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F.7S Physics Test 1 (07-08) Suggested Answers
1. The speed of the ion when reaching the anode v is given by
1 2
q
q
mv  qV  v  2V  v 
2
m
m
(1A)
For H+, q / m = 1 / 1 = 1 (assuming q = e = 1 unit of charge, and mass of 1 proton = 1 unit)
For He++, q / m = 2 / 4 = 0.5 (since He contains 2 protons and 2 neutrons)
For He+, q / m = 1 / 4 = 0.25
(1A for sub.1A for all correct)
Hence, H+ is the earliest, and then He++, and then He+.
(1A)
OR acceleration a = F / m = qE / m,
electric field can be regarded as the same for the ions.
2. (a) The slide-wire is uniform in both cross-section and resistivity.
(1A)
(b) No, the length L is measuring the p.d. across the resistance box or the terminal p.d. of the battery. OR
No, it is measuring E-Ir since there is internal resistance r.
(c) IR = kL where k is a constant because VPQ = IR = VAC  L
(1)
E = I( r + R)
rkL
E 1 1 1
 kL 
 
R
kr L R r
1
1 1
 constant   where constant = E / kr
R
L r
E
(1)
(1)
(d) The negative intercept on the (1/R) axis = 0.01 -1 (accept 0.0096 – 0.0104)
The internal resistance of the battery r = 1/ 0.01 = 100 
(2)
(e) Unchanged. The p.d. VAC corresponds to the same L.
(2)
3. (a) When the flame probe is brought near to a positively charged object, negative charges are induced in the
probe while positive charges are induced on the electroscope. However, the flame produces positive and
negative charges continuously. The positive charges neutralize the negative charges of the probe. The
electroscope remains positively charged and is at the same potential as the uncharged probe since they are
connected. The closer the probe to the charged object, the higher is its potential and the more the positive
charges induced in the electroscope and so the greater is the deflection. Hence the deflection is an indication of
the potential.
(b) Earthed objects have induced charges which would affect the original potential variation.
(c) E = dV/dx = (450 – 400) / 0.01 = 5000 V m-1. The direction is from A to B.
-----------------------------------------------------MC
1-5 A D B B D
6-10 A B C C B
11-13 C D D
1. Let potential at A be VA = 0 V, then VX = VY = 6 V, VB = 9 V,
hence p.d. across 6 k = 3 V, current through 6 k = 3 / 6k = 0.5 mA to the left
P.d. across 3 k = 6 V, so current through 3 k = 6 / 3k = 2 mA to the left
Hence current XY must be 2 mA – 0.5 mA = 1.5 mA and point downwards.
2. After connection, they must have the same potential.


Q
 = kQ/8r
 40 (2r ) 
Potential of S at surface = 

2Q 
 =kQ/2r
4

(
r
)
0


Potential of another sphere at surface = 
So charge Q’ will flow from the sphere to S. When equilibrium is reached,
 Q  Q'  2Q  Q'
 
4

(
2
r
)
0

 40 r
Potential of S at surface = 
Q’ = Q
Hence charge on S should be Q + Q = 2Q
3. Statement (1) is obvious. (Imagine a positive charge is placed at X)
Statement (2) is also obvious.
 Q   Q 
 + 
 = 0
 40 r   40 r 
Potential at X = 
Electric field at X is the smallest. (Try some numbers to get answer)
4. Statement (1) is correct, because terminal voltage is V = E – Ir, now a resistor is added in parallel, so
equivalent resistance decreases, so current increases, so terminal p.d. decreases.
Statement (2) is correct as explained above.
Statement (3) is incorrect because terminal p.d decreases, so current through R1 decreases.
5. Voltmeter has resistance of order of several thousand ohms, so it shares most of the battery’s voltage.
6. Statement (1) is true because the equipotential lines are more crowded near point P than at S, the potential
gradient, hence the electric field, is hence greater.
Statement (2) is incorrect because electric field points towards low potential region, and perpendicular to
equipotential lines, so electric field at S should be approximately south-west.
Statement (3) is incorrect. A negative charge tends to move towards high potential region.
7. vx = 2 m s-1,
vy = u + at = 0 + 10 (0.4) = 4 m s-1.
 = tan-1 (vy / vx) = tan-1 4/2 = 63.4o
8. The velocity increases until it reaches terminal velocity. The air resistance increases with object’s velocity
and the object finally reaches terminal velocity.
9. W = QV = (1/2) mv2
So to have the same velocity, QV/m must be the same.
10. There is no interaction between the sand and the cart, and so by inertia, the cart still moves with constant
speed 0.20 m s-1.
11. Statement (1). Power P= IV = 2 x 6 = 12 W, so statement (1) is wrong.
Statement (3) is the definition of p.d.
12. Consider the LHS portion,
Vertically, T sin  = mg/2 where T is the tension at the point of suspension on LHS
Horizontally, T cos  = T’
Dividing, we have T’ = mg / 2 tan
13. E is proportional to 1/r2. Inside the shell, there is no electric field.