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JEE MAIN
26.02.2021_S2
Question
A ball weighing 10 g is moving with a velocity of $90 \mathrm{~ms}^{-1}$. If the uncertainty in its velocity is $5 \%$, then the uncertainty in its position is $\_\_\_\_$ $\times 10^{-33} \mathrm{~m}$. (Rounded off to the nearest integer) [Given : $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$ ]
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Solution
$\begin{aligned} & \Delta v=90 \times \frac{5}{100} \\ & =4.5 \mathrm{~m} / \mathrm{s} \\ & \Delta v . \Delta x=\frac{\mathrm{h}}{4 \pi \mathrm{~m}} \\ & \Delta x=\frac{\mathrm{h}}{4 \pi \mathrm{~m} . \Delta \mathrm{v}} \\ & =\frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 0.01 \times 4.5} \\ & =1.17 \times 10^{-33}\end{aligned}$
Question Tags
JEE Main
Chemistry
Easy
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