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Question

Physics Question on Units and measurement

A physical quantity X is represented by X=[MηLθT!!!! ].X=[{{M}^{\eta }}{{L}^{-\theta }}{{T}^{-\text{o }\\!\\!|\\!\\!\text{ }}}]. The maximum percentage errors in the measurement of M, L and T, respectively are α\alpha %,\beta % and γ\gamma %. The maximum percentage error in the measurement of X will be

A

(ηαθβoγ)(\eta \alpha -\theta \beta -\text{o}|\gamma )%

B

(θβ+oγηα)(\theta \beta +\text{o}|\gamma -\eta \alpha \,\,)%

C

(αηβθγo)\left( \frac{\alpha }{\eta }-\frac{\beta }{\theta }-\frac{\gamma }{\text{o}|} \right)%

D

(ηα+θβ+oγ)(\eta \alpha +\theta \beta +o|\gamma )%

Answer

(ηα+θβ+oγ)(\eta \alpha +\theta \beta +o|\gamma )%

Explanation

Solution

Given, X=[MηLθTo]X=[{{M}^{\eta }}{{L}^{-\theta }}{{T}^{-\text{o}|}}] \therefore ΔXX×100\frac{\Delta X}{X}\times 100 =η(ΔMM×100)+θ(ΔLL×100)+!!!! (ΔTT×100)=\eta \left( \frac{\Delta M}{M}\times 100 \right)+\theta \left( \frac{\Delta L}{L}\times 100 \right)+\text{o }\\!\\!|\\!\\!\text{ }\left( \frac{\Delta \Tau }{T}\times 100 \right) \Rightarrow ΔXX×100=η(α)+θ(β)+o(γ)\frac{\Delta X}{X}\times 100=\eta (\alpha )+\theta (\beta )+o|(\gamma ) Thus, maximum percentage error in the measurement of X is given as ΔXX×100=(ηα+θβ+oγ)\frac{\Delta X}{X}\times 100=(\eta \alpha +\theta \beta +o|\gamma )%