calculate percentage error in determination of time period of a pendulum T=2π√l÷g where l and g are measured with ±1% and ±2% error.
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The correct relation is
T= 2π √l/g
So Δt/t * 100 = [1/2 Δl/l + 1/2 Δg/g]*100
( 2 π is constant so it will be ignored)
Hence by solving:
0.5% +1% = 1.5%
Hope it helps!!
T= 2π √l/g
So Δt/t * 100 = [1/2 Δl/l + 1/2 Δg/g]*100
( 2 π is constant so it will be ignored)
Hence by solving:
0.5% +1% = 1.5%
Hope it helps!!
mridul4:
thanks a lot can i get u some more ques. as i have got a lot of holiday homework
Yes.. Why not.. I'll try my best!!
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