Kalvisolai-plus Two Physics Compulsary Problems Em By Thamaraiselvan T 2v421v

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My hearty congratulations to +2 Students! If you want to get ‘200’ out of ‘200’ in Physics, you should be answered the compulsory problem as well as possible. The physics learners are advised to frame their own responses taking ideas from the help – points given for each problem. Here, All the public examination questions (Up to Sep – 2013) particularly, fifty nine problems are dealt in a detailed and simple manner, just like your attempts. The students are thus prepared in an effective way to face the compulsory problem confidently and come out with flying colours. With best wishes and regards,

Walk with educational service T.Thamaraiselvan, PGT Physics, G.B.H.S. School, Aranthangi – 614616 Cell: 9443645072 Email Id : [email protected]

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A Parallel Plate capacitor has plates of area 200 cm2 and separation between the plates 1mm. Calculate (i) The potential difference between the plates if 1 nc charge is given to the capacitor (ii) with the same charge if the plate separation is increased to 2mm, what is the new potential difference and ((iii) electric field between the plates.

[March

A=200 cm2 =200 x 10-4m2 d = 1 x 10-3 m

1 cm2 = 10-4 m2

q = 1 x 10-9 c σd v= ε0

(i)

]

The potential difference between the plates v =? qd = A. ε0

1 x 10-9 x 1 x 10-3 =

2 x 10-2 x 8.854x 10-12 v=5.65 v (ii) If d′′= 2 m m v′′=? If the distance increase by two times the P.D also will increase two times v′′= 2 x 5.65 =11.3 V (iii)

Electric Field E = v / d = 5.65 / 10-3 = 5650 v / m.

2. Three capacitors each of capacitance a 9 pF are connected in series (i) What is the total capacitance of the combination? (ii) What is the potential difference across each capacitor, if the combination is connected to 120V Supply? [June Given

, Sep

, June

]

C = 9 pF

v =120 v

n =3

v1= v2= v3=?

Cs = ?

v1= v2= v3=v / n = 120 / 3

CS = c/n = 9/3

v1= v2= v3=40 v

CS = 3 pF

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3.Two Capacitances 0.5 µF and 0.75 µF are connected in parallel and the combination to a 110v battery. Calculate the charge from the source and charge on each capacitor? Given: [June ] c1=0.5 µF ; c2=0.75 µF v = 110 v q1, q2=? Total Charge q = ? q1=c1 v = 0.5 x 10-6 x 110 q1= 55 µc q2=c2 v =0.75 x 10-6 x 110 q2 =82.5 µc q = q1+q2 =55+82.5 q = 137.5 µc 4. Calculate the electric potential at a point P, located at the centre of the square of point charges shown in the figure [June] q1=12 nc ; q2=-24nc; q3=31 nc ; q4=17 nc D C if a=1.3 m nc nc r= OA=OB =OC=OD= a / √2 o 12nc -24 nc V = 1 [q1+ q2+q3+ q4] A B πε r 1.3 m = 9 x 109 x √2 12-24+31+17 1.3 9 x 1.414 x 36 x 10-9 x 109 V = 1.3 V = 352.4 V

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5. Two positive charges of 12 µC and 8µC respectively are 10 cm apart. Find the work done in bringing them 4 cm Closer, so that, they are 6cm apart. [June 2008] Given :q1 =12 µc; q2 =8µc If r1=10 cm 9 x 109 x q1 q2 Intial Energy Ui = r1 9 x 109 x 12 x 10-6 x 8 x 10-6 = If r2 = 6 cm

10 x 10-2

= 8.64 J Final Energy 9 x 109 x 12 x 10-6 x 8 x 10-6

Uf

=

6 x 10-2

Uf = 14.4J Work Done = W = Uf-Ui =14.4 – 8.64 W= 5.76 J Two capacitors of unknown capacitances are connected in series and parallel if the net capacitances in the two combinations are 6 µ F and 25 µ F respectively find their capacitances [Sep ] Given :C1 C2 = 150 =25 µ F Cs = 6 µ F c1 , c2 =? C1 (25-C1) = 150 = c1 +c2 25C1 - C12 = 150 c1 +c2 = µ F C2 =(25 – C1)µ F 25C1 - C12 -150 = 0 Cs = C1 C2 / C1+C2 C12 - 25C1 + 150 = 0 C1 C2 (C1 -15)(C1 -10) = 0 =6 C1 = 15 (0r) 10 C1+C2 C1 =15µ F C1 C2 C2 = 10µ F =6 25

