JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 18)
A screw gauge of pitch $$0.5 \mathrm{~mm}$$ is used to measure the diameter of uniform wire of length $$6.8 \mathrm{~cm}$$, the main scale reading is $$1.5 \mathrm{~mm}$$ and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is :
[Screw gauge has 50 divisions on its circular scale]
6.8 cm2
3.4 cm2
3.9 cm2
2.4 cm2
Explanation
Least count $$ = {{0.5} \over {50}}$$ mm = 0.01 mm
$$\therefore$$ Diameter, d = 1.5 mm + 7 $$\times$$ 0.01
= 1.57 mm
$$\therefore$$ Surface area $$ = (2\pi r) \times l$$
$$ = \pi dl$$
$$ = 3.142 \times {{1.57} \over {10}} \times 6.8$$ cm2
= 3.354 cm2 = 3.4 cm2
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