Question

Two cylinders are made of the same material. Cylinder A is one-fourth (1/4) the length of cylinder B and it has a radius that is four times greater than the radius of cylinder B. What is the ratio of the mass of cylinder A to the mass of cylinder B?

Answer 1

Answer:

$M_{A}$ : $M_{B}$  = 4 : 1

Explanation:

Given that:

Volume of a cylinder = $\pi$$r^{2}$h

For cylinder A, $l_{A}$ = $h_{A}$ = $\frac{1}{4} l_{B}$ and $r_{A}$ = 4$r_{B}$.

Volume of cylinder A = $\pi$$(4r_{B}) ^{2}$ × $\frac{1}{4} l_{B}$

                                   = 4$\pi$$r_{B} ^{2}$ $l_{B}$

Volume of cylinder B =  $\pi$$(r_{B}) ^{2}$ $l_{B}$

                                  =  $\pi$$r_{B} ^{2}$$l_{B}$

To determine the ratio of their masses, density (ρ) is defined as the ratio of the mass (M) of a substance to its volume (V).

i.e    ρ = $\frac{M}{V}$

Thus, since the cylinders are made from the same material, they have the same density  (ρ). So that;

density of A = density of B

density of A = $\frac{M_{A} }{4\pi r_{B} ^{2}l_{B} }$

density of B = $\frac{M_{B} }{\pi r_{B} ^{2}l_{B} }$

⇒            $\frac{M_{A} }{4\pi r_{B} ^{2}l_{B} }$  =  $\frac{M_{B} }{\pi r_{B} ^{2}l_{B} }$

The ratio of mass of cylinder A to that of B is given as;

                       $\frac{M_{A} }{M_{B} }$  =  $\frac{4\pi r_{B} ^{2} l_{B} }{\pi r_{B} ^{2} l_{B} }$

⇒                        $\frac{M_{A} }{M_{B} }$ = $\frac{4}{1}$

Therefore, $M_{A}$ : $M_{B}$  = 4 : 1

Answer 2

Answer:

4.

Explanation:

Hello,

In this case, since we are talking about the same material, their densities are the same:

$\rho _A=\rho _B$

And each density is defined by:

$\rho _A=\frac{m_A}{V_A} \\\\\rho _B=\frac{m_B}{V_B}$

Thus, we also define the volume of a cylinder:

$V_{cylinder}=\pi r^2h$

Therefore, we obtain:

$\rho _A=\frac{m_A}{\pi r_A^2h_A}$

$\rho _B=\frac{m_B}{ \pi r_B^2h_B}$

Now, the given information regarding the the length and the radius is written mathematically:

$h_A=\frac{1}{4} h_B\\\\r_A=4 r_B$

So we introduce such additional equations in:

$\frac{m_A}{\pi r_A^2h_A}=\frac{m_B}{\pi r_B^2h_B}\\\\\frac{m_A}{\pi (4r_B)^2(\frac{1}{4}h_B)}=\frac{m_B}{\pi r_B^2h_B}\\\\\frac{m_A}{m_B} =\frac{\pi (4r_B)^2(\frac{1}{4}h_B)}{\pi r_B^2h_B}$

So we simplify for the radius and lengths:

$\frac{m_A}{m_B} =\frac{\pi (4r_B)^2(\frac{1}{4}h_B)}{\pi r_B^2h_B}\\\\\frac{m_A}{m_B} =16 *\frac{1}{4}\\ \\\frac{m_A}{m_B} =4$

So the ratio of the mass of cylinder A to the mass of cylinder B is 4.

Best regards.