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7. The plates of parallel plate capacitor have an area of 90cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it is a 400V supply, how much electrostatic energy is stored by the capacitor? [June 2009, Sep 13] A=90 cm2 = 90 x 10-4 m2 d=2.5 x 10-3 m v=400 v E=? E=cv2/2 C=ε A / d E =1 x ε A x v2 2 d 8.854 x 10 -12 x 90 x 10 -4x 400 x 400

1 E=

x 2 = 2.55 x 10 -6 J

2.5 x 10 -3

8. Three Charges -2x10-9C, 3x10-9 C, -4x10-9 C are placed at the vertices of an equilateral triangle ABC of side 20 cm Calculate the work done in shifting the charges A, B and C to A1, B1, and C1 respectively which are the mid points of the sides of the triangle. [June -2011] AB=BC=AC=20 X 10-2 m q1 = -2 x 10-9 C q2 = 3 x 10-9 C q3 = -4 x 10-9 C

U=?

A

B 3 x 10-9 C

B1

Initial Potential Energy Ui = 1 / πε d [q1 q2 + q2 q3+ q3 q1 ] 9 x 109 [ ( -2 x 10-9 x 3 x 10-9 ) + ( 3 x 10-9 x -4 x 10-9 ) + ( -4 x 10-9 x -2 x 10-9 ) ] 20 x 10-2

Ui =

9 x 109 =

20 x 10-2

[ ( -6 x 10-18-12 x 10-18+8 x 10-18]

-9 x 109 x 10 x 10-18 Ui = 20 x 10

-2

= 4.5x 10 -7 J

-2 x 10-9 C

C

-4 x 10-9 C

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If the Charges on A′, B′, and C′ distance final potential energy r′= 10 x 10-2 m 9 x 109 Uf=

10 x 10-2 =-9 x 10-7 J

[-10 x 10-18]

U=Uf-Ui = -9 x 10-7 –(-4.5 x 10-7) = -(9+4.5) x 10-7 U = -4.5 x 10-7 J In the given network, calculate the effective resistance between points A and B Find the effective resistance one unit [March ]

The network has three identical units C

D

R1, R2 are connected in series Rs1=R1+R2 =5+10= 15 5+10R1s R3, R4 are connected in series Rs2 = R3+R4 Rs2 = 5+10= 15 Rs1, Rs2 are connected in Parallel Rp1 = Rs / n = 15 / 2 =7.5 Each unit has a resistance of 7.5 Rs = nRp = 3 x 7.5 Rs = 22.5 The effective resistances are , when two resistors are connected in series and parallel. What are the resistances of individual resistors [March , March October ] Rs = 10 ; RP =2.4 ; R1,R2=? Rs= R1+R2 R1+R2= 10 R2= (10- R1) RP = R1 R2 /( R1+R2) R1R2 / (R1+R2) = 2.4 R1R2 / 10 =2.4 R1R2=24

R1 (10-R1)=24 10R1– R12 -24=0 R12 – 10 R1+24=0 (R1-6) (R1-4)=0 R1= 6 or 4 R1= 6 ; R2= 4

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11. A copper wire of 10-6m2 area of cross section carries a current of 2A, if the number of electrons per cubic metre is 8 x 10 28, calculate the current density and average drift velocity. e=1.6 x 10-19 c) [March ] -6 2 28 -19 A=10 m ; I=2A; n=8 x 10 ;e=1.6 x 10 c J=? vd=? J=I / A =2 / 10-6 =2 x 106A m-2 J 2 x 106 vd= = n.e 8 x 1028x1.6x10-19 =0.156 x 10-3 vd =1.56 x 10-4 ms-1 12. Three resistors are connected in series with 10V supply as shown in the figure. Find the voltage drop across each resistor [June March ] Effective resistance of series combination Rs= R1+R2+R3 Rs= 5+3+2 = 10 Current in Circuit I=v/ Rs=10/10 = 1 A Voltage drop across R1 v1 =I R1 = 1 x 5 =5 V Voltage drop acrosR2 v2 =I R2= 1 x 3 =3 V Voltage drop acrosR3 v3 =I R3 = 1 x 2 =2 V. 13. What is the drift velocity of an electron in a copper conductor having area 10 x 10-6 m2 carrying a current of 2A. Assume that there are 10 x 1028 electrons /m3 [March ] A = 10 x 10-6 m-2 ; I=2A; n=10 x 1028 electrons / m3 vd = ? I = n A e Vd Vd = I / n A e = 2 / [10 x 1028 x 10 x 10-6 x 1.6 x 10-19] Vd = 1.25 x 10-5 ms-1 14. Find the current flowing across the resistors 3 , 5 and 2 connected in parallel to a 15 v supply, Also find the effective resistance and total current drawn from the supply [Oct ] R1= 3 ; R2= 5 ; R3= 2 V=15 v I1,I2,I3 =? Total Current I =? Effective Resistance=? Current Through R1 I1 = V / R1 =15/3 = 5A Current Through R2 I2 = V / R2 =15/5 = 3A Current Through R3 I3 = V / R3 =15/2 = 7.5A Total Current I = I1+I2+I3 = 5+3+7.5=15.5A Effective Resistance R = v / I =15 / 15.5 = 0.967

