A duster is of dimension 5 × 5 × 7 all in cm. the duster is sinked into a cylindrical vessel of radius 5cm and height 20cm. After insertion, the level of water increased by 2.14cm. Then by what percentage the part of the duster is above the layer?
Answer
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Hint: Now the duster is sinked in a cylindrical vessel and the level of water rises. Now the volume of water risen is the volume of duster sinked. We know that volume of the cylinder is given by $\pi {{r}^{2}}h$ and the volume of the cuboid is given by $l\times b\times h$. Hence we can compare the volume of water risen and the volume of duster to find out the percentage of duster that is not submerged.
Complete step-by-step solution:
Now first consider the duster.
The dimensions of the duster is given as $5\times 5\times 7$.
Hence the volume of duster is $5\times 5\times 7 = 175$ sqcm ………………….. (1)
Now consider the cylinder.
Now we are given that the radius of the cylinder is 5cm. and after the duster is submerged the volume is risen by 2.14cm. Hence the height is 2.14cm.
Now we know that the volume of the cylinder is given by $\pi {{r}^{2}}h$
Hence the volume of water risen is given by
$\pi {{(5)}^{2}}(2.14)=25\pi (2.14)=53.5\pi $
Now the volume of water is nothing but the volume of duster submerged.
Hence the volume of duster submerged is $53.5\pi $sqcm ……………………….. (2)
Now the part of duster that is not submerged is $175-53.5\pi $
The percent of volume of duster that is not submerged is
$\begin{align}
& \dfrac{175-53.5\pi }{175}\times 100 \\
& =\dfrac{175-53.5(3.14)}{175}\times 100 \\
& =\dfrac{175-167.99}{175}\times 100 \\
& =\dfrac{175-168}{175}\times 100 \\
& =\dfrac{7}{175}\times 100 \\
& =\dfrac{700}{175} \\
& =4 \\
\end{align}$
Hence we get $4\%$ of duster is not submerged.
Note: Note that while calculating the percent of the water that is not submerged we will have to first find the volume of duster that is not submerged. Since we have found the volume of water that is submerged we will subtract it from the total volume to find the volume that is not submerged.
Complete step-by-step solution:
Now first consider the duster.
The dimensions of the duster is given as $5\times 5\times 7$.
Hence the volume of duster is $5\times 5\times 7 = 175$ sqcm ………………….. (1)
Now consider the cylinder.
Now we are given that the radius of the cylinder is 5cm. and after the duster is submerged the volume is risen by 2.14cm. Hence the height is 2.14cm.
Now we know that the volume of the cylinder is given by $\pi {{r}^{2}}h$
Hence the volume of water risen is given by
$\pi {{(5)}^{2}}(2.14)=25\pi (2.14)=53.5\pi $
Now the volume of water is nothing but the volume of duster submerged.
Hence the volume of duster submerged is $53.5\pi $sqcm ……………………….. (2)
Now the part of duster that is not submerged is $175-53.5\pi $
The percent of volume of duster that is not submerged is
$\begin{align}
& \dfrac{175-53.5\pi }{175}\times 100 \\
& =\dfrac{175-53.5(3.14)}{175}\times 100 \\
& =\dfrac{175-167.99}{175}\times 100 \\
& =\dfrac{175-168}{175}\times 100 \\
& =\dfrac{7}{175}\times 100 \\
& =\dfrac{700}{175} \\
& =4 \\
\end{align}$
Hence we get $4\%$ of duster is not submerged.
Note: Note that while calculating the percent of the water that is not submerged we will have to first find the volume of duster that is not submerged. Since we have found the volume of water that is submerged we will subtract it from the total volume to find the volume that is not submerged.
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