24c)
If the error in measuring momentum is +100%,then the error in measurement of kinetic energy is :
a>400% b>300% c>100% d>200%
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10 Answers
1do we have to assume that the error in measurement of mass is 0?in that way i'm getting 200%.
24looks liek i am confused with ur options
are they >100 and so on ???
or 100,200 ???
1no not greater than...option a is 400%,b is 300%,c is 100% and d is 200%.
24DOnt know I am right or not...but this may inspire others to solve it
p=m.v
Case1: Let error be in measuring mass..
=> Δp/p *100=Δm/m *100=100
K=p22m
=> ΔKKx100=2Δppx100+Δmmx100
=> ΔKKx100==2(100)+100=300
=>300%
Case2: If error is in velocity
=> Δp/p *100=Δv/v *100=100
K=pv2
=> ΔKKx100=Δppx100+Δvvx100
=> ΔKKx100==100+100=200
=>200%
1Yes.I totally agree.Actually i had taken error in velocity and proceeded.
Unless mentioned in the sum its difficult to understand.