If the error in measurement of momentum of a particle is 10% and mass is known exactly, the permissible error in the determination of kinetic energy is:A. 20%B. 21%C. 5%D. 10%â€‹

$\\ \text{We know that, Kinetic energy}, \mathrm{K}=\frac{1}{2} \mathrm{mv}^{2} \\ multiply by m both side, \mathrm{Km}=\frac{1}{2} \mathrm{m}^{2} \mathrm{v}^{2} or, \mathrm{Km}=\frac{1}{2}(\mathrm{mv})^{2} \ldots \ldots \ldots(1) \\ Also, momentum, \mathrm{p}=\mathrm{mv} \ldots \ldots \ldots \ldots \ldots \ldots . . (2) \\ from (1) and (2), we get, \mathrm{Km}=\frac{1}{2}(\mathrm{p})^{2} or, \mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{m}}$

According to the question, it is given that the mass is not changing.

$\begin{array}{l} K=\frac{p^{2}}{2 m} \Rightarrow \frac{\Delta K}{K}=2\left(\frac{\Delta P}{P}\right) \Rightarrow \frac{\Delta K}{K} \times 100=2\left(\frac{\Delta P}{P} \times 100\right) \\ \\ \Rightarrow \frac{\Delta K}{K} \times 100=2(10 \%) \Rightarrow \frac{\Delta K}{K} \times 100=20 \% \end{array}$

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