A golf ball has a mass of 40 g and a speed of $45 {ms}^{- 1}$ . If the speed can be measured with in accuracy of 2%, the uncertainty in position is

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$4.16 \times 10^{- 33} m$

$1.04 \times 10^{- 33} m$

$6.14 \times 10^{33} m$

$1.46 \times 10^{- 33} m$

answer is D.

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Detailed Solution

$∆ x = \frac{h}{4 πm ∆ v} ∆ x = \frac{6.626 \times 10^{- 34} kg m^{2} s^{- 1} \times 10^{3}}{4 \times 3.142 \times (\frac{2}{100} \times 45 {ms}^{- 1}) \times 40} = 10^{4} {ms}^{- 1} = 1.46 \times 10^{- 33} m$

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