Question

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 1

Let the 50 gf weight produce the anticlockwise moment about the middle point of metre rule i.e at 50 cm.

Let the 50 gf weight produce the anticlockwise moment about the middle point of metre rule i.e at 50 cm. let a weight of 100gf produce a clockwise moment about the middle point.

Let the 50 gf weight produce the anticlockwise moment about the middle point of metre rule i.e at 50 cm. let a weight of 100gf produce a clockwise moment about the middle point. Let its distance from the middle be d cm.

Let the 50 gf weight produce the anticlockwise moment about the middle point of metre rule i.e at 50 cm. let a weight of 100gf produce a clockwise moment about the middle point. Let its distance from the middle be d cm. Then be d cm.

Let the 50 gf weight produce the anticlockwise moment about the middle point of metre rule i.e at 50 cm. let a weight of 100gf produce a clockwise moment about the middle point. Let its distance from the middle be d cm. Then be d cm. Then according to the principle of moments

Anticlockwise moment = Clockwise moment

$\sf 50gf \times 50cm = 100gf \times d$

$\sf so \: d = \frac{50 \times 50 }{100} = 25 \: cm \: from \: the \: other \: end.$

$\sf @kímlílч$

Answer 2

Answer:

this i the correct Answer of question