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(a) 112 m

(b) 110 m

(c) 114 m

(d) 116 m

Answer

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We have iron cuboidal of dimensions \[4.4m\times 2.6m\times 1m.\]

As we all know that the volume remains fixed, so we can find the volume of the cuboidal block and compare it with the volume of the cylinder. We have the dimensions as \[4.4m\times 2.6m\times 1m\]

Therefore, the volume of the cuboid is

\[V=L\times B\times H\]

\[\Rightarrow V=4.4\times 2.6\times 1\]

\[\Rightarrow V=11.44{{m}^{3}}\]

So, we get the volume of the cuboidal block as \[11.44{{m}^{3}}.\]

Now, in the question, we are given that the cylinder is hollow inside. So,

The volume of Cylinder = Volume of Outer Cylindrical part – Volume of Inner Cylindrical part

We have inner radius, r = 30 cm = 0.3 m and thickness, d = 5 cm = 0.05 m.

So, Outer Radius, R = Inner Radius + Thickness

\[R=r+d\]

\[\Rightarrow R=0.3+0.05\]

\[\Rightarrow R=0.35m\]

Now, we get the volume of the cylinder as

\[\text{Volume of cylinder}=\pi {{R}^{2}}H-\pi {{r}^{2}}H\]

Here, the height H is the same for the outer and inner parts.

\[\Rightarrow \text{Volume of cylinder}=\dfrac{22}{7}\times {{\left( 0.35 \right)}^{2}}\times H-\dfrac{22}{7}{{\left( 0.3 \right)}^{2}}\times H\]

\[\Rightarrow \text{Volume of cylinder}=\dfrac{22}{7}\left[ {{\left( 0.35 \right)}^{2}}-{{\left( 0.3 \right)}^{2}} \right]H\]

Now, to find the height, we will compare the volume of the hollow cylinder with the volume of the cuboidal box. So, we get,

\[11.44=\dfrac{22}{7}\left[ {{\left( 0.35 \right)}^{2}}-{{\left( 0.3 \right)}^{2}} \right]H\]

\[\Rightarrow 11.44=\dfrac{22}{7}\left[ 0.0325 \right]H\]

Now, we will solve for H, we get,

\[H=\dfrac{11.44\times 7}{22\times 0.0325}\]

\[\Rightarrow H=\dfrac{80.08}{0.715}\]

\[\Rightarrow H=112m\]

Therefore, we get the height of the cylinder as 112m.

\[{{0.35}^{2}}-{{0.3}^{2}}=\left( 0.35+0.3 \right)\left( 0.35-0.3 \right)\]

\[\Rightarrow {{0.35}^{2}}-{{0.3}^{2}}=0.65\times 0.005\]

\[\Rightarrow {{0.35}^{2}}-{{0.3}^{2}}=0.0325\]