
A soft drink is available in two packs- (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10cm. Which container has greater capacity and by how much?
a). Cylinder has $85c{{m}^{3}}$ greater capacity.
b). Tin has $85c{{m}^{3}}$ greater capacity.
c). Cylinder has $65c{{m}^{3}}$ greater capacity.
d). Tin has $65c{{m}^{3}}$ greater capacity.
Answer
611.7k+ views
Hint: For solving this problem, first calculate the volume of cuboidal can by using the formula of volume of cuboid and volume of cylindrical can by using the formula of volume of cylinder. Once we obtain the numerical data, we can easily predict which volume is greater and by how much.
Complete step-by-step solution -
According to the problem statement we are given two packs. The first pack is a tin can in the shape of cuboid having the length 5 cm, width 4 cm and height 15 cm. The second pack is in the form of a plastic cylinder having circular base diameter 7 cm and height 10cm.
Volume of a cuboid is given as $l\times b\times h\ldots (1)$.
Volume of a cylinder is given as $\pi {{r}^{2}}h\ldots (2)$.
In equation (1), putting length as 5 cm, width as 4 cm and height is 15 cm, we get
$\begin{align}
& {{V}_{1}}=5\times 4\times 15 \\
& {{V}_{1}}=300c{{m}^{3}} \\
\end{align}$
In equation (2), putting base radius as 3.5 cm and height as 10 cm, we get
$\begin{align}
& {{V}_{2}}=\pi \times \dfrac{7}{2}\times \dfrac{7}{2}\times 10 \\
& {{V}_{2}}=\dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}\times 10 \\
& {{V}_{2}}=385c{{m}^{3}} \\
\end{align}$
Clearly, ${{V}_{2}} > {{V}_{1}}$ which means the volume of the cylinder is greater than cuboid.
Difference in the volume = volume of cylinder - volume of cuboid
$385-300=85c{{m}^{3}}$
So, the cylinder has $85c{{m}^{3}}$ greater capacity.
Hence, option (a) is correct.
Note: The key steps involved in solving this problem is the knowledge of volume of cuboid and cylinder. Students must assign clear denotation to each volume so that he must not get confused while comparing both the volumes.
Complete step-by-step solution -
According to the problem statement we are given two packs. The first pack is a tin can in the shape of cuboid having the length 5 cm, width 4 cm and height 15 cm. The second pack is in the form of a plastic cylinder having circular base diameter 7 cm and height 10cm.
Volume of a cuboid is given as $l\times b\times h\ldots (1)$.
Volume of a cylinder is given as $\pi {{r}^{2}}h\ldots (2)$.
In equation (1), putting length as 5 cm, width as 4 cm and height is 15 cm, we get
$\begin{align}
& {{V}_{1}}=5\times 4\times 15 \\
& {{V}_{1}}=300c{{m}^{3}} \\
\end{align}$
In equation (2), putting base radius as 3.5 cm and height as 10 cm, we get
$\begin{align}
& {{V}_{2}}=\pi \times \dfrac{7}{2}\times \dfrac{7}{2}\times 10 \\
& {{V}_{2}}=\dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}\times 10 \\
& {{V}_{2}}=385c{{m}^{3}} \\
\end{align}$
Clearly, ${{V}_{2}} > {{V}_{1}}$ which means the volume of the cylinder is greater than cuboid.
Difference in the volume = volume of cylinder - volume of cuboid
$385-300=85c{{m}^{3}}$
So, the cylinder has $85c{{m}^{3}}$ greater capacity.
Hence, option (a) is correct.
Note: The key steps involved in solving this problem is the knowledge of volume of cuboid and cylinder. Students must assign clear denotation to each volume so that he must not get confused while comparing both the volumes.
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