WBBSE Madhyamik Physical Science study materials
WBBSE Madhyamik Physical Science study resources

Corridor Measurement, Speed Claim and Pendulum Timing

WBBSE Madhyamik Physical Science · Put your knowledge to work in context.

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Corridor Measurement, Speed Claim and Pendulum Timing

During a school sports day, a student measures the length of a corridor with a metre scale whose first 2.0 cm has been worn away. She places the 2.0 cm mark at one end of the corridor and reads 86.4 cm at the other end. She then walks the corridor in 12.0 s and announces that her speed is 7.2 m/s. A second student times 10 complete oscillations of a simple pendulum with a stopwatch whose least count is 0.1 s; the total time is 18.4 s. The two students argue about whether the stopwatch is precise enough to compare their pendulum with a second pendulum whose period is shorter by 0.02 s. The corridor measurement and the speed claim are to be checked before the results are recorded in the school magazine.

1. What is the actual length of the corridor? Explain why the 86.4 cm reading cannot be used directly, and calculate the percentage error that would result from ignoring the worn zero.

Read the worked answer

Actual length = 86.4 cm − 2.0 cm = 84.4 cm = 0.844 m.

The scale is worn at the zero end, so placing the 2.0 cm mark at the start adds 2.0 cm to the reading. Using 86.4 cm directly would overestimate the length by 2.0 cm. Percentage error = (2.0/84.4)×100 ≈ 2.4%.

2. Evaluate the student's speed claim of 7.2 m/s. Calculate the correct average speed and identify the likely mistake.

Read the worked answer

Average speed = distance/time = 0.844 m / 12.0 s ≈ 0.0703 m/s.

The claim 7.2 m/s is about 100 times too large. It is obtained by treating 86.4 cm as 86.4 m and dividing by 12 s. A metre-scale reading in centimetres must be converted to metres before calculating speed.

3. For the pendulum, find the time period and its absolute uncertainty from the stopwatch. Decide whether the stopwatch can reliably distinguish the two periods differing by 0.02 s. Justify your answer.

Read the worked answer

Time period T = 18.4 s / 10 = 1.84 s. The stopwatch least count is 0.1 s, so timing 10 oscillations gives an uncertainty of about ±0.1 s in the total, i.e. ±0.01 s in T.

The difference to be detected is 0.02 s, which is only about twice the timing uncertainty and comparable to human reaction time. The stopwatch is therefore not reliable for this comparison unless more oscillations are timed, e.g. 50 oscillations, or a stopwatch with 0.01 s least count is used.