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15. In a metre bridge, the balancing length for a 10 resistance in left gap is 51.8 cm. Find the unknown resistance and specific resistance of a wire of length 108 cm and radius 0.2 mm. [Oct June ] P = 10 l1 = 51.8cm l2 = 48.2cm Q=? Q / P = l2 / l1 Q = P. l2 / l1 = 10 x 48.2 / 51.8 Q=9.3 L = 108 cm = 1.08m r = 0.2 x 10-3 m = 2 x 10-4 m. ρ=? ρ= Q.A / L =>Q.π r2 / L 9.3 x 3.14 x (2 x 10-4)2 =

1.08 ρ = 108.2 x 10-8 =1.082 x 10-6

m

16. An iron box of 400 W power is used daily for 30 minutes. If the cost per unit is 75 paise, find the weekly expense on using the iron box [June ] P=400 W; t=1/2 Hour. Rate /Unit = Paise Cost /Week = ? Energy consumed in one day = P x t = 400 x 1 / 2 =200 w h W =0.2 k wh W =0.2 Unit Energy consumed in one week =7 x 0.2 =1.4 Unit Cost / Week = Total units consumed x (rate / unit) = 1.4x0.75 = Rs.1.05 The resistance of a field coil measures 50 at 20oC and 65 at 70oC. Find the temperature coefficient of resistance.

t1=

º C; R1= , t2 = º C R2= =? = R2-R1 R1t2-R2 t1 = 65-50 = 15 (50x70) – (65x20) 3500-1300 = 15 2200

= 0.0068 / º C

[June

]

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18. A circular coil of radius 20cm has 100 turns wire and it carries a current of 5A. Find the magnetic induction at a point along its axis at a distance of 20cm from the centre of the coil. [March 2006, Oct -2006, March 2009] n =100 turns a = 20 x 10-2 m a2 = 4 x 10 -2 I=5A x = 20 x 10-2 m x2 = 4 x 10 -2 B =? B= =

nµ0 Ia2 2(a2 + x2 ) 3 / 2 102 x4 x 3.14 x 5 x 4 x10-2 x10-7 2(4 x 10-2 + 4 x 10-2) 3 / 2 4 x 3.14 x 10-6

=

-2 3 / 2

(8x 10 )

4 x 3.14 x 10-6 =

16√2 x 10-3

= 0.555 x 10-3 B = 5.55 x 10-4 T 19. A rectangular coil of 500 turns and area 6 x 10-4 m2 is suspended inside a radial magnetic field of induction 10-4T by a suspension wire of torsional constant 5 x 10-10 Nm/degree. Calculate the current required to produce a deflection of 100 [June 2006, Oct -09, March -13] -4 -4 N = 500; A=6 x 10 ;B=10 T C = 5 x 10-10 Nm / degree θ = 10°, I=? I = C θ / NBA 5 x 10-10x10 10-3 = = 2 -4 -4 5 x 10 x6 x 10 x10 6 I = 0.166 mA 20. A moving coil galvanometer of resistance 20 produces full scale deflection for a current of 50mA. How you will convert the galvanometer into (i) an ammeter of range 20A and (ii) a voltmeter of range 120V. [March ,March June -13] G = 20 ; Ig = 50 mA; (i) 20 A an ammeter range 20 A. I = 20 A = 20000 mA S=

