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A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold [given ³√2 = 1.2593] is approximately
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- 5.038 cm
- 5.190 cm
- 1.038 cm
- 0.518 cm
- 5.038 cm
Correct Option: D
| Volume of lead = | πr³ = | π × 2³ | ||
| 3 | 3 |
If the thickness of gold be x cm, then Volume of gold
| = | π[(2 + x)³ - 2³]cu.cm. | |
| 3 |
| ∴ | π[(2 + x)³ - 2³] | |
| 3 |
| = | π × 2³ | |
| 3 |
7rArr; (2 + x )³ – 2³ = 2³
⇒ (2 + x )³ = 8 + 8 = 16
⇒ (2 + x )³ = 2³.2
⇒ 2 + x = 2 × ³√2
⇒ 2 + x = 2 × 1.259 = 2.518
∴ x = 2.518 – 2 = 0.518 cm
