Physics 5054 · O Level · Physical quantities and measurement techniques

Physical quantities and measurement techniques — practice question

A student determines the volume of a small glass ball (marble) using two separate methods.
(a(i))[1]

Use the metre rule to read the positions of points A and B shown in Fig. 1.1. Record your values to the nearest $0.1\,\text{cm}$. Position of point A = [BLANK] cm. Position of point B = [BLANK] cm.

(a(ii))[2]

In Fig. 1.1, the length $l$ is the separation between points A and B. The mean diameter $d$ of one ball may be determined from the equation $l = 6d$. Use your responses to (a)(i) to work out the length $l$ and the diameter $d$. Give both answers to the nearest $0.1\,\text{cm}$. $l = [BLANK]\,\text{cm}$, $d = [BLANK]\,\text{cm}$.

(a(iii))[1]

The mean volume $V$ of one glass ball by this method is given by the equation $V = \frac{3.14d^3}{6}$. Calculate $V$.

(b(i))[1]

Figure 1.2 shows the volume of water in the measuring cylinder as $V_1$. Write down the reading $V_1$.

(b(ii))[1]

The six glass balls are then added carefully to the water in the measuring cylinder. Figure 1.3 shows the new reading on the measuring cylinder, $V_2$. The volume $V_T$ of the six balls is found from $V_T = V_2 - V_1$. Calculate $V_T$. Show your working.

(b(iii))[1]

Calculate the average volume $V$ of one ball by this method. Give your answer to the nearest $0.1\,\text{cm}^3$.

(c)[1]

Suggest whether method 1 or method 2 gives the more accurate value for the volume of the ball. Explain your answer.

(d)[2]

The student now uses the six glass balls to determine the average mass of one glass ball with a small beaker and a top pan (electronic) balance. Describe the method the student uses.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Point A = 10.2 cm and point B = 19.7 cm (cao)

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