## Tuesday, September 28, 2010

### H2_08_paper3

1. To show inverse proportionality plot p vs 1/V
ii) A straight line passing through the origin is obtained.

bi) Find the area under the graph for ever 1s interval . To plot the graph of W vs d , The cummulative area is the work done at each d on the x axis . A curve is obtained. After (1m, W = 5J) , (2m , W= 17.5J) , (3,37.5) and (4, 47.5)

c) the grad of the garph is -2
ii) hence g is proportional to r-2 . Then g is inversely proportional to r2.

2. ai) Magnetic flux linkage of the coil changes as it rotates. By Faraday's law emf is generated.
aii) The two factors are speed of rotation and the No of turns or B
iii) The flux linkage of the coil depends on the angle @ between the coil and field . This angle changes sinuisoidally with time . Hence emf is sinuisoidal.

bi) peak input emf = 72 x 1.414 where sq root of 2 = 1.414
peak input emf= 102 V
bii)rms output voltage = pd across the resistor
= 20x 72 V = 1440 v
iii) rms current = rms voltage / R
= 1440/160 =9 A
iv) mean power = I2R = 92(160) = 1.3 x 104 W
v) VI input = VI output
I input = 1.3 x 104/72 = 180 A

3. a)Wavefronts of stationary waves do not advance. Energy is trapped within the waves.

b) Diffraction is the spreading of waves when the waves pass through a slit or an obstacle.

c) Two waves have constant phase relationship are said to be coherent.
d) The oscillation of the wave is in one direction in a plane normal to the direction of transfer of energy.

4. The internal energy is the sum of ke and pe all all molecules of the gas .
ii) The increase in internal energy is the sum of the heat supplied to the gas and the work done on the gas.
b) i) Work done by the gas is the area under the graph .
W= p(change in vol) = 1.5 J
ii) A-B : 0 / 4.2 / 4.2
B-C : -8.5 / 0 / -8.5
C-A : 5.8 / -1.5 / 4.3

5a) force acting on per unit positive charge placed at that point.
ii) Work done = F d but F = qE
W = qEd.
iii) V = W/q , sub W = qEd
V = Ed
bi) The number of electrons per s = N/t
I = Ne/t , N /t = I/e =5.4 x 1016 s-1
qV = 1/2 mv2
v2 = 2eV /m
v=1.45 x 10 8 ms-1
iii) power = N( ke) of electrons
P= N/t ( KE) = 5.2 x 105 W
or Power = IV = 8.6 mA ( 60kV) = 516 W
c) power supplied by electrons= rate of heat removed by coolant.
516 W = mc@/t , m/t = 516 /c@= 4.9 x 10-3 kgs-1
d) The lines appear to radiate from the centre. So the charge is concentrated at the centre of the sphere.
ii) E = kQ/r2= 5.4 x103 NC-1
iii) The Va and Vb and Vc are to be calculated separately . potential difference is the same . Hence Va - Vb = Vb - Vc.
Cancel Q and k . we have (1/0.4 -1/0.5 ) = (1/0.5 - 1/r)
r = 0.67 m

6a) freq is no of cycles per s but the angular freq has units in radians per s. Therefore angular freq is w= (2x3.142xf) rad s-1
bi) loss in gpe = mgh = 0.4x 9.81x 0.2=0.785J
ii) epe = 1/2 kx2 = 1/2 x19.62x(0.2)2= 0.392 J
c) loss in gpe is not equal to gain in epe as energy is loss as heat due to the stretching of the spring.
d) T-mg = ke - mg = 19.62(0.4) - mg = 3.93 N
(Note ; Important to draw a free body diag at the bottom of the motion. The positive is taken to be up (towards the eqm) T-mg = ma
ii) a = w2Xo
w2 = a/xo = F/mxo= 3.93 /0.4x0.2.
iii) vo=wxo = 7.0x 0.2 = 1.4 ms-1
e) lowest point : 0/1.57/0/1.57
equilm point : 0.78/0.39 /0.39/1.57
highest point: 1.57/ 0/0/1.57

7. This expt involves alpha particle and gold foil. The particles travel without deflection through the atom (not nucleus and not between the atoms) . The nucleus is small in comparision with the atom. Use the correct words for description.
b) x= 138, y=40
ii) E released = (235.0439+1.0087-137.9603-97.9197)x 1.66x10-27xc2= 2.6x 10-11J
iii) The remaining energy is released as gamma photon.
iv) Com: Initial p = final p
0=m1v1 +m2v2 where m1 is zirconium and m2 is tellerium , hence ratio of their speed
v1/v2=m2/m1=138/98
=1.4

2. The ratio of the ke = 1/2 m1(v1)2 divide by 1/2 m2(v2)2
ratio = m1/m2(1.4)2 = 98/138(1.4)2= 1.4
v) ke of zirconium = (1.4 /2.4 ) (2.3 x 10-11) =1.34 x 10-11J
v2 = 2x1.34x 10-11/98u
v = 1.3 x 10 7 ms-1
vi) Assumption one : The initial momentum assumed zero as the neutron speed is negligible.
Assumption 2: No other external forces are acting and hence conservation momentum is true.