Question

$\text{If a computed value is obtained by dividing two measurements as s =}\frac{620.5}{24.15},\text{the maximum error and most probable error will be}$

• A. 0.045, 0.0082
• B. 0.0321, 0.0061
• C. 0.0085, 0.0032
• D. 0.0073, 0.0285

Answer 1

The correct option is D 0.0073, 0.0285

x = 620.5

y = 24.15

$max.errorinx,\left({\delta }_x\right)=0.05$

$probableerrorinx,\left(e_x\right)=\pm 0.025$

$max.erroriny,\left({\delta }_y\right)=0.005$

$probableerroriny,\left(e_y\right)=\pm 0.0025$

Maximum error in s

${\delta }_s=\frac{1}{y}\delta x+\frac{x}{y^2}\delta y$

${\delta }_s=\frac{0.05}{24.15}+\frac{620.5}{(24.15{)}^2}\times 0.005$

${\delta }_s=0.002+0.0053$

${\delta }_s=0.0073$

Probable error,

$e_s=s\times \sqrt{{\left(\frac{e_x}{x}\right)}^2+{\left(\frac{e_y}{y}\right)}^2}$

$e_s=\left(\frac{620.5}{24.15}\right)\times \sqrt{{\left(\frac{0.025}{620.5}\right)}^2+{\left(\frac{0.0025}{24.15}\right)}^2}$

$e_s=0.00285$