Ig I-Ig

G

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50 =

20 20000-50 50

=

x 20 19950

S = 0.05 A shunt of 0.05

should be connected in parallel

(ii) 120 v a voltmeter range v=120 R=? R= =

V Ig

-G

120 -20 -3 50 X10

= (120000 / 50 ) -20 =2380 A resistance of 2380

should be connected in series with the galvanometer

A long straight wire carrying current produces a magnetic induction of at a point, 15 cm from the wire. Calculate the current through the wire [Oct ] a=15 x 10-6 m; B=4 x 10-6 T; I=? B=2 x10-7 x I / a I= B x a / 2 x 10-7 =4 x 10-6 x15 x 10-2 / 2 x 10-7 I=3A

x 10-6 T

22. In a hydrogen atom electron moves in an orbit of radius A° making 1016 revolutions per second. Determine the magnetic moment associated with orbital motion of the electron [June- 2008] -6 radius r= 0.5 x 10 m n=1016 revolution / Second e=1.6 x 10-19 c µl=? µl =en x πr2 =1.6 x 10-19 x 1016 x 3.14 x (0.5 x 10-10)2 µl = I x A I=e/T =e.n µl =1.256 x 10-23Am-2 A=πr2

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23. Two parallel wires each of length 5cm are placed at a distance of 10cm apart in air. They carry equal currents along the same direction and experience a mutually attractive force of 3.6 x 10-4 N Find the current through the conductors. [Oct 2009; June 2010, March 2013] l=5m; F=3.6 x 10-4 N

-2

a = 10 x 10 m; I1 = I2 = I =? 2 x 10-7 x I1I2 F= a

xl

2 x 10-7 x I2 x l a

F= I2 = =

I2 = 36 I = 6A

Fxa 2 x 10-7 x l 3.6 x 10-4x10x10-2 2 x 10-7 x 5

24. A galvanometer has a resistance of 40 It shows full scale deflection for a current of 2mA. How you will convert the galvanometer into a voltmeter of range o to 20v? [Oct – 2010] G = 40 ; Ig = 2 x 10-3A; V = 20V R=? V R= -G Ig 20 =

-3

- 40

2 x 10

20000 = 2 -40 = 10000-40 R = 9960 A resistance of 9960

should be connected in series

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25. Two straight infinitely long parallel wires carrying equal currents and placed at a distance of 20cm apart in air experience a mutually attractive force of 4.9 x 10-5 N per unit length of wire, calculate the current [Oct ] -5 -2 F = 4.9 x 10 N; a=20 x 10 m; l=1m I1 = I2= I=? F = 2 x 10-7 x I1I2 xl a F = 2 x 10-7 x I2 x l a Fxa l2 = 2 x 10-7x l 4.9 x 10-5x20 x 10-2 = 2 x 10-7x 1 2 l = 49 I = 7A 26. The deflection in a galvanometer falls from 50 divisions to 10 divisions when 12 resistance is connected across the galvanometer. Calculate the galvanometer resistance [Sep ] S=12 ; θ=50 divisions θg =10 divisions G=?

G=

s

I

θ

θ-θg G=

s θg 50 – 10

=

x 12 10 40

=

x 12 10

G = 48 27. An A.C generator consists of a coil 10,000 turns and of area 100cm2 The coil rotates at an angular speed of 140 rpm in a uniform magnetic field of x 10-2 T Find the maximum value of the emf induced. [June 2009] N =104 turns; A=100 x 10-4 m2 B= x 10-2 T; =140 rpm E0 = ? E0 = N A B ω = N A B 2 =140 / 60 rps E0 =104 x100 x10-4 x x 10-2 x2 x3.14x140/60 E0 = 52.752 V

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A Soap film of retractive index 1.33, is illuminated by white light incident at an angle 30 . The reflected light is examined by spectroscope in which dark band corresponding to the wave length 6000 Å is found. Calculate the smallest thickness of the film. [Oct , Sep -2013] µ= ; i=30°; n=1 (smallest thickness) λ=6000 Å, t=? If dark band µt cosr = nλ t=nλ / 2 µcos r 0

µ=sin i / sin r sin i sinr =

sin30° =

µ

=0.373 1.33

∴ r = sin -1 0.3759 r = 22°04′

1 x 6000 x 10-10 t= 2 x 1.33 x cos 22°04′ 6000 x 10-10 t= 2 x 1.34 x 0.9267 t= 2.434 x 10-7 m In young’s experiment of a light of frequency 6 X 1014 Hz is used distance between the centres of adjacent fringes is 0.75 mm. Calculate this distance between the slits, if the screen is 1.5m away. [Oct ] 14 -3 γ 6 X 10 HZ; β = X 10 m D = 1.5 m d=? β = λD /d [λ=c/γ ] β = C D / γ .d d=CD/γ.β 3 x 1 08 x1. 5 = 6 x1014 x x 10-3 d = 1 x 10-3 = 1m m

