Direction: In a machining experiment, tool life was found to vary with the cutting speed in the following manner:
| Cutting speed (m/min) | Tool life (minutes) |
| 60 | 81 |
| 90 | 36 |
-
What is the percentage increase in tool life when the cutting speed is halved?
-
- 50%
- 200%
- 300%
- 400%
Correct Option: C
VTn = C
| ∴ Tn = | |
| V |
| or T = | ![]() | ![]() | 1/n | = | ![]() | ![]() | 2 | ||
| V | V |
| or T ∝ | |
| V2 |
| ∴ | = | ||
| T | V02 |
| ∴ | = | ||
| T | V02 |
| or | – 1 = 4 – 1 | |
| T |
Hence percent increase = 300%
Alternately:
When cutting speed is halved
| 60 × (8.1)0.5 = | = (T2)0.5 | |
| 2 |
| or | = (2)2 | |
| 81 |
or T2 =4 × 81
| % change in tool life = | |
| T1 |
| = | × 100 | |
| 81 |
| = 3 × | × 100 = 300% | |
| 81 |

