A Tip to Multiple Choice Questions:

Challenge: Can you get the answer in 15 seconds for the following question?

Q1. Earth of mass M_{1} and moon of mass M_{2} are a distance d apart. An object lying on the line joining the centre of the Earth and centre of the moon, what is the distance of this object from the Earth if the object experiences zero resultant gravitational force?

A. d(M_{2}/M_{1) }

B. d(M_{1}/M_{1}+M_{2})

C. d(M_{2}/M_{1})^{1/2}

D. dM_{1}^{1/2}/(M_{1}^{1/2}+M_{2} ^{1/2})

**Method In short: whenever you think assigning a particular value for a unknown constant can help, please assign a nice value for the unknown. (No worry, nobody will check your working)**

For example shown above, you can assign both M1 and M2 to be equal to 1kg ! So the object must be at the centre, hence, answer is D.

We use example below (Nanyang Junior College H2 Physics Preliminary Examination 2013) below for illustration.

Q2.

In this example, many students know that they will need to draw the corresponding velocity-time graph for the three trucks and then find the area under the graph. However, p and q are unknown constants, this makes it very difficult to sketch a velocity velocity-time graph.

But wait, this is MCQ, nobody will check your working. So it is perfectly alright to choose some nice values for p and q.

In this example, we can choose p and q both to be 5 (there is no specific requirement of p and q indicated in the questions) I choose p and q to be the same so that p – q = 0. This further simplifies the calculation.

Lastly I will assign a value of 10 for t_{1}.

With this: the velocity time graph for P, Q and R will be as followed:

It is obvious that the area under the graph for P is the largest. Hence, answer is A.

Time taken should be around 2 minutes.

Let us now compare with standard way of solving this problem in which you do not substitute a value for p and q.

For P: velocity after t_{1} seconds = u +at = 0 + (p + q)t_{1}

Velocity at t = 2t_{1} = u + at = (p + q)t_{1 }+ (p – q)t_{1} = 2pt_{1}

For Q, R, going through the same procedure, we can

get the following graphs for your comparision.

If you have nice graph like the one shown above, it may be obvious that the area under the graph for P is the largest. Chance is that your graph will not as nice as the one shown above and if that is the case, you have to find the area under the graph one by one.

Estimate time taken in this method will be 8 minutes. Do not forget this is just a MCQ, allocated time should be 2 minutes.

Another example for illustration:

Q3. Three identical lamps M, N and P are connected to a battery as shown. What happen to the brightness of lamp N and P when the lamp M breaks?

For those who have good mental calculation and a perfect understanding in the circuit, the answer will be A and the reasoning will be like this:

When M breaks, total resistance of the circuit will be increased, hence, total current will decreases, therefore currently passing P will decreases, therefore its power decreases.

As current passing through P decreases, potential difference across it decrease, hence potential difference across N will increases, therefore power = V^{2}/R should increase.

For those who cannot do it this way, you can assume the resistance of each lamp to be 4 ohms (or any other convenient values) and e.m.f of the battery to be 12V and carry out the power passing through the lamp P and N.

Before it breaks:

After it breaks: