Higher November 2021 Paper 2 Q21
21 The time period, \(T\) seconds, of a simple pendulum of length \(l\) cm is given by the formula
\[T = 2\pi\sqrt{\frac{l}{g}}\]
Katie uses a simple pendulum in an experiment to find an estimate for the value of \(g\).
Here are her results.
\(l = 52.0\) correct to 3 significant figures.
\(T = 1.45\) correct to 3 significant figures.
Work out the upper bound and the lower bound for the value of \(g\).
Use \(\pi = 3.142\)
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 984.(3677853) and 969.(0181643) | B1 | stating bound of 51.95 or 52.05 or 1.445 or 1.455 |
| P1 | for process to rearrange formula to give \(g\) as the subject, eg \(g = \dfrac{4\pi^2 l}{T^2}\) oe | |
| P1 | for process to use LB of \(l\) and UB of \(T\) in formula for \(g\) or \(T\) or process to use UB of \(l\) and LB of \(T\) in formula for \(g\) or \(T\) eg \(\dfrac{4\pi^2[\text{LB of } l]}{[\text{UB of } T]^2}\) or \(\dfrac{4\pi^2[\text{UB of } l]}{[\text{LB of } T]^2}\) | |
| A1 | for upper bound = 984.(3677853) or 984.(1125639..) and lower bound = 969.(0181643) or 968.(7669227..) |
Additional guidance
Accept \(52.04\dot{9}\) or 52.0499… for 52.05
Accept \(1.454\dot{9}\) or 1.4549… for 1.455
Rearrangement may occur after substitution, in this case correct bounds are not needed for this mark
\(51.95 \leqslant [\text{LB of } l] \lt 52.0\)
\(1.45 \lt [\text{UB of } T] \leqslant 1.455\)
\(52.0 \lt [\text{UB of } l] \leqslant 52.05\)
\(1.445 \leqslant [\text{LB of } T] \lt 1.45\)
Rearrangement may not be correct
NB: correct answer without supportive working gets 0 marks
Accept answers rounded or truncated to 3sf or better