Refer to Problem 16 and Figure P 14.16 . A hydrometer is to...

Refer to Problem 16 and Figure $\mathrm{P} 14.16 .$ A hydrometer is to be constructed with a cylindrical floating rod. Nine fiduciary marks are to be placed along the rod to indicate densities of $0.98 \mathrm{~g} / \mathrm{cm}^{3}, 1.00 \mathrm{~g} / \mathrm{cm}^{3}, 1.02 \mathrm{~g} / \mathrm{cm}^{3}, 1.04 \mathrm{~g} / \mathrm{cm}^{3}, \ldots$ $1.14 \mathrm{~g} / \mathrm{cm}^{3} .$ The row of marks is to start $0.200 \mathrm{~cm}$ from the top end of the rod and end $1.80 \mathrm{~cm}$ from the top end. (a) What is the required length of the rod? (b) What must be its average density? (c) Should the marks be equally spaced? Explain your answer.

Refer to Problem 16 and Figure $\mathrm{P} 14.16 .$ A hydrometer is to be constructed with a cylindrical floating rod. Nine fiduciary marks are to be placed along the rod to indicate densities of $0.98 \mathrm{~g} / \mathrm{cm}^{3}, 1.00 \mathrm{~g} / \mathrm{cm}^{3}, 1.02 \mathrm{~g} / \mathrm{cm}^{3}, 1.04 \mathrm{~g} / \mathrm{cm}^{3}, \ldots$
$1.14 \mathrm{~g} / \mathrm{cm}^{3} .$ The row of marks is to start $0.200 \mathrm{~cm}$ from the top end of the rod and end $1.80 \mathrm{~cm}$ from the top end.
(a) What is the required length of the rod? (b) What must be its average density? (c) Should the marks be equally spaced? Explain your answer.

Key Concepts

Calibration of Hydrometer Scales

The process of calibration involves marking specific reference points on the hydrometer that correspond to predetermined density values. This ensures that as the hydrometer floats differently in fluids of varying densities, the scale accurately reflects those changes. Considerations such as the spacing of marks often require adjustments because the relationship between displacement and density is not strictly linear with simple geometry.

Density Measurement

Density is defined as mass per unit volume, and knowing the density of a fluid can be determined by observing the depth to which a calibrated instrument sinks. In hydrometry, the level at which the device floats correlates with the fluid’s density, making precise calculations and adjustments essential to obtain accurate measurements.

Buoyancy and Floating Equilibrium

This concept refers to the balance between the weight of a floating object and the buoyant force exerted by the displaced fluid. Achieving equilibrium is critical in the design of instruments such as hydrometers, where the submerged volume must adjust so that the object's average density matches the density of the fluid in which it floats.

Archimedes’ Principle

This principle states that a body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced. It is fundamental for understanding how and why objects float and is the basis for designing devices like hydrometers that measure fluid density by floating at different levels in fluids of different densities.

Transcript

00:01 Hi there, so for this problem, we have the situation that is shown here in this figure.

00:08 So the first question for this is what is the required length of the rot? so what we are going to use is the result from the previous exercise, the problem 14 .30, and yes, 14 .16.

00:29 16.

00:31 So from there we know that the density, for the rod floating in the liquid density, the density of this liquid is equal to the initial density.

00:42 This times the length l, this divided by the length l minus the height.

00:47 Now with that set, what we need to do is to substitute the values that we are given for this.

00:54 So the density is 0 .98.

00:56 Then this is equal to.

01:04 The initial density times the length l, this divided by l the length l minus the height that is 0 .2.

01:24 So solving here we will have 0 .90l and 98 times l, this minus 0 .98 times 0 .2 so we will obtain a value of 0 .196.

01:46 And then this equals to the density rows of zero times the length l.

01:56 Now, for the floating in the dense liquid, we have something similar, but in this case, with a density of 1 .14.

02:05 So we will have 1 .14 times l, this minus, in this case, the floating height is 1 .8.

02:16 Now that 1 .8 multiplies with 1 .14, we obtain a value of 2 .052.

02:25 And then this is equal to the density l times l.

02:30 So as you can see, we can just equal these two expressions.

02:36 So we'll have 0 .98 times l.

02:39 This minus 0 .196 is equal to 1 .14 times l minus 2 .052...

Please give Ace some feedback