2020 A/L Physics Past Paper · Question 26 (English Medium)

Copper block on a spring balance — upthrust and reading at three positions

A copper block is suspended from a spring balance above a beaker of water, as shown in the figure. As the beaker of water is slowly raised, consider the following positions of the copper block.

Position 1 : the block is partly immersed.

Position 2 : the block is fully immersed.

Position 3 : the block rests on the bottom surface of the beaker.

At these positions 1, 2, 3 the upthrust forces are $B_1, B_2, B_3$ respectively, and the spring balance readings are $W_1, W_2, W_3$ respectively. Which of the following about them is correct?

(1) $B_1 < B_2 < B_3$ ; $W_1 > W_2 > W_3$
(2) $B_1 = B_2 < B_3$ ; $W_1 = W_2 > W_3$
(3) $B_1 = B_2 < B_3$ ; $W_1 > W_2 = W_3$
(4) $B_1 < B_2 = B_3$ ; $W_1 > W_2 = W_3$
(5) $B_1 < B_2 = B_3$ ; $W_1 > W_2 > W_3$

Answer and explanation

Answer: (5) $B_1 < B_2 = B_3$ ; $W_1 > W_2 > W_3$

What the question asks

We must find: the comparison of the upthrusts $B_1, B_2, B_3$ and the readings $W_1, W_2, W_3$ at the three positions.

Relevant theory

This happens due to an upward force acting on the ball. This force is known as the 'upthrust' or the force of buoyancy.

When a body is completely or partially immersed in a fluid at rest, it experiences an upthrust equal to the weight of the fluid displaced by the body.

$U = V\rho g$

Required equations

$B = V_{\text{immersed}}\,\rho g$

$W + B = mg$

$W + B + R = mg$

Reading the options

The options combine two orderings: the order of the upthrusts, and the order of the readings.

We find each separately, then match.

Reasoning and calculation

Common to all three: the same block (weight $mg$ , volume $V$ ), the same water ( $\rho$ ).

What changes: the immersed volume, and whether the bottom is touched at position 3.

What changes: at position 1, $V_{\text{immersed}} < V$ .

At positions 2 and 3, $V_{\text{immersed}} = V$ .

Only at position 3 is there a bottom reaction $R > 0$ .

Position 1

$B_1 < V\rho g$

Partly immersed.

$B_1 = V_1\rho g < V\rho g$

$W_1 = mg - B_1$

Position 2

$B_2 = V\rho g$

Fully immersed.

$W_2 = mg - B_2 < W_1$

Position 3

$B_3 = B_2,\ W_3 < W_2$

Fully immersed, on the bottom.

$B_3 = V\rho g = B_2$

Since the bottom reaction $R$ also supports it:

$W_3 = mg - B_3 - R < W_2$

Putting the pairs together:

$B_1 < B_2$ (the immersed volume grows)

$B_2 = B_3$ (the volume stays at $V$ )

$W_1 > W_2$ (since $B$ grows)

$W_2 > W_3$ (since $R$ is added)

B_1

B_2 = B_3

$B_1 < B_2 = B_3$ and $W_1 > W_2 > W_3$ — option (5).

Checks

$8.9$

$mg = 0.89 \times 10$

$1.0$

$B_2 = 10^{-4} \times 1000 \times 10$

$7.9$

$W_2 = 8.9 - 1.0$

Pairwise check: the order of $B$ alone eliminates (1), (2), (3).

The order of $W$ eliminates (4) — only (5) remains.

Conclusion

The upthrust stops growing once fully immersed ( $B_1 < B_2 = B_3$ ); the reading falls at every position ( $W_1 > W_2 > W_3$ ) — the answer is (5).

Thinking the reading stays the same once it rests on the bottom.

The bottom's reaction supports part of the weight, so the string tension falls further.

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