The implicit assumption in the question is that the density of sugar is invariant. Whether that is a physical fact I do not know. I am embarrassingly ignorant of physical chemistry. But if it is a fact (or a reasonanle approximation, then mass and volume are in a linear relationship.The mass of sugar the drums can hold are their capacities? How does their capacities affect their volume?
View attachment 35631
(Capacity is really just another word for volume; they are often used interchangeably.)The mass of sugar the drums can hold are their capacities? How does their capacities affect their volume?
To put it a little differently, for any given substance (under given conditions, so that the density is constant), mass or weight is proportional to volume, just as, for similar shapes, volume is proportional to the cube of linear dimensions.The mass of sugar the drums can hold are their capacities? How does their capacities affect their volume?
Noted.To put it a little differently, for any given substance (under given conditions, so that the density is constant), mass or weight is proportional to volume, just as, for similar shapes, volume is proportional to the cube of linear dimensions.
So this problem is about proportionality twice. That's all it is.
So here I am. Ratio of height of small drum to big drum = [math]\frac{2}{3}[/math]Because I am dealing with volumes, I will have to cube the ratio and increase 800 gram in that ratio:So please now show us your working to prove that "C" is the correct answer (before this thread runs onto several more pages too!).
That's correct.\(\displaystyle \frac{3^3}{2^3}×800=2700\) and it becomes visible that the big drum holds 2700 grams sugar. ✔
Yes. I never took my time to look at that instruction too. [math]1 kg= 1000~ g[/math] So, [math]2700~g=\frac{2700~g}{1000~g}=2.7~kg[/math] Though, I wondered why the 2700 g I obtained was included in the option hiddenly -(don't mind the English as long as the message is passed across). I can now clearly see the solution as 2.7 kg. Thanks for that eye opener.(Although it did say the answer should be in kg, ie 2.7; just a minor adjustment required there ?)
Yes!I trust you are getting the hang of these now and should be able to tackle any future problems like these confidently yourself. ?
Please see here.Try the classic approach and let us see your final answer!
The classic we go.Try the classic approach and let us see your final answer!
@chijioke,The classic we go.
H= height of big cylinder
h=height of small cylinder
d=diameter of small cylinder
D=diameter of big cylinder
V=volume of big cylinder
v=volume of small cylinder
R=radius of big cylinder
r=radius of small cylinder
[math]\frac{h}{H}=\frac{d}{D}[/math][math]v=\pi{r}^2h=\pi{\left(\frac{d}{2}\right)}^2h[/math]We are given the height and volume of small cylinder, so we can find its diameter.
[math]d=\sqrt{\frac{4v}{\pi h}}[/math]Back to the classic formula
[math]\frac{h}{H}=\frac{ \sqrt{ \frac{4v}{\pi h} } }{D}[/math]Now we can find the diameter of the big cylinder.
[math]D=\frac{ H\sqrt{\frac{4v}{\pi h} } }{h}[/math][math]= \frac{ 3\sqrt{\frac{4×800}{2\pi} } }{2}=33.85137501 ~m[/math]We can now find the volume of bigger cylinder.
[math]V=\pi R^2H=\pi{\left(\frac{D}{2}\right)}^2H[/math] [math]= \pi{\left(\frac{33.85137501}{2}\right)}^2×3[/math] [math]\approx 2700g[/math]Since the solution is required in kg. 2700g = 2.7 kg