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Chapter 7: Problem 1034

A piece of solid weighs \(120 \mathrm{~g}\) in air, \(80 \mathrm{~g}\) in water and \(60 \mathrm{~g}\) in liquid the relative density of the solid and that of the solid and that of the liquid are respectively. (A) 3,2 (B) \(2,(3 / 4)\) (C) \((3 / 4), 2\) (D) \(3,(3 / 2)\)

Short Answer

Expert verified
The relative density of the solid is 3, and the relative density of the liquid is 2. The answer is (A) 3, 2.

Step by step solution

01

Understanding Archimedes' principle

Based on Archimedes' principle, when an object is submerged in a fluid, it experiences a buoyant force equal to the weight of the fluid it displaces. This force acts opposite to the force of gravity and reduces the apparent weight of the object when submerged in the fluid. The buoyant force can be calculated using the formula: \(Buoyant \ force = weight_{object \ in \ fluid} - weight_{object \ in \ air}\).
02

Calculate the buoyant force in water

We will calculate the buoyant force experienced by the solid in water using the given weights: Buoyant force in water = apparent weight of the solid in water - weight of the solid in air Buoyant force in water = \(80 \mathrm{~g} - 120 \mathrm{~g} = -40 \mathrm{~g}\)
03

Calculate relative density of solid

Relative density or specific gravity is the ratio of the density of a substance to that of a reference substance (water in this case). Using the calculated buoyant force in water, we can find the relative density of the solid: Relative density of solid = \(\frac{Weight \ of \ solid \ in \ air}{Weight \ of \ solid \ in \ air - Weight \ of \ solid \ in \ water}\) Relative density of solid = \(\frac{120 \mathrm{~g}}{40 \mathrm{~g}} = 3\)
04

Calculate the buoyant force in liquid

Now we will calculate the buoyant force experienced by the solid in the liquid using the given weights: Buoyant force in liquid = apparent weight of the solid in liquid - weight of the solid in air Buoyant force in liquid = \(60 \mathrm{~g} - 120 \mathrm{~g} = -60 \mathrm{~g}\)
05

Calculate relative density of liquid

We can find the relative density of the liquid in the same way as we found for the solid: Relative density of liquid = \(\frac{Weight \ of \ solid \ in \ air}{Weight \ of \ solid \ in \ air - Weight \ of \ solid \ in \ liquid}\) Relative density of liquid = \(\frac{120 \mathrm{~g}}{60 \mathrm{~g}} = 2\) So the answer is (A) relative density of the solid = 3, and relative density of the liquid = 2.

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Most popular questions from this chapter

Radius of a soap bubble is \(\mathrm{r}^{\prime}\), surface tension of soap solution is \(\mathrm{T}\). Then without increasing the temperature how much energy will be needed to double its radius. (A) \(4 \pi r^{2} T\) (B) \(2 \pi r^{2} T\) (C) \(12 \pi r^{2} T\) (D) \(24 \pi r^{2} T\)

When liquid medicine of density \(\mathrm{S}\) is to be put in the eye. It is done with the help of a dropper as the bulb on the top of the dropper is pressed a drop forms at the opening of the dropper we wish to estimate the size of the drop. We dirst assume that the drop formed at the opening is spherical because the requires a minimum increase in its surface energy. To determine the size we calculate the net vertical force due to surface tension \(\mathrm{T}\) when the radius of the drop is \(\mathrm{R}\). When this force becomes smaller than the weight of the drop the drop gets detached from the dropper. If $\mathrm{r}=5 \times 10^{-4} \mathrm{~m}, \mathrm{p}=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}=10 \mathrm{~ms}^{-2} \mathrm{~T}=0.11 \mathrm{~N} \mathrm{~m}^{-1}$ the radius of the drop when it detaches from the dropper is approximately (A) \(1.4 \times 10^{-3} \mathrm{~m}\) (B) \(3.3 \times 10^{-3} \mathrm{~m}\) (C) \(2.0 \times 10^{-3} \mathrm{~m}\) (D) \(4.1 \times 10^{-3} \mathrm{~m}\)

A material has poisson's ratio \(0.50\). If uniform rod of it suffers longitudinal strain of \(2 \times 10^{-3}\). Then what is percentage change in volume ? (A) \(0.6\) (B) \(0.4\) (C) \(0.2\) (D) 0

Melting point of ice (A) Increases with increasing pressure (B) Decreases with increasing pressure (C) Is independent of pressure (D) is proportional of pressure

Two wires \(A \& \mathrm{~B}\) of same length and of the same material have the respective radius \(\mathrm{r}_{1} \& \mathrm{r}_{2}\) their one end is fixed with a rigid support and at the other end equal twisting couple is applied. Then what will we be the ratio of the angle of twist at the end of \(\mathrm{A}\) and the angle of twist at the end of \(\mathrm{B}\). (A) \(\left(\mathrm{r}_{1}^{2} / \mathrm{r}_{2}^{2}\right)\) (B) \(\left(\mathrm{r}_{2}^{2} / \mathrm{r}_{1}^{2}\right)\) (C) \(\left(\mathrm{r}_{2}^{4} / \mathrm{r}_{1}^{4}\right)\) (D) \(\left(\mathrm{r}_{1}^{4} / \mathrm{r}_{2}^{4}\right)\)
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