Question

Measure of two quantities along with the precision of respective measuring instrument is

$\mathrm { A } = 2.5 \mathrm {~ms} ^ { - 1 } \pm 0.5 \mathrm {~ms} ^ { - 1 } , \mathrm {~B} = 0.10 \mathrm {~s} \pm 0.01 \mathrm {~s}$

The value of AB will be

• A. (a) $( 0.25 \pm 0.08 ) \mathrm { m }$
• A. (b) $( 0.25 \pm 0.5 ) \mathrm { m }$
• A. (c) $( 0.25 \pm 0.05 ) \mathrm { m }$
• A. (d) $( 0.25 \pm 0.135 ) \mathrm { m }$

Answer 1

(a)

Sol. Here , $\mathrm { A } = 2.5 \mathrm {~ms} ^ { - 1 } \pm 0.5 \mathrm {~ms} ^ { - 1 }$

$B = 0.10 s \pm 0.01 s$

$\mathrm { AB } = \left( 2.5 \mathrm {~ms} ^ { - 1 } \right) ( 0.10 \mathrm {~s} ) = 0.25 \mathrm {~cm}$ $\frac { \Delta \mathrm { AB } } { \mathrm { AB } } = \left( \frac { \Delta \mathrm { A } } { \mathrm { A } } + \frac { \Delta \mathrm { B } } { \mathrm { B } } \right) = \left( \frac { 0.5 } { 2.5 } + \frac { 0.01 } { 0.10 } \right) = 0.3$ $\Delta \mathrm { AB } = 0.3 \times 0.25 \mathrm {~m} = 0.075 \mathrm { n }$

$= 0.08 \mathrm {~m}$ (Rounded off to two significant figures)

The value of AB is $( 0.25 \pm 0.08 ) \mathrm { m }$