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Chapter 7: Problem 1031

The fraction of floating object of volume \(\mathrm{V}_{0}\) and density \(\mathrm{d}_{0}\) above the surface of a Liquid as density \(\mathrm{d}\) will be (A) \(\left(\mathrm{d}_{0} / \mathrm{d}\right)\) (B) $\left[\left\\{\mathrm{dd}_{0}\right\\} /\left\\{\mathrm{d}+\mathrm{d}_{0}\right\\}\right]$ (C) \(\left[\left\\{d-d_{0}\right\\} / d\right]\) (D) $\left[\left\\{\mathrm{dd}_{0}\right\\} /\left\\{\mathrm{d}-\mathrm{d}_{0}\right\\}\right]$

Short Answer

Expert verified
The short answer is: \(\left[\left\\{d-d_{0}\right\\} / d\right]\).

Step by step solution

01

Find the weight of the object

We can find the weight of the object (W_object) by multiplying its volume (V0) by its density (d0) and gravitational acceleration (g). So, \(W_\text{object} = V_0 d_0 g \).
02

Find the weight of the liquid displaced

The weight of the liquid displaced (W_liquid) can be found by the volume of the object submerged in the liquid (V_submerged) times the density of the liquid (d) and gravitational acceleration (g). Since we want the fraction of the object above the surface, we find the fraction submerged f_submerged, which is equal to the object's density (d0) divided by the liquid's density (d) i.e., \(f_\text{submerged} = \dfrac{d_0}{d}\). Now, the volume submerged (V_submerged) is the fraction submerged (f_submerged) times the total volume (V0) i.e., \(V_\text{submerged} = V_0 f_\text{submerged}\). So, \(W_\text{liquid} = V_\text{submerged} d g = V_0 f_\text{submerged} d g\).
03

Find the fraction of the object above the surface

According to Archimedes' principle, the weight of the liquid displaced (W_liquid) is equal to the weight of the object (W_object). So, \(V_0 d_0 g = V_0 f_\text{submerged} d g\). We can cancel out V0 and g from this equation, leaving us with \(d_0 = f_\text{submerged} d\). Now, we can solve for the fraction above the surface (f_above) by noting that it's equal to 1 minus the fraction submerged (f_submerged), so \(f_\text{above} = 1 - \dfrac{d_0}{d}\). Comparing the expression for the fraction above the surface with the given options, we find that the correct answer is: (C) \(\left[\left\\{d-d_{0}\right\\} / d\right]\)

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Most popular questions from this chapter

The upper end of a wire of radius \(4 \mathrm{~mm}\) and length $100 \mathrm{~cm}$ is clamped and its other end is twisted through an angle of \(30^{\circ}\). Then what is the angle of shear? (A) \(12^{\circ}\) (B) \(0.12^{\circ}\) (C) \(1.2^{\circ}\) (D) \(0.012^{\circ}\)

The excess of pressure inside a soap bubble than that of the outer pressure is (A) \((2 \mathrm{~T} / \mathrm{r})\) (B) \((4 \mathrm{~T} / \mathrm{r})\) (C) \((\mathrm{T} / 2 \mathrm{r})\) (D) \((\mathrm{T} / \mathrm{r})\)

Read the assertion and reason carefully and mark the correct option given below. (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If the assertion and reason both are false. Assertion: The water rises higher in a capillary tube of small diametre than in the capillary tube of large diameter. Reason: Height through which liquid rises in a capillary tube is inversely proportional to the diameter of the capillary tube. (A) a (B) b (C) c (D) d

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When more than \(20 \mathrm{~kg}\) mass is tied to the end of wire it breaks what is maximum mass that can be tied to the end of a wire of same material with half the radius? (A) \(20 \mathrm{~kg}\) (B) \(5 \mathrm{~kg}\) (C) \(80 \mathrm{~kg}\) (D) \(160 \mathrm{~kg}\)
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