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In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while 46th division the circular scale coincide with the reference line. The diameter of the wire is _______ \times 10^{-2}mm

Option: 1

220


Option: 2

__


Option: 3

__


Option: 4

__


Answers (1)

best_answer

Pitch =0.5 \mathrm{~mm} \\ L.C. =\frac{\text { pitch }}{\text { circular division }}=\frac{0.5 \mathrm{~mm}}{100}=0.005 \mathrm{~mm}

\text { Zero error }=6 \times \text { L.C. }=6 \times(0.005) \mathrm{mm}

\text { Reading }=\text { main linear scale reading }+\mathrm{n}(\mathrm{L} . \mathrm{C} \text {. })-\text { zero error }

                    \begin{aligned} & =4(0.5 \mathrm{~mm})+46(0.005)-6(0.005) \\ & =2 \mathrm{~mm}+40 \times 0.005 \mathrm{~mm} \end{aligned}

                    \begin{aligned} & =2 \mathrm{~mm}+\frac{200}{1000} \mathrm{~mm} \\ & =2.2 \mathrm{~mm} \end{aligned}

\text { Rading }=220 \times 10^{-2} \mathrm{~mm}        

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