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30. A parallel beam of monochromatric light is allowed to incident normally on a plane transmission grating having 5000 lines per centimetre. A second order spectral line is found to be diffracted at an angle 300 Find the wave length of the light. [March 2008, March 2010, June -12] N = 5000 lines / cm = 5 x 105 lines / m m=2 θ = 30° λ=? sinθ = Nmλ λ = sinθ / Nm = sin 30° / (5 x105 x 2) = 1 = 2x5x105x2 = 0.5 x 10-6 = 5000 x 10-10 m λ = 5000 A° 31. A monochromatic light of wavelength A° is incident on a water surface having refractive index 1.33 Find the velocity frequency and wavelength of light in water [Oct – 2008, March -11] µ =4/3 λ = 5893 A° Velocity of Cw = ? λ w = ? γ = ? Cw = C / µ 3 x 108 =

4 /3 9 Cw = x 108 4 = 2.25 x 108 ms-1 λw = λ / µ 5893 x 10-10 =

=

4 /3 5893 x 10-10 x 3

4 = 4419.75 x 10-10m γ =C/λ =

3 x 108 -10

5893 x 10 γ = 5.09 x 1014 HZ

[ frequency in water = frequency in air]

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32.In a Newton’s rings experiment the diameter of the 20th dark ring was found to be 5.82 mm and that of the 10th ring 3.36mm. If the radius of the plano convex lens is 1 m. Calculate the wavelength of light used. [March ] -3 -3 d20 = 5.82 x 10 m ; d10 = 3.36 x 10 m n + m = 20; n = 10; m = 10 ; R=1m λ=? d2n+m - d2n d220 - d210 λ= = 4mR 4mR (5.82 x 10-3)2 - (3.36 x 10-3)2 = 4 x 10 x 1 (5.82 +3.36) (5.82-3.36 )x 10-6 = 40 λ = 9.18 x 2.46 x 10-6 /40 λ = 0.56457 x 10-6 = 5.645 x 10-10 m λ = 5645 A° 33. A plano – convex lens of radius 3m is placed on an optically flat glass plate and is illuminated by monochromatic light. The radius of the 8th dark ring is 3.6mm. Calculate the wave length of light used. [March ] -3 R= 3 m; r8=3.6 x 10 m ; n=8; λ=? λ = r2n / nR 3.6 x 3.6 x 10-6 = 8x3 = 0.54 x 10-6 =5400 x 10-10 m λ = 5400 A°

34. A Plane transmission grating has 5000 lines / cm. Calculate the angular separation in second order spectrum of red line 7070Å and blue line 5000Å [June ] 5 N = 5000 lines / cm (or) N = 5x10 lines per meter m= R = 7070 Å; V = 5000 Å v=? R– Sin = Nm Sin R = 5x105 x 2x7070x10-10 Sin R = 0.707 = R =45

0

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5x105x2x5000x10-10 v = 0.5 = v = 30 º R – v = 45 º -30 º 0 R – v = 15

Sin Sin

v=

35. A 300 mm long tube containing 60 cc of sugar solution produces a rotation of 9o

when placed in a polarimeter. If the specific rotation is 60o, calculate the quantity of sugar contained in the solution

l=300 mm = 3dm v=60 CC, = 9º, S=60º, m =?

s=

[June

]

=

s= m=

36. line.

=

= 3g

Wave length of Balmer second line is λ 2= 4861 A° λ 1= ? For Balmer II Line n1 =2; n2 =4 γ2 = R ( 1 _ 1 n12 - n22) 1/λ 2 = R ( 1 1) 2 2 - 42 =

4–1 16

R

1/λ 2 = 3R / 16

1

For Balmer I line n1 =2; n2 = 3 γ2 = R 1 _ 1 22 - 32

=

A°. Calculate the wave length of first [March ]

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=

9 – 4R 36

1/λ 1 = 5 / 36 R 1

2

2

λ 1/λ 2 =

λ 1/λ 2 =27 /20

λ1=

λ 1= λ 1= 6562 A° In Bragg’s Spectrometer the glancing angle for first order spectrum was observed to be 8º. Calculate the wave length of x-ray. If d =2.82 x 10-10m. At what angle will the second maximum occur? [June ] -10 θ1 = 8° ; d= 2.82 x 10 m; n=1 λ=? 2d sinθ = n λ 2d sinθ λ = n 2 x 2.82 x10-10 x sin 8° = 1 = 2 x 2.82 x10-10 x0.139 λ = 0.7849 x 10-10 m n = 2 when θ2 = ? 2d sin θ2 = n λ sin θ2 = n λ / 2d 2 x 0.7849 x10-10 = 2 x 2.82 x10-10 sin θ2 = 0.2783 θ2 = sin-1 0.2783 θ2 = 16° 9′ 38. An α- particle is projected with an energy on 4 MeV directly towards a gold nucleus, Calculate this distance of its closest approach (Given atomic number of gold 79 and atomic number of a particle =2) [March ] (Atomic number of Gold = 79, Atomic number of particle = 2 K.E = 4 x 10-6 x 1.6 x 10-19 J Z=79 r0 =?

