Physics, asked by tiara5, 1 year ago

a uniform metre scale of weight 50 GF is balanced at 10 cm mark when a weight of 100 GF is suspended at 5 cm mark.Where must a weight of 80gf be suspended to balance the metre scale.

Answers

Answered by misa3

anti clockwise moment = 100gf × (10-5)cm
= 100gf ×5 cm
= 500gf×cm
clockwise moment = 50gf×(50-10)cm
= 50×40gf×cm
= 2000gf×cm
for equilibrium,
anti clockwise moment =clockwise moment
500gf×cm + 80 gf × d cm= 2000gf×cm
80 gf×d cm =( 2000-500)gf×cm
d = 1500/80 cm
d = 18.75cm
therefore the weight of 80 gf must be suspended at a distance of 18.75cm to the left of 10 cm mark.

Attachments:

tiara5: I appreciate your effort but the answer is wrong

misa3: what's the correct answer

tiara5: I did the same 4 my exam but it was wrong

tiara5: I don't know the correct answer

misa3: is the anwer 6.25 cm

misa3: ??????

tiara5: No

misa3: then what is the correct answer

misa3: as soon as u know the correct answer then u report it to me

misa3: ok

Answered by panjwarhiaapannu

(60)15+50(2)=900+1000=1900anwer

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