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A cylindrical road roller made of iron is 1 m long. Its internal diameter is 54 cm and the thickness of the iron sheet used in making the roller is 9 cm. Find the mass of the roller, if \[1c{{m}^{3}}\] of iron has 7.8 g mass.

Answer Verified Verified
Hint: Calculate the inner radius of sheet and use it to calculate the outer radius of sheet by adding the thickness of sheet to the inner radius of sheet. Use the formula for calculating the volume of a cylinder whose inner and outer radius is given, which is given by \[\pi \left( {{R}^{2}}-{{r}^{2}} \right)h\], where R represents the outer radius, r represents the inner radius of cylinder and h represents the height of cylinder. Use a unitary method to calculate the total mass of the cylinder.

Complete step-by-step solution -
We have a cylindrical roller which is 1m long, has an inner diameter 54cm and thickness of iron sheet used to make a cylindrical roller is 9cm. We have to calculate the mass of the roller given that\[1c{{m}^{3}}\] of iron has 7.8 g mass.
We will calculate the inner and outer radius of the cylinder and then use it to calculate the volume of the cylinder.
The inner radius of the cylinder is 54cm. We know that \[diameter=2\times radius\]. Thus, the inner radius of the cylindrical part \[=r=\dfrac{54}{2}=27cm\].
We will now calculate the outer radius of the cylinder. We know that outer radius of cylinder is the sum of inner radius of cylinder and thickness of iron sheet. Thus, the outer radius of cylinder \[=R=27+9=36cm\].
We will now evaluate the volume of the cylinder. We know that when inner and outer radius is given, volume of cylinder \[\pi \left( {{R}^{2}}-{{r}^{2}} \right)h\], where R represents the outer radius, r represents the inner radius of cylinder and h represents the height of cylinder.
We know that \[1m=100cm\]. Thus, the height of the cylinder \[=h=1m=100cm\].
Substituting the values in the formula for volume of cylinder, we have, the volume of cylinder \[=\pi \left( {{36}^{2}}-{{27}^{2}} \right)100\].
Further simplifying, the volume of cylinder \[=\pi \left( {{36}^{2}}-{{27}^{2}} \right)100=3.14\times 567\times 100=1780.63c{{m}^{3}}\].
We will now find the weight of iron using the volume of the cylinder. We know that \[1c{{m}^{3}}\] weighs 7.8g. So, \[xc{{m}^{3}}\] will weigh \[x\times 7.8\] g.
Thus, \[x=1780.63c{{m}^{3}}\] will weigh \[1780.63\times 7.8=13,888.914\] g.
We can convert grams into kilograms by dividing it by 1000.
Thus, we have \[13,888.914g=\dfrac{13888.914}{1000}kg=13.88kg\].
Hence, the weight of a given cylindrical sheet is 13.88kg.

Note: One needs to be extremely careful about the units of quantities while performing calculations. Also, if inner and outer radius of cylinder are equal, i.e., thickness of the cylinder is negligible, we use the formula \[\pi {{r}^{2}}h\] to calculate the volume of the cylinder, where r represents the radius of cylinder and h represents the height of cylinder.

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