F.7 Physics Test 8 (07-08) Suggested Answers
1. (a) (i) Energy required = area under the graph from 3 x 10-10 m to 9 x 10-10 m
= 0.5 x 9.6 x 10-10 (9-3) x 10-10 =
2.88 x 10-19 J
Alternatively,
Young modulus 
breaking stress
S

breaking strain a / a
4 x 1011 = breaking stress / (3.08-3.00)/3.00 => breaking stress = 1.07 x 1010 Pa
2.
(c) Work done per unit mass of petrol

Fs PAs P


V As 
Work done on petrol at entrance = P1 / 
Work done on petrol at exit = - P2 / 
Net work done on petrol = (P1 – P2) / 
Let W be the required work done per unit mass of petrol by pump.
Total input energy to petrol = Increase in KE of petrol + Increase in PE of petrol
mW  m( P1  P2 ) /   108m
W  108 
P2  P1

 108 
(2.8  0.1)  10 5
 468 Jkg 1
750
(Increase in PE = 0)
(d) Power = m W / t = W  A1v1 = 468 x 750 x 4 x 10-3 x 3 = 4212W
(e) It is because work is done against viscous force and in producing turbulence. Energy is also dissipated
as heat in the pump.
MC
1-5 C D C B B
6-10 A B B A A
11-14 B/C A D C
-------------------------------------------------------------------------------------1. (1) The magnetic force is perpendicular to the B field, the path should be a horizontal circle.
(2) The magnetic force does no work on the electron, so its speed remains unchanged.
(3) qvB = m v2 / r => r B = constant => r is inversely directly proportional to B.
=> +3 and –3 for u will give the same value of v, so symmetric about V axis.
2. eV = (1/2) mu2
3. Length l  2l, width w  2w, height h 2h, then
contact area A = lw  4 A.
Volume V = lwh  8V
If density is unchanged, m  8 m.
Stress = F / A = 8 mg / 4A = 2 S
4.
E
F/A

e/l
mgl
2
d 
e  
2
E m g  l  e
d


 
2
E
m
g
l
e
d
Try the percentage error one by one and get your result.
A) % error = 0.51 %
B) 2 x 1.64 % = 3.28% (because of the factor 2)
C) 0.1%
D) 2.56%
5. From figure 1, 100 : f1 = 2:5, => f1 = 250 Hz
From figure 2, the frequencies are the same, so f2 = 100 Hz.
The phase difference is  where 0o <  < 90o.
6. Thin wire should be used so as to increase the stress, thereby producing a larger extension.
7. The PE is a minimum at a, not b.
8.
9. The acceleration of water when passing through a constriction is a consequence of continuous flow, not
Bernoulli’s principle.
10. Refer to notes.
11. (1) The increase in energy of the positive charge is qV joules.
(2) qV = (1/2) mv2, if V is doubled, v is increased to 2 v.
(3) The increase in speed obviously depends on mass m, if m is smaller, the increase in speed is larger.
12. For (1), eE = evB => v = E / B = 1.5 x 105 / 0.1 = 1.5 x 106 m/s
For (2), P1 is at a higher potential.
For (3), period T = 2r/v, but qvB = mv2 /r.
So T = 2  m / qB = constant, independent of v.
13. For (1), when the accelerating p.d. is increased, the velocity of the electron beam entering the deflecting
plates is increased, hence the time of flight is decreased, so deflection decreases.
For (2), If l is increased, time of flight increases, so deflection increases.
For (3), If L is increased, the time of flight outside the plates is increased, so vertical deflection on the screen
increases.
14. I = n A q v
A. Doubling the p.d. could increase the no.of electrons produced, but it is not sure that the current must be
doubled.
B. Halving the area will half the current.
C. Doubling the electron speed will double the current.
D. Doubling the ke will increase the speed by 2 .