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Standard XII

Physics

Significant Figures in Product

A screw gauge...

Question

A screw gauge of pitch $1$ mm has a circular scale divided into $100$ divisions. The diameter of a wire is to measured by above said screw gauge. The main scale reading is $1$ mm and $47^{t h}$ circular division coincides with main scale. Find the curved surface area of wire in true significant figures. (Given the length of wire is equal to $5.6$ cm and there is no zero-error in the screw gauge)

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Solution

Least count= $\frac{1 m m}{100} = 0.01 m m$
Diameter=M.S.+no of division coinciding with main scale $\times$ least count
$= 1 m m + 47 \times 0.01 m m$
$= 1.47 m m = 0.147 c m$
curved surface area= $π d l$
$= \frac{22}{7} \times 0.147 \times 5.6$
$= 2.6 c m^{2}$

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Least count=1mm100=0.01mmDiameter=M.S.+no of division coinciding with main scale×least count=1mm+47×0.01mm=1.47mm=0.147cmcurved surface area=πdl=227×0.147×5.6=2.6cm2

Least count= $\frac{1 m m}{100} = 0.01 m m$
Diameter=M.S.+no of division coinciding with main scale $\times$ least count
$= 1 m m + 47 \times 0.01 m m$
$= 1.47 m m = 0.147 c m$
curved surface area= $π d l$
$= \frac{22}{7} \times 0.147 \times 5.6$
$= 2.6 c m^{2}$