Hydrostatics CBSE Class 9 Physics Worksheets Set 08

Questions with Answers

Question 1. Two graduations on the stem of a hydrometer indicate specific gravities 0.6 and 0.9. What is the specific gravity at a mark midway between them?
Answer:
Let W be the weight of the hydrometer, V its volume immersed at the 0.9 mark and a the area of its stem (volume per unit length), taking a unit length of stem.
At the 0.9 mark: 0.9V = W. At the 0.6 mark the immersed volume is V + a: 0.6(V + a) = W.
So 0.9V = 0.6V + 0.6a, giving V = 2a. The mark midway between the two lies at immersed volume \( V + \tfrac{a}{2} = 2.5a \), so its specific gravity = \( \dfrac{W}{2.5a} = \dfrac{0.9 \times 2a}{2.5a} = 0.72 \).
The specific gravity at the midway mark is 0.72 (not 0.75, because the scale is not linear).
In simple words: Equal lengths of the stem correspond to equal volumes, but specific gravity varies inversely with volume.

Teacher's Note:
1. Use the law of flotation at each mark: weight = specific gravity x immersed volume.
2. The scale of a hydrometer is not uniform.

Question 2. An ice cube floats on water with 1/12 of its volume projecting above the surface. Calculate the specific gravity of ice.
Answer:
Immersed fraction = \( 1 - \tfrac{1}{12} = \tfrac{11}{12} \).
Specific gravity of ice = immersed fraction = \( \tfrac{11}{12} \approx 0.917 \).
The specific gravity of ice is about 0.92.
In simple words: The fraction below the surface equals the density ratio of ice to water.

Teacher's Note:
1. For a body floating on water, fraction immersed = specific gravity.
2. This is why most of an iceberg is hidden.

Question 3. A flat-bottomed test-tube loaded with lead shot floats in water in a jar. Some of the lead shot inside is now put into the water (it sinks to the bottom). What happens to the level of the water in the jar?
Answer:
While all the shot is inside the floating tube, the tube displaces water of weight equal to the weight of tube plus shot, so the volume displaced is (total weight) / (density of water).
After some shot is dropped into the water, the tube floats displacing less water (its weight is less), while the shot displaces only its own small volume. Lead is denser than water, so the shot's volume is smaller than the water previously displaced by its weight.
The water level in the jar decreases.
In simple words: A floating object displaces water equal to its weight, but a sunk object displaces only its own volume, which is smaller for a dense material.

Teacher's Note:
1. Compare the volume displaced before and after.
2. The same reasoning shows that the level also falls when a heavy stone is thrown from a floating boat into the water.

Question 4. A body of mass 0.1 kg and density 10 000 kg/m³ is submerged in water. What is the buoyant force on the body?
Answer:
Volume = \( \dfrac{0.1}{10000} = 10^{-5} \) m³.
Buoyant force = weight of water displaced = \( 10^{-5} \times 1000 = 10^{-2} \) kgf = 0.01 kgf.
In newtons: 0.01 x 9.8 = 0.098 N.
In simple words: The body displaces only one hundred-thousandth of a cubic metre of water.

Teacher's Note:
1. Buoyant force = V d g.
2. kgf multiplied by 9.8 gives newtons.

Question 5. A gold jewel suspected to contain copper weighs 110 g in air and 103.7 g in water. Find the ratio of copper to gold by weight. (Relative density of gold 19.3, of copper 8.9.)
Answer:
Volume of the jewel = 110 - 103.7 = 6.3 cm³ (loss of weight in water).
Let the weight of gold be w g and of copper (110 - w) g: \( \dfrac{w}{19.3} + \dfrac{110 - w}{8.9} = 6.3 \).
Solving: \( 8.9w + 19.3(110 - w) = 6.3 \times 19.3 \times 8.9 \Rightarrow w \approx 100 \) g. Copper = 10 g.
Copper : gold = 10 : 100 = 1 : 10.
In simple words: The volume from the loss of weight must match the volumes of the two metals.

Teacher's Note:
1. Volume = mass / density for each metal.
2. Check: 100/19.3 + 10/8.9 = 5.18 + 1.12 = 6.3 cm³.

Question 6. A compound block of 10 kg of wood and 11.4 kg of lead just sinks in water. If the density of lead is 11 400 kg/m³, calculate the density of the wood.
Answer:
Volume of lead = 11.4/11400 = 10⁻³ m³. Let the volume of wood be V.
Just sinking means the average density equals that of water: total mass = (V + 10⁻³) x 1000, so \( 21.4 = 1000V + 1 \) and \( V = 0.0204 \) m³.
Density of wood = 10 / 0.0204 = 490 kg/m³.
In simple words: The wood and lead together just match the density of water.

Teacher's Note:
1. Average density = total mass / total volume.
2. Wood is lighter than water; the lead brings the average up.

Question 7. A boat weighing 100 kg has an overall volume of 3 m³. Keeping a safety margin of 20% in its maximum weight, how much weight of cargo and passengers can the boatman accommodate?
Answer:
Maximum weight that can float = 3 x 1000 = 3000 kgf. With a 20% margin only 80% of this is used: 2400 kgf.
Load = 2400 - 100 = 2300 kg.
In simple words: Keep 20% of the buoyancy as spare and subtract the boat's own weight.

Teacher's Note:
1. The safe total weight is 80% of the full buoyant force.
2. The boat's weight counts as part of the total.

Question 8. A trough full of water is placed on a spring balance. A hand is put into the water without touching the trough and the water overflows into a vessel placed outside the balance. How does the reading of the balance change?
Answer:
The hand experiences a buoyant force equal to the weight of the water displaced, and it pushes down on the water with the same force.
The displaced water, equal in weight to this force, overflows off the balance.
The net change is zero: the reading remains unchanged. (If the beaker were not full and nothing overflowed, the reading would increase by the weight of the water displaced.)
In simple words: What the hand adds is exactly what the overflowing water removes.

Teacher's Note:
1. The reaction to buoyancy equals the weight of the displaced water.
2. Compare with the case where there is no overflow.