Physics Paper 2, May/June. 2011
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
Weakness/Remedies
Strength

Question 5

 In fig. 1a and fig. 1b above, Wa and Wb represent the respective loads on a spring placed near a 30cm rule, when in air and when in water. (a) Identify the force causing shrink in the spring in fig. 1(b). (b) Given that the force constant of the spring is 2.0 x 1011 Nm-1, calculate the workdone by the force in causing the shrink. In response, the force to be identified is the upthrust. Many candidates correctly identified upthrust however, they failed to put down the correct formula. Those that were able to write the correct formula missed the arithmetic because of conversion from cm to m while some got the arithmetic but omitted the unit/put down wrong units. The expected answers are: (a) The force is the upthrust of the water (b) Workdone = ½ Kx2 = ½ (2.0 X 1011) X (2.0 X 10-2)2 = 4.0 X 107 J
Observation

In response, the force to be identified is the upthrust. Many candidates correctly identified upthrust however, they failed to put down the correct formula. Those that were able to write the correct formula missed the arithmetic because of conversion from cm to m while some got the arithmetic but omitted the unit/put down wrong units.