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Two solid cylinders of radii 4 cm and 5cm and length 6 cm and 4 cm respectively are recast into cylindrical disc of thickness 1 cm. The radius of the disc is
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- 7 cm
- 14 cm
- 21 cm
- 28 cm
- 7 cm
Correct Option: B
Sum of the volume of two cylinders = πr1²h1 + πr2²h2
| = | (4 × 4 × 6 + 5 × 5 × 4) | |
| 7 |
| = | (96 + 100) | |
| 7 |
| = | × 196 = 616 cm³ | |
| 7 |
Let the radius of the disc be r cm.
∴ πr² × 1 = 616
| ⇒ r² = | × r² = 616 | |
| 7 |
| ⇒ r² = | = 196 | |
| 22 |
⇒ r = √196 = 14 cm
