| Given: Mass of the sinker | = | 40 g |
| Density of the sinker (lead) | = | 11.0 g cm–3 |
| (i) Volume of sinker | = | Mass / Volume |
| = | 40 g/11.0 g cm–3 | |
| = | 3.64 cm3 | |
| (ii) Loss in weight of sinker when immersed in water | = | Weight of water displaced by the sinker |
| = | Volume of sinker × density of water × g | |
| = | 3.64 × 1 | |
| = | 3.64 gf | |
| (iii) Weight of sinker in water | = | Weight of sinker in air – Loss in weight |
| = | 40 – 3.64 | |
| = | 36.36 gf | |
| (iv) Weight of cork in water | = | Weight of cork together with the sinker in water – Weight of sinker in water |
| = | 11.5 – 36.36 | |
| = | – 24.86 gf | |
| (v) Loss in weight of cork in water | = | Weight of cork in air – Weight of cork in water |
| = | 6.5 – (– 24.86) | |
| = | 31.36 gf | |
| (vi) R.D. of cork | = | (Weight of cork in air) / (Loss in weight of cork) |
| = | 6.5/31.36 | |
| = | 0.21 |
A spring balance reads 220 gf when carrying a lump of lead in air. If the lead is now immersed with half of its volume in brine, what will be the new reading of the spring balance? R.D. of lead is 11.4, and R.D. of brine is 1.1?
| Mass of lump of lead | = | 220 g |
| Density of lead | = | 11.4 g cm–3 |
| Volume of lump of lead | = | (mass) / (density) |
| = | 220/11.4 | |
| = | 19.3 cm3 | |
| Volume of lump of lead submerged in brine | = | 1/2 its volume |
| = | (1/2) × 19.3 | |
| = | 9.65 cm3 | |
| Weight of brine displaced | = | Volume of lump of lead submerged * density of brine * g |
| = | 9.65 × 1.1 gf | |
| = | 10.62 gf | |
| But loss in weight | = | weight of brine displaced |
| ∴ Loss in weight | = | 10.62 gf |
| Now the apparent weight (or the new reading of spring balance) | ||
| = | true weight – loss in weight | |
| = | 220 gf – 10.62 gf | |
| = | 209.38 gf. |
A solid of density d has weight W. Show that its apparent weight will be W [1 – dL/d)] when it is completely submerged in a liquid of density dL?
| Volume of the solid | = | Mass / Density = W/d |
| Volume of liquid displaced | = | Volume of the solid |
| = | W/d | |
| Upthrust on the solid | = | Volume of liquid displaced × density of liquid. |
| = | (W/d) × dL | |
| ∴ Apparent weight | = | True weight – upthrust |
| = | W – (W/d) × dL | |
| = | W (1 – (dL/d)) |
| Given: Volume of iron piece | = | 100 cm3 | |
| = | 100 × 10–6 m3 | ||
| = | 10–4 m3 | ||
| (i) | Mass in air | = | Volume × density |
| = | 10–4 × (7.8 × 103) | ||
| = | 7.8 × 10–1 | ||
| = | 0.78 kg | ||
| ∴ Weight in air | = | 0.78 kgf = 0.78 × 10 = 7.8 N | |
| (ii) | Upthrust | = | (Volume of water displaced) × ρ × g |
| But volume of water displaced | = | volume of iron piece when it is | |
| completely immersed | = | 10–4 m3 | |
| Upthrust | = | 10–4 × 1000 × 10 = 1 N | |
| (iii) | Apparent weight | = | True weight – Upthrust |
| = | 7.8 – 1 | ||
| = | 6.8 N |