JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 3)
A silver wire has a mass (0.6 $$\pm$$ 0.006) g, radius (0.5 $$\pm$$ 0.005) mm and length (4 $$\pm$$ 0.04) cm. The maximum percentage error in the measurement of its density will be :
4%
3%
6%
7%
Explanation
$$\rho = {m \over V} = {m \over {\pi {r^2} \times l}}$$
$$\therefore$$ % error in $$\rho = \left( {{{0.006} \over {0.6}} + 2 \times {{0.005} \over {0.5}} + {{0.04} \over 4}} \right) \times 100$$
$$ = 4\% $$
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