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Q. A uniform half metre scale of weight 100 gf is balanced at 30 cm point. If 80 gf is suspended from one end where should we suspend a weight of 55 gf to bring it in equilibrium?

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Solution

$D e a r s t u d e n t, b y p r i n c i p l e o f m o m e n t s w e h a v e, 80 \times 30 = 100 \times 20 + 55 \times x 55 x = 2400 - 2000 = 400 \Rightarrow x = 7.27 c m s o t h e 55 g f w e i g h t s h o u l d b e a t 7.27 c m f r o m t h e b a l a n c e d p o i n t i n t h e o t h e r s i d e o f t h a t o f 80 g f w e i g h t . R e g a r d s$

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Solve this: Q. A uniform half metre scale of weight 100 gf is balanced at 30 cm point. If 80 gf is suspended from one end where should we suspend a weight of 55 gf to bring it in equilibrium?

Dear student,by principle of moments we have,80×30=100×20+55×x55x=2400-2000=400⇒x=7.27cmso the 55gf weight should be at 7.27cm from the balanced point in the other side of that of 80gf weight.Regards

Solve this:
Q. A uniform half metre scale of weight 100 gf is balanced at 30 cm point. If 80 gf is suspended from one end where should we suspend a weight of 55 gf to bring it in equilibrium?