MCQOPTIONS
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| 1. |
When a 4 kg mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by 2cms. The work required to be done by an external agent in stretching this spring by 5cms will be \[(g=9.8\text{ }m/se{{c}^{2}})\] |
| A. | 4.900joule |
| B. | 2.450joule |
| C. | 0.495 joule |
| D. | 0.245 joule |
| Answer» C. 0.495 joule | |