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Kinetic energy of a particle depends on the square of speed of the particle. If error in measurement of speed is $$40 \%$$ then error in the measurement of kinetic energy will be

Solution
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Hint :
Error in kinetic energy is $$\frac{\Delta K}{K}$$.
$$\textbf{Step1:Initial Kinetic Energy}$$
Kinetic energy of a particle depends on square of speed of a particle.
$$\mathrm{K}=1 / 2 \mathrm{mv}^{2} \ldots .(1)$$
Let initial velocity of a particle is v and initial kinetic energy is $$K$$
$$\textbf{Step2:Final Speed of particle}$$
If error in speed is $$40 \%$$
then new speed of particle $$v^{\prime}=v+40 \% \text { of } v$$
$$=v+40 v / 100$$
$$=1.4 \mathrm{v}$$
$$\textbf{Step3:Final Kinetic Energy}$$
Final kinetic energy will be $$\mathrm{K}^{\prime}=1 / 2 \mathrm{mv}^{\prime 2}$$
$$=1 / 2 m(1.4 v)^{2}$$
$$=1 / 2 \mathrm{~m}\left(1.96 \mathrm{v}^{2}\right)$$
$$=(1.96) 1 / 2 m v^{2}$$
$$=\text { (1.96) K }$$ [ form equation (1), ]
$$\textbf{Step4:Error in Kinetic Energy}$$
Now, % error in Kinetic energy $$=$$ final kinetic energy$$-$$ initial kinetic energy$$/$$initial kinetic energy$$\times 100$$
$$=\left(\mathrm{K}^{\prime}-\mathrm{K}\right) / \mathrm{K} \times 100$$
$$=(1.96 \mathrm{~K}-\mathrm{K}) / \mathrm{K} \times 100$$
$$=0.96 \mathrm{~K} / \mathrm{K} \times 100$$
$$=96 \%$$
Therefore maximum error in kinetic energy should be $$96 \%$$.

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