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Question 324: [Forces > Hooke’s law]
The graph shows the relationship between load F and extension x for a certain spring.
A load of 6.0N is placed on the spring.

What additional strain energy is will be stored in the spring if it is then extended a further 0.01m?
A 0.010 J                     B 0.060 J                     C 0.070 J                     D 0.160 J

Reference: Past Exam Paper – J94 / I / 22



Solution 324:
Answer: C.
When a load of 6.0N is placed on the string, the extension x = 0.03m.

If the spring is extended a further 0.01m, the total extension is 0.04m, which corresponds to a load of 8.0N.

The additional strain energy stored in the spring when it is extended the further 0.01m is given by the area under the graph between x = 0.03m and x = 0.04m. This is the area of a parallelogram with parallel sides being 6.0N and 8.0N and a height of 0.01m.

Additional strain energy stored = ½ (sum of parallel sides) (height)
Additional strain energy stored = 0.5 (6.0 + 8.0) (0.01) = 0.070 J









Question 325: [Pressure]
Diagram represents a sphere under water. P, Q, R, and S are forces acting on the sphere, due to the pressure of the water.

Each force acts perpendicularly to sphere’s surface. P and R act in opposite directions vertically. Q and S act in opposite directions horizontally.
Which information about magnitudes of the forces is correct?
For June 2009 Paper 1 Q11:
A P < R ; S = Q
B P > R ; S = Q
C P = R ; S = Q
D P = R = S = Q

For June 2013 Paper 11 Q15:
A P < R and S = Q
B P > R and S = Q
C P = R and S = Q and P ≠ S
D P = R and S = Q and P = S

Reference: Past Exam Paper – June 2009 Paper 1 Q11 & June 2013 Paper 11 Q15



Solution 325:
Answer: A.
It is obvious that the sphere will not move in a diagonal direction under the water surface. So, S = Q.

For an object submerged under water, there will be a resultant upthrust force acting on it, causing it to move upwards. So, R > P

Choice B implies that there is a net downthrust (that the sphere would sink) while C and D imply that the sphere would remain stationary.










Question 326: [Physics of Materials ~ Matter > Hooke’s law]

The diagram (a) shows a simple apparatus for investigating the extension of a wire W under tension.
The pointer P is freely pivoted at H. Masses M are hung from a loop of thread L, and the scale reading opposite the end of the pointer is recorded. When the scale reading is plotted against load M, a good straight line is obtained except for the, last point K (Fig (b)).

Which one of the following statements is a possible explanation for the fact that K lies above the straight line?
A The value recorded for M was less than its true value.
B The scale reading was measured at less than its true value.
C The loop L has been inadvertently moved towards H.
D The wire W has not been firmly fixed at Q.
E The wire W has been loaded beyond its limit of proportionality.

Reference: Past Exam Paper – N80 / II / 40



Solution 326:
Answer: C.
The load M causes a torque about the point at which the pointer is fixed to the vertical section of the apparatus. By increasing the load M, this torque becomes greater.

From the way the apparatus has been set up, an increase in the value of the load M causes the scale reading to decrease.

The scale reading obtained for point K is more than expected. This means that the torque produced by load M is less than expected. Choice D “The loop L has been inadvertently moved towards H” is correct since this is the only one that causes the torque to be less.

Consider choice A: “The value recorded for M was less than its true value.” This means that the actual load M is greater. Looking at the graph, this would mean that the scale reading for point K  would be even greater than expected (since for a bigger load M, the scale reading should be less).

Choice B “The scale reading was measured at less than its true value” is also incorrect since the scale reading has actually been measured greater than its actual value.


If choice D “the wire W has not been firmly fixed at Q”, it would snap, causing the scale reading to be less than the actual value, not greater.

If choice E “the wire W has been loaded beyond its limit of proportionality,” then its extension would be greater than expected. This would cause the scale reading to less even less, not greater than the expected value (as displayed by the position of point K).










Question 327: [Current of Electricity > Resistance]

Student has available some resistors, each of resistance 100Ω.

(a) Draw circuit diagrams, one in each case, to show how a number of these resistors may be connected to produce combined resistance of

(i) 200Ω

(ii) 50Ω

(iii) 40Ω

(b) Arrangement of resistors shown in Fig is connected to a battery.

Power dissipation in the 100Ω resistor is 0.81W. Calculate

(i) current in the circuit,

(ii) power dissipation in each of the 25Ω resistors

Reference: Past Exam Paper – June 2002 Paper 2 Q8



Solution 327:

(a)

(i) 2 resistors in series

(ii) 2 resistors in parallel

(iii) Any correct combination

(in the circuit below, each side has an equivalence of 20Ω.)

or

(Equivalent resistance = [(1/200) + (1/100) + (1/100)]^-1 = 40Ω)

(b)

(i)

(Consider the 100Ω resistor)

Power dissipated P = I²R

0.81 = 100 I²

Current I = 0.090A

(ii)

(The current is divided equally in the 25Ω resistors since they are connected in parallel and have the same resistance.)

Current in the 25Ω resistor (= 0.090 / 2) = 0.045A

Power dissipated = (I²R = (0.045² (25) =) 0.051W