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Hint :In this question we need to find out the error in volume of sphere which means $ \dfrac{{\Delta v}}{v} \times 100 = ? $
And the error in radius is already given to us, that is $ \dfrac{{\Delta r}}{r} \times 100 = 3\% $ . Here we will use the formulae of volume of sphere to solve this question and we know that volume of sphere $ = $ $ \dfrac{4}{3}\pi {r^3} $ , from this formula we can clearly observe that v is directly proportional to r.
volume of sphere $ = $ $ \dfrac{4}{3}\pi {r^3} $ and percentage error $ \delta = \left| {\dfrac{{\left( {{\nu _A} - {\nu _E}} \right)}}{{{\nu _E}}}} \right| \cdot 100\% $
Complete Step By Step Answer:
It is given that $ \dfrac{{\Delta r}}{r} \times 100 = 3\% $ where r is the radius $ (1) $
We have to find percentage error in volume that is $ \dfrac{{\Delta v}}{v} \times 100 = ? $ where v is the volume $ (2) $
We know that volume of the sphere is $ \dfrac{4}{3}\pi {r^3} $ which implies $ v \prec {r^3} $ (v is directly proportional to r)
$ \Rightarrow $ $ \dfrac{{\Delta v}}{v} \times 100 = 3\left( {\dfrac{{\Delta r}}{r}} \right) \times 100\% $
Therefore, by using equation $ (1) $ and $ (2) $ we have the equation
$ \dfrac{{\Delta v}}{v} \times 100 = 3\left( {\dfrac{{\Delta r}}{r}} \right) \times 100\% $ $ = $ $ 3 \times 3\% = 9\% $
Hence, the error in volume of the sphere will be $ 9\% $ .
Note :
In the solution the terms like $ \Delta v $ and $ \Delta r $ are used which simply means change in velocity and change in radius. The formula for percentage error is $ \delta = \left| {\dfrac{{\left( {{\nu _A} - {\nu _E}} \right)}}{{{\nu _E}}}} \right| \cdot 100\% $ where, $ \delta $ is assumed to be percentage error, $ {\nu _A} $ is assumed to be actual value observed and $ {\nu _E} $ is assumed to be expected value.
Here, we have to subtract the actual value and expected value and then divide the error by the exact value which you will get in decimal form, and hence to convert it into percentage we multiply the decimal number by 100.
And the error in radius is already given to us, that is $ \dfrac{{\Delta r}}{r} \times 100 = 3\% $ . Here we will use the formulae of volume of sphere to solve this question and we know that volume of sphere $ = $ $ \dfrac{4}{3}\pi {r^3} $ , from this formula we can clearly observe that v is directly proportional to r.
volume of sphere $ = $ $ \dfrac{4}{3}\pi {r^3} $ and percentage error $ \delta = \left| {\dfrac{{\left( {{\nu _A} - {\nu _E}} \right)}}{{{\nu _E}}}} \right| \cdot 100\% $
Complete Step By Step Answer:
It is given that $ \dfrac{{\Delta r}}{r} \times 100 = 3\% $ where r is the radius $ (1) $
We have to find percentage error in volume that is $ \dfrac{{\Delta v}}{v} \times 100 = ? $ where v is the volume $ (2) $
We know that volume of the sphere is $ \dfrac{4}{3}\pi {r^3} $ which implies $ v \prec {r^3} $ (v is directly proportional to r)
$ \Rightarrow $ $ \dfrac{{\Delta v}}{v} \times 100 = 3\left( {\dfrac{{\Delta r}}{r}} \right) \times 100\% $
Therefore, by using equation $ (1) $ and $ (2) $ we have the equation
$ \dfrac{{\Delta v}}{v} \times 100 = 3\left( {\dfrac{{\Delta r}}{r}} \right) \times 100\% $ $ = $ $ 3 \times 3\% = 9\% $
Hence, the error in volume of the sphere will be $ 9\% $ .
Note :
In the solution the terms like $ \Delta v $ and $ \Delta r $ are used which simply means change in velocity and change in radius. The formula for percentage error is $ \delta = \left| {\dfrac{{\left( {{\nu _A} - {\nu _E}} \right)}}{{{\nu _E}}}} \right| \cdot 100\% $ where, $ \delta $ is assumed to be percentage error, $ {\nu _A} $ is assumed to be actual value observed and $ {\nu _E} $ is assumed to be expected value.
Here, we have to subtract the actual value and expected value and then divide the error by the exact value which you will get in decimal form, and hence to convert it into percentage we multiply the decimal number by 100.
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