2024 · M/J · 5054/31 | Q1

One correct answer. This part can’t be checked automatically yet, so compare your own working with it.

Line up the six balls beside the fixed metre rule so that each touches the next, with no gaps. At each end of the row, stand a set square upright with its bottom edge across the rule at right angles to the rule, and slide it until its flat face just touches the outermost point of the end ball without pushing the row apart, as at A and B in Fig. 1.1.

Read the scale where that face of each set square meets the rule, to the nearest 0.1 cm (1 mm, the smallest division on the rule), and record both positions in cm. Look straight down at that point, not from the side, to avoid parallax. The mark is for two readings whose separation is the length of your row of six touching balls, so check the balls have not rolled apart before you read.

One correct answer. This part can’t be checked automatically yet, so compare your own working with it.

Find l by subtracting the smaller reading from the larger: l = position of B − position of A. Write it to the nearest 0.1 cm, with the unit cm. Use your own two readings from (a)(i); the first mark is for l agreeing with them.

Six touching balls in a row span six diameters, so l = 6d and d = l ÷ 6. Divide your l by 6 and round to the nearest 0.1 cm, as the question asks. The second mark needs both the correct division of your l and the answer written to 0.1 cm, so do not leave extra decimal places or round to a whole centimetre.

One correct answer. This part can’t be checked automatically yet, so compare your own working with it.

Substitute your own d from (a)(ii), in cm, into V = 3.14d³/6. Cube d first (d × d × d), then multiply by 3.14 and divide by 6. The answer is in cm³ because d is in cm. Make sure you cube d rather than multiplying it by 3.

This is the volume of a sphere written in terms of its diameter: (4/3) × 3.14 × (d/2)³ simplifies to 3.14d³/6. The mark is for your d put into the formula and V worked out correctly, so show the substitution and check the arithmetic.

One correct answer. This part can’t be checked automatically yet, so compare your own working with it.

Add the six balls carefully, for example one at a time, so that no water splashes out; lost water would make V₂ too small. Check that all six balls are below the surface, because the rise in the water level equals their volume only when they are fully under water.

Read the new level V₂ in cm³ and write it down: the mark needs V₂ shown as well as V_t. Then calculate V_t = V₂ − V₁, the volume of water pushed up by the six balls, in cm³, using your own V₁ from (b)(i).

One correct answer. This part can’t be checked automatically yet, so compare your own working with it.

V_t is the combined volume of all six balls, so the average volume of one ball is V = V_t ÷ 6. Divide your own V_t from (b)(ii) by 6 and give the answer in cm³. The mark is for this division done correctly with your value.

One correct answer. This part can’t be checked automatically yet, so compare your own working with it.

Choose method 1. The scheme accepts it with either of two reasons. One: the measuring cylinder can only be read to the nearest 0.5 cm³ or 1 cm³, depending on its scale, which is coarse for the volume of six small balls. Two: the rule measures the length, and so the diameter, to the nearest millimetre, which is more accurate.

The single mark is for method 1 together with a reason like one of these, so write both the method and the reason.

One correct answer. This part can’t be checked automatically yet, so compare your own working with it.

Either of two routes earns both marks. Route one: put the small beaker on the top-pan balance and press tare so the balance reads zero. Add all six dried balls, read the mass in g, and divide it by 6 to get the mass of one ball.

Route two: put the beaker on the balance and record its mass. Add the six balls and record the new mass. Subtract the beaker's mass from the total and divide the answer by 6. Write the method as steps, list each reading in g, and give the mass of one glass ball in g.