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18 x 109x Ze2 r0 =

K.E 18 x 109 x 79 x 1.6 x 10-19x1.6 x 10-19

=

4 x 106 x 1.6 x 10-19 = 568.8 x 10-16 r0 = 5.688 x 10-14 m 39. An electron beam es through a transverse magnetic field of 2 x 10-3 T and an electric field E of 3.4 x 104 V / m, acting simultaneously. If the path of the electrons remain undeviated calculate the speed of electrons. If the electric field is removed what will be the radius of the electron path..? [Oct ] E = 3.4 x 104 v / m ; B = 2 x 10-3 T v=? r=? v=E/B 3.4 x 104 = 2 x 10-3 +7 v = 1.7 10 ms-1 Bev = mv2 / r Be = mv / r r = mv / Be 9.11 x 10-31 x 1.7 x107 = 2 x 10-3x1.6 x 10-19 r = 4.839 x 10-2 m 40. How fast would a rocket have to go relative to an observer for its length to be corrected to 99% of its length at rest [Oct -2007, Oct -11, March -12] 99 when l = lo ; = ? 100 99

l = lo√ 1-v2 / c2 lo= lo √ 1-v2 / c2

100 0.99 = √ 1-v2 / c2 (0.99)2 = 1-(v2 / c2) (c2-v2) / c2 = 0.992 c2-v2 = (0.99c)2

v2 = c2 -0.9801 c2 = c2 (1-0.9801)

v2 = 0.0199 c2 v = 0.141 c = 0.141 x 3 x 108 v = 0.423 x 108 ms-1

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41. At what speed is a particle moving if the mass is equal to three times its rest mass? [Oct ] m = 3 m0 v=? m0 m= √ 1 -v2 / c2 m0

3 m0 =

√ 1 -v2 / c2 3 √1 -v2 / c2 = 1 9(1 -v2 / c2) = 1 9 (c2-v2) /c2 9c2 - 9v2

=1 = c2

9v2

= 9c2 - c2

9v2

= 8c2

v2

= 8 c2 / 9

v

= 2√2 c / 3 = 2√2 x 3 x 108 / 3 = 2√2 x108 ms-1 =2.828 ms-1

v v

42. The time interval measured by an observer at rest is 2.5 x 10-8 s What is the time interval as measured by an observer moving with velocity v=0.73c [June ] to = 2.5 x 10-8 s v = 0.73c t =? t = to / √1 -v2 / c2 v2 / c2 = 0.5329 c2 / c2 1 –(v2 / c2 )= 1 - 0.5329 = 0.4671 √1 -v2 / c2 = √ 0.4671 = 0.6834 t = t0 / 0.6834 = 2.5 x 10-8 / 0.6834 = 3.658 x 10-8 s

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Work function of Iron is 4.7eV Calculate the cut off frequency and the corresponding cut off wavelength for this metal [June 09, March 2012] w = 4.7 e V w = 4.7 x 1.6 x 10-19 J γo = ? λo =? w = h γo γo = w / h 4.7 x 1.6 x 10-19 = 6.626 x 10-34 γo = 1.134 x 1015 λo = c / γo =

3x 108

1.134 x 1015 = 2.645 x 10-7 λo = 2645 A° A metallic surface when illuminated with light of wavelength A° emits electrons with energies up to 0.6ev. Calculate the work function of the metal [Oct 2009, June 2012, March 2013] λ = 3333A° K.E = 0.6 e v w=? hγ = w + k.E ∴w = hγ -k.E w = h.c / λ - k.E 6.626 x 10-34 x 3 x 108 = - 0.6 3333 x 10-10 x 1.6x 10-19 w = 3.727 – 0.6 w = 3.127 eV A proton is moving at a speed of 0.900 times the velocity of light Find the kinetic energy in joules and Mev. [March ] -27 v = 0.9 c; mo = 1.67 x 10 kg (m- mo)c2 = ? m = mo/ √1 -v2 / c2 1.67 x 10-27 = √ c2 / c2 1.67 x 10-27 = √ mo = 3.831 x 10-27 kg

