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A stone weighs 70 N in air and 50 N when fully submerged in water. Calculate its relative density and explain the apparent loss of weight.

Physics

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Answers

4 parts
a(i)

70 N70 \text{ N}

a(ii)

50 N50 \text{ N}

b(i)

20 N20 \text{ N}

b(ii)
3.5

Step-by-step solution

5 steps

Part a(i)· 1 step

  1. Read the value from the spring balance W1W_1.

    The pointer for W1W_1 is at the 70 N70 \text{ N} mark.

    W1=70 NW_1 = \boxed{70 \text{ N}}

Part a(ii)

50 N50 \text{ N}

  1. Read the value from the spring balance W2W_2.

    The pointer for W2W_2 is at the 50 N50 \text{ N} mark.

    W2=50 NW_2 = \boxed{50 \text{ N}}

Part b(i)

20 N20 \text{ N}

  1. Calculate the apparent loss of weight.

    The apparent loss of weight is the difference between the weight in air and the weight in water.

    Apparent loss of weight=W1−W2\text{Apparent loss of weight} = W_1 - W_2 Apparent loss of weight=70 N−50 N\text{Apparent loss of weight} = 70 \text{ N} - 50 \text{ N} Apparent loss of weight=20 N\text{Apparent loss of weight} = \boxed{20 \text{ N}}

Part b(ii)3.5

  1. Calculate the relative density.

    Relative density is the ratio of the weight of the stone in air to the apparent loss of weight in water.

    Relative density=W1Apparent loss of weight\text{Relative density} = \frac{W_1}{\text{Apparent loss of weight}} Relative density=70 N20 N\text{Relative density} = \frac{70 \text{ N}}{20 \text{ N}} Relative density=3.5\text{Relative density} = \boxed{3.5}

Part c· 1 step

  1. Explain the apparent loss of weight.

    The apparent loss of weight is due to the upthrust (or buoyant force) exerted by the water on the submerged stone. According to Archimedes' principle, this upthrust is equal to the weight of the water displaced by the stone, which acts upwards, making the stone appear lighter.

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