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Q.

The percentage error in the mass and speed are 2% an 3% respectively. What is the maximum error percentage in kinetic energy calculated using these quantities

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Detailed Solution

Given
Δmm×100=2%Δvv×100=3%
But K.E=12mϑ2
 i.e., KE αmϑ2
⇒ΔKEKE×100=Δmm×100+2Δϑϑ×100⇒ΔKEKE×100=2%+6%⇒ΔKEKE×100=8%

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