1.67 x 10-27 = √0.19

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m - mo = (3.831 x 10-27-1.67 x 10-27) = 2.161 x 10-27 (m - mo) c2 = 2.161 x 10-27 x 9 x 1016 m - mo) c2 = 19.449 x10-11 J 19.449 x 10-11 E = 1.6 x 10-19 = 12.155 x10-11 x 10-19ev = 12.155 x108 E = 1215 Mev

What is the de Braglie wave length of an electron of Kinetic energy 120ev? [Oct ] If. k.E = 120 ev v = 120 v λ =? λ = (12.27 / √v) Å = (12.27 / √120) = (12.27 / 10.95) λ = 1.121 A° A piece of bone from an archaeological site is found to give a count rate of 15 counts per minutes A similar sample of fresh bone gives a count rate of 19 counts per minute. Calculate the age of the specimen (Given T1/2= 5570 Years [Oct , March Sep -2013,] N = 15 counts / minute T1/2=0.693 / λ N0 = 19 counts minute t=? N = N0 e-λ t N / N0 = e-λ t N0 / N = eλ t loge(N0 / N) = λ t t= t=

loge

=

x2.303x loge

x 2.303 x log 1.266 5570

t= 0.693 t = 1899 Years

x 2.303 x .1026

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Calculate the energy released when 1Kg of U undergoes nuclear fission. Assume, energy fission is 200 Mev. Avagadro Number =6.023 x Express your answer in Kilowatt hour also [March , Sep -2011, Sep 2013] Energy produced fission = 200 Mev = 200 x x1.6 x J = 3.2 x Avagadro Number N = 6.023 x 1 kg U Energy released f 1kg 92U235 = ? 235 g U Number of atoms in 235 g of uranium = 6.023 x Number of atoms in 1Kg (or ) 1000 g of uranium U – = 6.023 x x 1000 = 2.562 x 1024 ∴ Energy produced by 1 kg of uranium during fission = 2.562 x 1024 x 3.2 x E = 8.2016 x J 1 k w h = 3.6 x J 8.2016 x ∴E= = 2.2782 x 1 k w h 3.6 x Find the energy released when two H nuclei fuse together to form a single He nucleus. Given the binding energy per nucleon of H a single He nucleus, Given the binding energy per nucleon of H and He are 1.1 Mev and 7.0 Mev respectively, [June 2006, October ] H BE/A of = Mev He BE/A of = Mev H He + Q Energy released Q = ? BE/A of H = Mev BE of H = 2 x Mev = 2.2 Mev BE/A of He = Mev BE of He = 4 x 7=28 Mev H He + Q 2 x 2.2 28 + Q 4.4 28 + Q Q = 28 – 4.4 = 23.6 Mev A reactor is developing energy at the rate of 32 MW. Calculate the required number of fissions per second of U Assume that energy per fission is 200 Mev. [June ,2008, ] Energy / fission = 200 MeV = 200 x x 1.6 x Total Energy = MW No of fission of 92U23 = x N= N=

200 x x 1.6 x fission / Second

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Show that the mass of radium

Ra

gram [T ½ - 1600 Years ;

curie = 3.7 x T ½ =1600 Years =1600 x 365.25 x 24 x 60 x 60 dN / d t = 3.7 x 1 disintegration / Second m =? T ½ = 0.6931 / λ dN / d t = λN N = 1 dN λ dt =

T½

x 0.6931

with an activity of 1 curie is almost a [June ,March , March

]

disintegration / second]

dN dt

1600 x 365.25 x 24 x 60 x60 N=

X 3.7 x 1 0.6931

= 2.695 x 1 According to Avagadro’s Principle, Radium, x1 atoms 226 m= x 2.695 x 1 x 1

x 1

atoms =

g

of

= 1.011 g m ≈1g Calculate the mass of coal required to produce the same energy as that produced by the fission of 1 Kg of U [October ] Energy fission = Mev = 200 x 1.6 x 10-19 x 106 Avagadro Number =6.023 x 1023 Heat of combustion of Coal = 33.6 x 106 J / kg Mass of Coal= ? Number of Atoms in g U –= 6.023 x 1023 Number of Atoms in g ( or ) 1 kg U 23 6.023 x 10 N= x 103 = 2.562 x 1024 235 Energy/fission =3.6 x1011 J Energy produced by the fission of 1Kg of U235 = 2.562 x 1024 x3.6 x10-11 E = 8.2016 x 1013J

