Let \(X\) and \(Y\) be the heights of the big and small cylinder respectively.
In figure B height of water in the small cylinder is inadvertently \(X-20\).
So, the Volume occupied by water \(=\pi3^2X+\pi 1^2(20-X)\)
\(\Rightarrow V=20\pi+8\pi X\)
In figure \(C\) height of water in small cylinder is simply \(Y\) and that in the bigger one is\(28-Y\)
In figure \(C\),
\(V=\pi Y +9\pi (28-Y)\)
\(\Rightarrow V=252\pi-8\pi Y\)
Equating the two volume equations, gives us:
\(20\pi+8\pi X=252\pi-8\pi Y\)
\(\Rightarrow X+Y=29 cm\) which is the height of the bottle.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestLet \(X\) and \(Y\) be the heights of the big and small cylinder respectively. In figure B height of water in the small cylinder is inadvertently \(X-20\). So, the Volume occupied by water \(=\pi3^2X+\pi 1^2(20-X)\) \(\Rightarrow V=20\pi+8\pi X\) In figure \(C\) height of water in small cylinder is simply \(Y\) and that in the bigger one is\(28-Y\) In figure \(C\), \(V=\pi Y +9\pi (28-Y)\) \(\Rightarrow V=252\pi-8\pi Y\) Equating the two volume equations, gives us: \(20\pi+8\pi X=252\pi-8\pi Y\) \(\Rightarrow X+Y=29 cm\) which is the height of the bottle.

Log in to reply

(A) 29 cm of water. Using that volume of water is the same in figure b and figure c.

Log in to reply