Elin You have a bucket containing an unknown liquid. You also have a cube-shaped

Question

Elin You have a bucket containing an unknown liquid. You also have a cube-shaped wooden block that you measure to be 8.0 cm on a side, but you don't know the mass or density of the block. To find the density of the liquid, you perform an experiment. First you place the wooden block in the liquid and measure the height of the top of the floating block above the liquid surface. Then you stack various numbers of U.S. quarter-dollar coins onto the block and measure the new value of h. The straight line that gives the best fit to the data you have collected is shown in the figure. The mass of one quarter is 5.670 g. Use this information and the slope and intercept of the straight-line fit to your data. The lowest point on the graph corresponds to 1.2 cnm

Explanation / Answer

Let,

Density of the liquid be d m3,

Mass of the block be m kg,

Number of coins in the wooden block be n,

Height above the water surface be h m.

Given, side of the block is 8 cm = 0.08 meters

Mass of one quarter coin (mc) = 5.67 * 10-3 kg

Therefore, the height of the block that is immersed inside the water = (0.08 - h) meters

Using, Archimedes principle, the weight of the block is equal to the weight of the amount of fluid displaced by the block.

=> (m + n * mc) * g = d * A * h * g

(m + n * 5.67 * 10-3) * g = d * (0.08 * 0.08) * (0.08 - h) * g

(m + n * 5.67 * 10-3) = d * 6.4 * 10-3 * (0.08 - h)

(m + n * 5.67 * 10-3) = (d * 6.4 * 10-3 * 0.08) - (d * 6.4 * 10-3 * h)

(n * 5.67 * 10-3) + (d * 6.4 * 10-3 * h) = (d * 6.4 * 10-3 * 0.08 - m) ..............I

In the above equation, d and m are constants (considering the given graph), our variables are n and h.

Using analogy, x -> n and y -> h, the above equation is in the form of ax+by = c

Slope of the above equation is (-a/b) = - 5.67 * 10-3 / d * 6.4 * 10-3 = - 0.8859375 / d

Slope, from the given graph using points (25, 0.012) and (0, 0.03) = (0.03 - 0.012) / (0 - 25) = - 0.00072

Equating the above slopes, we get

- 0.8859375 / d = - 0.00072

d = 0.8859375 / 0.00072 = 1230.46875 kg/m3

We know that at point (0, 0.03) in equation I, we get

(n * 5.67 * 10-3) + (d * 6.4 * 10-3 * h) = (d * 6.4 * 10-3 * 0.08 - m)

0 + 1230.46875 * 6.4 * 10-3 * 0.03 = 1230.46875 * 6.4 * 10-3 * 0.08 - m

=> m = 1230.46875 * 6.4 * 10-3 * 0.08 - 1230.46875 * 6.4 * 10-3 * 0.03 = 1230.46875 * 6.4 * 10-3 * (0.08 - 0.03)

m = 0.39375 kg,

Therefore, mass of the block in kg = 0.394 kg

Density of the liquid = 1230.47 kg/m3

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