Answer:
[tex]M_{A}[/tex] : [tex]M_{B}[/tex] = 4 : 1
Explanation:
Given that:
Volume of a cylinder = [tex]\pi[/tex][tex]r^{2}[/tex]h
For cylinder A, [tex]l_{A}[/tex] = [tex]h_{A}[/tex] = [tex]\frac{1}{4} l_{B}[/tex] and [tex]r_{A}[/tex] = 4[tex]r_{B}[/tex].
Volume of cylinder A = [tex]\pi[/tex][tex](4r_{B}) ^{2}[/tex] × [tex]\frac{1}{4} l_{B}[/tex]
= 4[tex]\pi[/tex][tex]r_{B} ^{2}[/tex] [tex]l_{B}[/tex]
Volume of cylinder B = [tex]\pi[/tex][tex](r_{B}) ^{2}[/tex] [tex]l_{B}[/tex]
= [tex]\pi[/tex][tex]r_{B} ^{2}[/tex][tex]l_{B}[/tex]
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 = [tex]\frac{M}{V}[/tex]
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 = [tex]\frac{M_{A} }{4\pi r_{B} ^{2}l_{B} }[/tex]
density of B = [tex]\frac{M_{B} }{\pi r_{B} ^{2}l_{B} }[/tex]
⇒ [tex]\frac{M_{A} }{4\pi r_{B} ^{2}l_{B} }[/tex] = [tex]\frac{M_{B} }{\pi r_{B} ^{2}l_{B} }[/tex]
The ratio of mass of cylinder A to that of B is given as;
[tex]\frac{M_{A} }{M_{B} }[/tex] = [tex]\frac{4\pi r_{B} ^{2} l_{B} }{\pi r_{B} ^{2} l_{B} }[/tex]
⇒ [tex]\frac{M_{A} }{M_{B} }[/tex] = [tex]\frac{4}{1}[/tex]
Therefore, [tex]M_{A}[/tex] : [tex]M_{B}[/tex] = 4 : 1
Answer:
4.
Explanation:
Hello,
In this case, since we are talking about the same material, their densities are the same:
[tex]\rho _A=\rho _B[/tex]
And each density is defined by:
[tex]\rho _A=\frac{m_A}{V_A} \\\\\rho _B=\frac{m_B}{V_B}[/tex]
Thus, we also define the volume of a cylinder:
[tex]V_{cylinder}=\pi r^2h[/tex]
Therefore, we obtain:
[tex]\rho _A=\frac{m_A}{\pi r_A^2h_A}[/tex]
[tex]\rho _B=\frac{m_B}{ \pi r_B^2h_B}[/tex]
Now, the given information regarding the the length and the radius is written mathematically:
[tex]h_A=\frac{1}{4} h_B\\\\r_A=4 r_B[/tex]
So we introduce such additional equations in:
[tex]\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}[/tex]
So we simplify for the radius and lengths:
[tex]\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[/tex]
So the ratio of the mass of cylinder A to the mass of cylinder B is 4.
Best regards.
