Experimental Physics

JEE Main level, single correct, medium

31 of 70

Question 31 of 70

A screw gauge of pitch 0.5mm0.5mm is used to measure the diameter of uniform wire of length 6.8cm6.8cm, the main scale reading is 1.5mm1.5mm and circular scale reading is 77. The calculated curved surface area of wire to appropriate significant figures is
[Screw gauge has 5050 divisions on the circular scale]

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Answer: (B) 3.4cm23.4{cm}^{2}

Least Count (L.C)=PN=0.5mm50=0.01mm=\dfrac{P}{N}=\dfrac{0.5mm}{50}=0.01mm
Length of wire =6.8cm=6.8cm
Diameter of wire =1.5mm+7×=1.5mm+7\times L.C
=1.5mm+7×.01=1.57mm=1.5mm+7\times .01=1.57mm
Curved surface area =πDl=\pi Dl
=3.14×6.8×1.57×10−1cm2=3.14\times 6.8\times 1.57\times {10}^{-1}{cm}^{2}
=3.352cm2=3.4cm2=3.352{cm}^{2}=3.4{cm}^{2}

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