- #1
brotherbobby
- 618
- 152
- Homework Statement
- A plastic "boat" with a ##25\; \text{cm}^2## square cross section floats in a liquid. One by one, you place lead pellets of ##m = 50\; \text{g}## each inside the boat and measure how far the boat extends (##x## cm) below the surface. Your data are as follows: (1) m = 50 g : x = 2.9 cm, (2) m = 100 g : x = 5.0 cm, (3) m = 150 g : x = 6.6 cm and (4) m = 200 g : x = 8.6 cm.
Draw an appropriate graph of the data and, from the slope and intercept of the best-fit line, determine (1) the mass of the boat and the (2) density of the liquid.
- Relevant Equations
- Law of flotation : Mass of the body = Mass of liquid displaced (##m_B = \Delta m_L##)
Using ##m_B = \Delta m_L\Rightarrow m_B = \rho_L A_B\, x(B)## where ##\rho_L## is the density of the liquid, ##A_B## is the cross-sectional area of the boat and ##x(B)## is the depth to which the boat submerge into the liquid due to its own weight without the pellets(##m = 0##). As we can see from the graph, ##x(B) = 0.8\; \text{cm}##, the y-intercept in the graph. See ##x(B)## as "x due to B only".
From above, ##\frac{m_B}{x(B)} = \rho_L A_B##. The graph has an incline of ##26^{\circ}##. Hence, from mathematics, ##\frac{\Delta x}{\Delta m} = \tan 26^{\circ}\Rightarrow \frac{\Delta m}{\Delta x} = \cot 26^{\circ} = 2.05##. Thus, the density of the liquid ##\rho_L = \frac{2.05}{A_B} = \frac{2.05}{25\times 10^-4} = \boxed{820\; \text{kg/m}^3}##.
The mass of the boat ##m_B = \rho_L A_B\, x(B) = 820\times (25\times 10^{-4})\times (8\times 10^{-3}) = \boxed{16.4\; \text{g}} ##.
My answers are way different from those in the text : (1) Mass of the boat = ##\boxed{m_B = 29\; \text{g}}##, (2) Density of the liquid = ##\boxed{\rho_L = 1100\; \text{kg/m}^3}##.
Any hint as to where am going wrong?
(A possible error could be in the incline and hence the slope of the graph. I have obtained conflicting answers using the "ruler" (slope = 20.5)in the software and finding the slope manually (slope = 2.44). Still I am mistaken, the slope from the book is 2.67. Is my logic mistaken? Any help would be welcome).