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A ball weighing $10g$ is moving with a velocity of $90m{s^{ - 1}}$. If the uncertainty in its velocity is $5$ percent, then the uncertainty in its position is _______$ \times {10^{ - 33}}m$.

Answer
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Hint: In order to solve this question, we will first find the value of uncertainty in the velocity of the ball, and then using the Heisenberg uncertainty principle we will solve to find the uncertainty in the position of the ball.

Formula used:
The Heisenberg uncertainty principle is related by $\Delta x.\Delta {p_x} \geqslant \dfrac{h}{{4\pi }}$ where, $\Delta x,\Delta {p_x}$ is the uncertainty in the position and momentum, $h = 6.626 \times {10^{ - 34}}Js$ is known as the Planck’s constant.

Complete answer:
We have given that mass of the ball is $m = 10g = 0.01kg$
velocity of the ball is $v = 90m{s^{ - 1}}$ with an uncertainty of $5$ percent which means
$
  \Delta v = \dfrac{5}{{100}} \times v \\
  \Delta v = 4.5m{s^{ - 1}} \\
 $
Planck’s constant value is $h = 6.626 \times {10^{ - 34}}Js$ so , using these values and the Heisenberg uncertainty principle formula $\Delta x.\Delta {p_x} \geqslant \dfrac{h}{{4\pi }}$ we have,
$\Delta x.\Delta {p_x} \geqslant \dfrac{{6.626 \times {{10}^{ - 34}}}}{{4(3.14)}}$
as we know the momentum of a body is the product of mass and its velocity so uncertainty in momentum can be written in terms of mass multiplied by the uncertainty in the velocity so, we can write $\Delta {p_x} = m.\Delta v$ so, on putting the values we get,
$
  \Delta x.0.01(4.5) \geqslant \dfrac{{6.626 \times {{10}^{ - 34}}}}{{4(3.14)}} \\
  \Delta x \geqslant 1.17 \times {10^{ - 33}}m \\
 $
On comparing this value with the given form we see that the blank place of ___$ \times {10^{ - 33}}m$ has a value of $1.17$

Hence, the uncertainty in the position of the ball is $1.17 \times {10^{ - 33}}m$


Note: In Heisenberg, uncertainty principle formula the direction in which uncertainty in position is measured then the uncertainty in the momentum is also measured in the same direction, and the basic unit of conversion here is used as $1kg = 1000g$.