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3188078493's Question
Physics M
Posted 4 months ago

U1-2022-01-15

1 A hollow brass cylinder with closed ends is floating on the surface of water.
The eylinder has a length of 4.0 cm and an external diameter of 2.1 cm as shown.
(ii) Deduce whether an identical hollow cylinder made of gold would also float.
density of gold =19.3×103 kg m−3=19.3 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}
density of brass =8.7×103kgm−1=8.7 \times 10^{3} \mathrm{kgm}^{-1}
63%63 \% of the volume of the cylinder is submerged. The cylinder contains negligible weight of air
(a) Explain why the brass cylinder floats.
(2)
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(b) The density of water is 1.0×109 kg m−11.0 \times 10^{9} \mathrm{~kg} \mathrm{~m}^{-1}
(i) Show that the mass of the cylinder is about 9×10−1 kg9 \times 10^{-1} \mathrm{~kg}.
V=(0.0212)2π×0.04=1.385×10−5 m3M=PV=1.0×103×63%×1.385×00−1=8.7255×10−3
\begin{array}{l}
V=\left(\frac{0.021}{2}\right)^{2} \pi \times 0.04=1.385 \times 10^{-5} \mathrm{~m}^{3} \\
M=P V=1.0 \times 10^{3} \times 63 \% \times 1.385 \times 00^{-1}=8.7255 \times 10^{-3}
\end{array}

m=ρV=19.3×103×1.385×10−5=0.267 kgm=ρV=8.7×103×1.385×10−5
\begin{array}{l}
m=\rho V=19.3 \times 10^{3} \times 1.385 \times 10^{-5}=0.267 \mathrm{~kg} \\
m=\rho V=8.7 \times 10^{3} \times 1.385 \times 10^{-5}
\end{array}

m=PV=8.7×103×1−385×10−5=0.12 kg
m=P V=8.7 \times 10^{3} \times 1-385 \times 10^{-5}=0.12 \mathrm{~kg}

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Answer from Sia
Posted 4 months ago
Solution
a
The brass cylinder floats because the buoyant force acting on it is equal to the weight of the water displaced. According to Archimedes' principle, the buoyant force FbF_b can be calculated as: Fb=ρwaterVdisplacedgF_b = \rho_{water} V_{displaced} g where ρwater=1.0×103 kg/m3\rho_{water} = 1.0 \times 10^{3} \, \text{kg/m}^3, Vdisplaced=0.63×VcylinderV_{displaced} = 0.63 \times V_{cylinder}, and g≈9.81 m/s2g \approx 9.81 \, \text{m/s}^2. The weight of the cylinder must be equal to the buoyant force for it to float
b
To find the mass of the cylinder, we first calculate its volume using the formula for the volume of a cylinder: V=πr2hV = \pi r^2 h. The external radius rr is 2.1 cm2=0.0105 m\frac{2.1 \, \text{cm}}{2} = 0.0105 \, \text{m} and the height hh is 0.04 m0.04 \, \text{m}. Thus, V=π(0.0105)2(0.04)≈1.385×10−5 m3V = \pi (0.0105)^2 (0.04) \approx 1.385 \times 10^{-5} \, \text{m}^3. The mass can then be calculated using the density of brass: m=ρbrassV=8.7×103×1.385×10−5≈0.120 kgm = \rho_{brass} V = 8.7 \times 10^{3} \times 1.385 \times 10^{-5} \approx 0.120 \, \text{kg}
c
To determine if an identical hollow cylinder made of gold would float, we compare the densities. The density of gold is 19.3×103 kg/m319.3 \times 10^{3} \, \text{kg/m}^3, which is greater than that of brass. Since the weight of the gold cylinder would exceed the buoyant force acting on it, it would not float
Answer
The brass cylinder floats because the buoyant force equals its weight, while the gold cylinder would sink due to its higher density.
Key Concept
Archimedes' Principle: The buoyant force on an object is equal to the weight of the fluid displaced by the object. Equation: Fb=ρVgF_b = \rho V g (Buoyant force equals density times volume times acceleration due to gravity).
Explanation
The brass cylinder floats because the buoyant force equals its weight, while the gold cylinder would sink due to its higher density.

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