Question

Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)

Answer 1

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Complete Question

Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)

A) 1/3 times

B) 3 times

C) 9 times

D) The same amount of aluminum will be required.

Answer:

B) 3 times

Step-by-step explanation:

From the question, we are told that we are to find the amount of aluminum to cover Dewar A and B , we are also told that the base and the lid are not covered with aluminum.

From this information, we can determine that we are to find the curved or Lateral surface area of the cylinder because the base and lid are not included

The curved surface area of a cylinder is calculated as 2πrh

We told that both cylinders have the same height. Hence,

For Dewar A

We have Diameter of 30 cm, radius = Diameter ÷ 2 = 30cm ÷ 2 = 15cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 15 ×h

= 30πhcm²

For Dewar B

We have Diameter of 10 cm, radius = Diameter ÷ 2 = 10cm ÷ 2 = 5cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 5 ×h

= 10πhcm²

When we compare the curved surface Dewar A to the curved surface area of Dewar B

Dewar A : Dewar B

30πhcm² : 10πhcm²

3(10πhcm²) : 10πhcm²

From the above comparison, we can see that 3 times more aluminum is required to cover Dewar A compared to Dewar B.

Answer 2

3 times aluminum will be used to cover Dewar A compared to that required for Dewar B

The given parameters are:

• $D = 30cm$ --- the diameter of Dewar A
• $d = 10cm$ --- the diameter of Dewar B

Start by calculating the radii of both Dewar.

The radius of a shape is half its diameter.

So, we have:

$R =0.5 \times 30cm = 15cm$

$r =0.5 \times 10cm = 5cm$

Because the aluminum does not cover the base and the lid of both Dewar, we simply calculate the curved surface area of both Dewar using:

$C = 2\pi rh$

Where the height (h) is the same for both Dewar

So, we have:

$C_A = 2\pi \times 15h$

$C_A = 30\pi h$

$C_B = 2\pi \times 5h$

$C_B = 10\pi h$

Divide CB by CA

$n = \frac{C_B}{C_A}$

This gives

$n = \frac{30\pi h}{10\pi h}$

$n =3$

Hence, 3 times aluminum will be used to cover Dewar A compared to that required for Dewar B

Read more about surface areas at:

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