InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A uniform metre rule is pivoted at its mid-point. A weight of 50 gf is suspended at one end of it. Where should a weight of 100 gf be suspended to keep the rule horizontal? |
|
Answer» Let the 50 gf weight produce anticlockwise moment about the middle point of metre rule .i.e, at 50cm. Let a weight of 100gf produce a clockwise moment about the middle point. Let its distance from the middle be d cm. Then, according to principle of moments, Anticlockwise moment = Clockwise moment 50gf × 50 cm = 100gf × d So, d = 50 x 50/100 = 25cm from the other end |
|