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  • A screw gauge of pitch is used to measure the diameter of uniform wire of length <img alt="\mat

A screw gauge of pitch \mathrm{0.5 \mathrm{~mm}} is used to measure the diameter of uniform wire of length \mathrm{6.8 \mathrm{~cm}, } the main scale reading is \mathrm{1.5 \mathrm{~mm} } and circular scale reading is \mathrm{7 }. The calculated curved surface area of wire to appropriate significant figures is :
[Screw gauge has 50 divisions on its circular scale]

Option: 1

\mathrm{6.8 \mathrm{~cm}^{2}}


Option: 2

\mathrm{3.4 \mathrm{~cm}^{2}}


Option: 3

\mathrm{3.9 \mathrm{~cm}^{2}}


Option: 4

\mathrm{2.4 \mathrm{~cm}^{2}}


Answers (1)

best_answer

\mathrm{pitch=0.5\: mm}

\mathrm{l=6.8\: cm}

\mathrm{MSR=1.5\: mm}

\mathrm{CSD=7\: mm}

\mathrm{Reading \: (Diameter)=MSR+CSD\times LC}

                                        \mathrm{=1.5\: mm+7\times \left ( \frac{0.5\: mm}{50} \right )}

                                        \mathrm{\because Least\: count=\frac{pitch}{No\: of\: circular\: scale\: division}}

\mathrm{=1.5\: mm+0.07\: mm}

\mathrm{Diameter (D)=1.57\: mm=1.57\times 10^{-1}cm}

\mathrm{Curved\: surface\: area=2\pi Rl}

                                       \mathrm{=\pi Dl}

                                       \mathrm{=3.14\times 1.57\times 6.8\times 10^{-1}}

                                       \mathrm{=3.4\: cm^{2}}

Hence (2) is correct option

 

Posted by

sudhir.kumar

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