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The weight of a hollow sphere is directly dependent on its [#permalink]

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12 Nov 2012, 17:16

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The weight of a hollow sphere is directly dependent on its surface area. The surface area of a sphere is 4π·R^2, where R is the radius of the sphere. If a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams, a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams?

If the large sphere is L and the small one is S, then the surface area of S = 4π·0.15^2, and the surface area of L = 4π·0.3^2. Thus the surface area of L is larger than that of S by a factor of 0.3^2/0.15^2 = (0.3/0.15^)2 = 2^2 = 4.

Could someone explain to me the reasoning behind putting the .3 over the .15 in the proportion part?

I think I understand the reason why they throw away the 4π, is it because they are both equal therefore not necessary in solving the problem? (because we are dealing with proportion in weight)

Re: The weight of a hollow sphere is directly dependent on its s [#permalink]

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12 Nov 2012, 17:29

This is how I worked it out. Since the question states that Weight is dependent (proportional) to SA, therefore, weight = k (constant) X SA, then find k, use the same formula and you will obtain the answer.

The weight of a hollow sphere is directly dependent on its surface area. The surface area of a sphere is 4π·R^2, where R is the radius of the sphere. If a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams, a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams?

If the large sphere is L and the small one is S, then the surface area of S = 4π·0.15^2, and the surface area of L = 4π·0.3^2. Thus the surface area of L is larger than that of S by a factor of 0.3^2/0.15^2 = (0.3/0.15^)2 = 2^2 = 4.

Could someone explain to me the reasoning behind putting the .3 over the .15 in the proportion part?

I think I understand the reason why they throw away the 4π, is it because they are both equal therefore not necessary in solving the problem? (because we are dealing with proportion in weight)

The surface area of a sphere of radius 0.15 cm is \(4\pi{r^2}=4\pi{*0.15^2}\); The surface area of a sphere of radius 0.3=2*0.15 cm is \(4\pi{R^2}=4\pi{*2^2*0.15^2}=4*(4\pi{*0.15^2})\).

Since the surface are of the bigger sphere is 4 times the surface area of the smaller sphere and since the weight of a sphere is directly proportional to surface area, then the weight of the bigger sphere is 4 times the weight of the smaller sphere, so 4*8=32 grams.

Re: The weight of a hollow sphere is directly dependent on its [#permalink]

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13 Nov 2012, 12:55

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When its directly proportional, it's all depends on the power of the variable that we are manipulating. if A = pi r^2 then area increases by the square of the multiple by which you increase the radius. if r becomes 2r then the area becomes 4A and if r becomes 1/2 r then area becomes 1/4A. Same thing can be applied if you are manipulating multiple variables.

For e.g. SA = 2 pi r^2 h r has powe of two and h has power of 1 if r increase to 2r and h increases to 2h, SA will increases by 2^2 * 2^1 = 8 times.

For e.g. V = 4/3 pi r^3 now the volume increase by the power of 3 of the mulitplier of r. if r becomes 2r volume goes up by 2^3 = 8 times.

Re: The weight of a hollow sphere is directly dependent on its [#permalink]

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30 Sep 2013, 05:51

anon1 wrote:

The weight of a hollow sphere is directly dependent on its surface area. The surface area of a sphere is 4π·R^2, where R is the radius of the sphere. If a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams, a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams?

If the large sphere is L and the small one is S, then the surface area of S = 4π·0.15^2, and the surface area of L = 4π·0.3^2. Thus the surface area of L is larger than that of S by a factor of 0.3^2/0.15^2 = (0.3/0.15^)2 = 2^2 = 4.

Could someone explain to me the reasoning behind putting the .3 over the .15 in the proportion part?

I think I understand the reason why they throw away the 4π, is it because they are both equal therefore not necessary in solving the problem? (because we are dealing with proportion in weight)

Thinking simply: Weight is propotional to Surface Area .....(1) info given

Weight1 = (some constant) x Surface Area 1 (4π·R1^2)....(2)

Weight2 = (some constant) x Surface Area 2 (4π·R2^2)....(3)

Dividing equation (3) by equation (2) we get

Weight2 /8 gms = (0.30)^2 / (.15)^2

Weight2 = 8 x 4 = 32 gms _________________

Thanks & Regards Yodee

‘A good plan violently executed now is better than a perfect plan executed next week.’ - General Georg S. Patton

Re: The weight of a hollow sphere is directly dependent on its [#permalink]

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04 Aug 2015, 12:47

anon1 wrote:

The weight of a hollow sphere is directly dependent on its surface area. The surface area of a sphere is 4π·R^2, where R is the radius of the sphere. If a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams, a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams?

If the large sphere is L and the small one is S, then the surface area of S = 4π·0.15^2, and the surface area of L = 4π·0.3^2. Thus the surface area of L is larger than that of S by a factor of 0.3^2/0.15^2 = (0.3/0.15^)2 = 2^2 = 4.

Could someone explain to me the reasoning behind putting the .3 over the .15 in the proportion part?

I think I understand the reason why they throw away the 4π, is it because they are both equal therefore not necessary in solving the problem? (because we are dealing with proportion in weight)

Plug in easy values. Instead of 0.15 use 1 for the small radius and instead of 0.3 use 2 for the small radius. This makes the calculation much easier and it is still correct because the proportions are the same.

So 4pi * 1 = 4pi, and 4pi*4 = 16pi. Therefore the weight of the large sphere is 4 times more than the small one. _________________

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Re: The weight of a hollow sphere is directly dependent on its
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04 Aug 2015, 12:47

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