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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
$\begin{aligned} \frac{Δm}{m} \times 100 = 2 % \\ \frac{Δv}{v} \times 100 = 3 % \end{aligned}$
But $K . E = \frac{1}{2} {mϑ}^{2}$
$i.e., KE {αmϑ}^{2}$
$\begin{array}{l} \Rightarrow \frac{ΔKE}{KE} \times 100 \\ = \frac{Δm}{m} \times 100 + 2 \frac{Δϑ}{ϑ} \times 100 \\ \Rightarrow \frac{ΔKE}{KE} \times 100 = 2 % + 6 % \\ \Rightarrow \frac{ΔKE}{KE} \times 100 = 8 % \end{array}$

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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