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Let M be the mass of the coal required to produced equivalent to energy produced by 1Kg of U M x Heat of combustion of Coal =8.2016x 1013 M x 33.6 x 106 = 8.2016 x 1013 8.2016 x 1013 M = = 2.441 x 106 kg = 2441x103 Kg 6 33.6 x 10 M = 2441 ton. Mass defect of 6C12 is 0.098 amu. Calculate the B.E /A of 6C12 nucleus. [June ] ∆ m of - C = 0.098 amu; A=12 BE / A = ? ∆ m of - C = 0.098 amu BE of - C = ∆ m x 931 Mev = 0.098 x 931 MeV = 91.238 MeV BE/A Of - C = 91.238 /12 = 7.603MeV Calculate the time required for 60% of a sample of radon to undergo decay. Given T1/2 redon =3.8 days [March ] Half life of Rn = days Original Amount N0 = 100% ; Amount of Sample disintegrated N0 – N )= 60% Amount of Sample present N = 40% time required t=? T1/2 = 0.6931 λ 0.6931 λ= T1/2 N = N0 e-λ t N / N0 = eλ t N0/ N = eλ t log (N0/ N) = λt t = loge = x2.303x loge t=

x 2.303 x log 2.5 3.8

t= 0.6931 3.8 t=

2.303 x 0.3979 0.6931

t = 5.022 days

2.303 x log 10 2.5

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The binding energy per nucleon for C nucleus is MeV and that for C is MeV. Calculate the energy required to remove a neutron from C nucleus [March ] BE/A of - C = Mev BE/A of - C = Mev BE/A of Q = ? C C + n +Q. BE/A of - C = Mev Total BE of C13= 13 x 7.47 MeV = 97.11 MeV C BE/A = Mev Total BE of 6 C 12 = 12 x 7.68 Mev = 92.16 Mev Binding Energy of a neutron = = BE of C --BE of C = 97.11 – 92.16 = 4.95 MeV Calculate the energy released in the following reaction Li + n He4+ H Given Mass of Li –= 6.015126 amu Mass of n = 1.008665 amu Mass of He4 = 4.002604 amu Mass of H = 3.016049 amu Mass of the reactants =? Mass of Li = 6.015126 amu Mass of n = 1.008665 amu + 7.023791 amu Mass of He4 = 4.002604 amu Mass of H –= 3.016049 amu 7.018653 Mass Defect ∆m = Mass of the reactants Mass of the products ∆m = 7.023791 - 7.018653 ∆m = 0.005138 amu Energy released in the reaction = ∆m x 931 MeV = 0.005138 x 931 MeV E= 4.783 MeV

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57.

Calculate the energy released in the reaction AI + H

Mg + He

Mass of AI –= 26.981535 amu Mass of Mg = 24.98584amu Mass of H = 2.014102 amu Mass of He = 4.002604 amu Energy related in the reaction = ? Mass of reactants AI –+ H = 26.981535 +2.014102 = 28.995637 amu Mass of products Mg + He – = 24.98584 + 4.002604 = 28.988444 amu Mass of defect = Mass of reactions - Mass of Products ∆m = 0.007193 Energy released in the reaction E = ∆m x 931 MeV = 0.007193 X 931 MeV E = 6.696 Mev 58. A transistor is connected in CE configuration. The voltage drop across the load resistance Rc k Ω is V. Find the basic current. The current gain α of the transistor is 0.97 [June ] α = 0.97 Rc = 3K Ω VCE= 6V IB =? Ic = VCE / Rc β = Ic / IB = 6 / 3 x 103 IB = Ic / β = 2x 10-3 A = 2 x 10-3 / 32.33 β = α / (1 - α) = 0.0618 x 10-3 = 0.97 / (1 - 0.97) IB = 61.8 x 10-6A = 0.97 /0.03 IB = 61.8 µA β = 32.33 59.. A 10 MHz Sinusoidal carrier wave of amplitude mv is modulated by Kz sinusoidal audio signal wave of amplitude mV. Find this frequency components of the resultant modulated wave and their amplitude [March ] Frequency of the carrier ( fc) = 10MHz Frequency of signal ( fs) = 5 KHz= 0.005 MHz Amplitude of the signal = 6 mV Amplitude of the carrier Signal = 10 mV Frequency components of Modulated Wave = ? Amplitude of the components in the modulated wave = ? Upper side band frequency fc + fs = 10+0.005 = 10.005MHz Lower side band frequency = fc - fc = 10 - 0.005 = 9.955MHz The Modulation Factor m = Es / Ec = 6 / 10 = 0.6 Amplitude of USB = Amplitude of LSB = mEc / 2 = (0.6 x 10 ) / 2 = 3 mV

T.Thamaraiselvan,

Cell: 9443645072