation (1) gives

∂L ∂V ∂L and = – = mr 2 sin2 θφɺ ɺ ∂φ ∂φ ∂φ

…(4)

With these substitutions equation (4) becomes d ∂V 2 2 (m r sin θɺφ) + =0 dt ∂φ

52(i).(1) We have E = Exˆ and we seek φ ( x,t ) such that E = – ∇φ

…(1)

Since B = 0, we seek a gauge in which A = 0. Intergr Intergrat atii ng (1) we o btain φ(x) = - εx + c, where c is a con stant of integrati on. Let us choo ch oo se c = 0; then then

φ ( x) = − ε x

…(2)

The total Hamiltonian is H=

p2 1 + mω 2 x 2 − εx 2m 2

…(3)

The first term on the right-hand side of (3) is the standard kinetic term, the second term is the ha rmoni rmoni c osci osci llator potent ot entii al energy, energy, an d t he thi thi rd term term i s th the e el el ectri ectri cal cal po ten tial energy. 52(ii).(2) We will now write part (i) eq.(3) in the following form: Hy =

p y2

1 2m 2

+ m ω y2 y2 + H0

…(4)

Where H0 i s a con c ons sta nt and and y = ax ax + b. Con Con si der the kineti kineti c term. term . We We see see tha t py = p x, so a = 1. Now we can substitute y = x + b into (4) and obtain p x2 1 p2x 1 2 1 2 Hy = + m ω ( x + b ) + H0 = + mω 2 bx + mω2 b2 + H0 2m 2 2m 2 2 2

…(5)

2

2

From Eq. (3) of part (i) we see that Hx = Hy only if b = – ε /mω and H0 = – ε / 2mω . 52(iii).(3) To conclude, if we perform the coordinate transformation y = x – ε /mω , we get a one-dimensional harmonic oscillator with no charge, and the energy eigenvalues of a one-dimensional harmonic oscillator are 2


Page 43


En =

1  1 ℏw  n +  2  2

..(6)

Corresponding to the eigenstate | ψ n . We have a shifted harmonic oscillator; thus, the energy eigenvalues are now. En =

1  1  1 ε2 ℏω  n +  − 2  2  2 mω 2

53.(2) f(x) = x3 – x2 + 4x – 4 f’(x) = 3x2 – 2x + 4 f(2) = 8 – 4 + 8 – 4 = 8 f’(2) f’(2) = 12 – 4 + 4 = 12 xn +1 = xn – xn + 1 = 2 –

f ( xn ) f '( '( xn )

8 = 2 – (2/ 3) = 4 / 3 12

54.(1) T aking aking earth as refe rence l evel evel for zero potential energy, we have V(q) V(q ) = mgz

..(1)

T he Hamilton-Jacobi amilton-Jacobi equation for Ham Hamii lton ’s princi princi pal pal function i s

 1  ∂S 2  ∂S V ( q )  + + =0    2m  ∂q   ∂t

..(2)

T he fifi rst rst term in brack bracket et i s ffun un ction ction of q only, while the secon secon d te rm is fu nction nction o f t only, the refore each term must be equal to the same constant with opposite signs.

 1  ∂S  + = α  i.e. V ( q )    2m  ∂q  2

an d

∂S = – α ∂t

  

...(3)

T hen we have have S = W(q, α)– αt, ...(4) W being a constant of integration.


Page 44


As

∂s ∂W , therefore the Hamilton-Jacobi equation for Hamilton’s characteristic function W takes = ∂q ∂ q

the form form 2

1  ∂W    + V(q) = α 2m  ∂q  T his gives

...(5)

∂W = 2m {α – V (q )}  ∂q

...(6)

Integrating above expression, we get W=

1 /2

∫ 2m {α – V( q)} 

dq

Now

p=

1 /2 ∂S ∂W = =  2m {α – V (q)}  ∂q ∂ q

an d

β=

∂S ∂W = – t ∂α ∂α

=

m 2

dq

∫ [α – V (q )

β +t=

∴

T he Hamiltoni Hamiltonian an o f system ystem i s

1/ 2

p2 + m gz = E = α 2m

From (9) β + t =

=

Thus

...(8)

1/2 ∂   α 2 m – V ( q ) { } ∫  dq – t ∂α 

i.e.

H=

...(7)

...(9)

...(10)

dz m 2 m = ⋅ (E – mg z)1/2  2  ∫ [E – m gz 1/2 g z ] 2 m g   1 2 (E – mg z) z)1/2 g m ( β + t)2 =

...(11)

1 2 (E – m gz) ⋅ g2 m

Solving for z, we we get z = –

1 E 2 g(β + t) + 2 mg

...(12)

Initial conditions are ɺ = 0 and so p = mz ɺ = 0 At t = 0, z = h, v = z

...(13)

Using (12), equations (10) and (12) yield


Page 45


E = mgh, mgh, h = –

1 2 E gβ + 2 mg

Solving these equations, we get β = 0 Now

p=

∂W gt u sin g (1 (1 2) = mzɺ = m gt ∂z

Thus we have from (12) and (14)

...(14) 1

z=h–

2

2

  

gt 

p = mg t

...(15) ...(15)

T hese hese are requi requi red red equ ation s with wi th S = W – Et where

W =

∫

( 2m) ⋅ (E – mg mgz )1 /2 d z

55.(3) From M axwell-Boltzm axwell-Boltzmann ann di stribution law, the num num ber of m ole cules in the νth state relative to that in the th e ν = 0 (lowest) state at T Kelvin temperature is given by N ν N0

= e –G

0

( ν )h c/ kT

= e –G( ν ) – G( 0 )hc /kT ,  

where G( ν ν) = ωe  ν +

∴

Nn = e –(ω n – ω N0 e

1  – ωe xe 2

2

 1 ν + 2   

2

e

x en – ωe x e n)hc /kT

For the n = 1 level, we have

N1 = e –(ω N0

e

– 2ω e x e )hc / kT

Here

N1 1 = , ω e = 21 4.6cm –1 a nd ω e xe = 0. 6 cm–1 ( give n) N0 10

∴

1 = e –( 214.6 cm 10

or

10 = e (213 .4 c m

–1

–1

–1

– 1 .2 cm

)hc /k /kT

)hc /kT

0m = e (2134 0m

–1

)hc / kT kT

T aking aking na tural ural l ogarithm: ogarithm: –1 –1

Lo ge 10 = (21340 m ) hc/kT.


Page 46


∴

T =

2 13 40 m – 1 hc 2 . 30 3

= (9 26 6 m –1 )

k (6 .6 3 × 10 –34 Js)(3 .0 × 10 8 ms–1 ) = 133.5 K. 1 . 38 × 1 0 – 23 JK –1


Page 47

Csir Ugc Net_physics_ Free Solved Paper

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