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This is second of our many quizzes on Density. Click on this to access more such competitive quizzes on Mechanics.

Density Quiz 2

physics

During winter, the surface of the Pangong Lake sees sheets of ice floating on top, while water flows under it. This happens because the relative density of ice with respect to water is ___________ than 1. If a ship is taken through the water of the lake and the anchor is dropped somewhere in between, the anchor sinks while the ship floats because the average density of the ship and the anchor is ______ than water and  _______ than water respectively, since the hull of the ship is usually hollow while the anchor is solid throughout. If the ship experiences a buoyant force of $F_B$ against gravity $g$ and the ship displaces a volume $V$ of water as it floats in the lake, the density of the water of the lake is given as ___________.  

Determine the average density by considering the amount of air in the ship and the lack of it in the anchor. Use Archimedes’ principle to find the lake water density.  

  $\text{ The ice floats on water because the density of ice}\,\rho_{ice}\,\text{ is less than that of water}\,\rho_{water}.\\ \text{ In such a case, the relative density of ice:}\\ \rho_{r} = \dfrac{\rho_{ice}}{\rho_{water}}\\ \text{ But we have}\,\rho_{ice} < \rho_{water},\\ \Rightarrow \rho_{r} < 1.\\ \text{ In the context of the question:}\\ \text{ “This happens because the relative density of ice with respect to water is less than 1”.}\\ \text{ “…the anchor sinks while the ship floats because the average density of the ship and}\\ \text{ the anchor is less than water and more than water respectively, …”.}\\ \text{ From Archimedes’ principle, we know that the buoyant force acting on a body will be}\\ \text{ equivalent to the weight of the fluid displaced by the body, i.e.}\\ F_{B} = W_{water} = m_{water}g = \rho_{water}Vg\\ \Rightarrow \rho_{water} = \dfrac{F_B}{Vg}\\ \text{ In the context of the question:}\\ \text{“…the density of the water of the lake is given as}\,\, \dfrac{F_B}{Vg}.”$

A typical mercury thermometer contains $0.7\,g$ of mercury. The height of the mercury is $5\,cm$ at the triple point of water and at the steam point of water, the height of the mercury is $6.84\,cm$, occupying a tube of length $0.1\,cm$ and width $0.1\,cm$. Determine the average density of mercury.  

First find the volume of mercury and its density in each case and then calculate the mean of the two densities. 

  $ \text{ We have the mass of mercury as: m = 0.7 g.}\\ \text{ The volume of mercury at triple point will be:}\\ V_1 = 5 \times 0.1 \times 0.1 \\ \Rightarrow V_1  = 0.05\,cm^3\\ \text{ The density of mercury at this point will be:}\\ \rho_1 = \dfrac{m}{V_1} \\ \Rightarrow\rho_1= \dfrac{0.7}{0.05} \\<br>\Rightarrow\rho_1= 14gcm^{-3}\\ \text{ The volume of mercury at steam point will be:}\\ V_2 = 6.84 \times 0.1 \times 0.1 \\ \Rightarrow V_2 = 0.0684\,cm^{3}\\ \text{ The density of mercury at this point will be:}\\ \rho_2 = \dfrac{m}{V_2} \\ \Rightarrow\rho_2 = \dfrac{0.7}{0.0684} \\ \Rightarrow\rho_2 = 10.23\,gcm^{-3}\\ \text{ The average density of mercury will thus be:}\\ \rho_{avg} = \dfrac{\rho_1 + \rho_2}{2} \\ \Rightarrow \rho_{avg}= \dfrac{14+ 10.23}{2} \\ \therefore\rho_{avg}= 12.11\,gcm^{-3}$

Rocky beaches protect the beach soil from erosion. This is why rocky structures are sometimes artificially placed on shores to preserve the sand. These are called groynes. Groynes are usually made from basalt rocks which have a typical density of $3.2\,gcm^{-3}$. Suppose this rock needs to weigh at least three tons in order to stay in place and not be moved around by tides, what would be the approximate dimensions of such a rock? $(\text{Take 1 ton = 900 kg})$ 

 Find the volume of the rock from the given mass and density and see which of the options agrees with the calculated volume.  

 $ \text{ We are given that the density of a basalt rock is,}\\ \rho = 3.2\,gcm^{-3} = 3.2 \times 10^3\,kgm^{-3}\\ \text{ The mass of this rock will be,}\\ 3tons = 3 \times 900 = 2700\,kg\\ \text{ The volume of such a rock can be found to be:}\\ V = \dfrac{m}{\rho} \\ \Rightarrow V= \dfrac{2700}{3.2 \times 10^3} \\ \Rightarrow V= 843.75 \times 10^{-3}  \\ \Rightarrow V= 0.84\,m^3\\ \text{ Of all the options given to us,}\\ \text{ the option that comes closest to our above-calculated volume is:}\\ (d) 3 \times 1 \times 0.3 = 0.9\,m^3\\  \text{ Therefore, the dimensions of our rock would be:}\\ 3\,m \times 1\,m \times 0.3\,m$

 Fill in the blank: Moissanite is a naturally occurring form of silicon carbide and is often used to make faux diamonds. A worker at a moissanite quarry uncovers a rock that seems to be dispersing light a lot lesser than usual, so he hands it over to a geologist who takes it back to his lab for examination. He measures the mass of the rock to be $7\,kg$ with dimensions of $20\,cm \times 10\,cm \times 10\,cm$.He finds that the density of the rock is ________$\,gcm^{-3}$, which is different from the density of moissanite, which is usually around $3\,gcm^{-3}$, indicating that the rock that the worker unearthed, was in fact, a diamond.  

First find the volume of the rock from its dimensions, following which the density can be calculated. Maintain consistency of units.  

 $\text{ We have the mass of the rock to be m = 7 kg = 7000 g.}\\ \text{ The volume of the rock will be:}\\ V = 20 \times 10 \times 10 = 2000\,cm^{3}\\ \text{ Therefore, the density of the rock will be:}\\ \rho=\dfrac{m}{V} \\ \Rightarrow\rho= \dfrac{7000}{2000} \\ \Rightarrow\rho= 3.5\,gcm^{-3}\\ \text{ In the context our question, we get:}\\ \text{“He finds that the density of the rock is $3.5\,gcm^{-3}$, which is different from the density of moissanite,…”.}$

The following figure shows the cross-section of an earth-like planet. If the mass of the inner core is $10^{23}\,kg$, the outer core is $1.8 \times 10^{24}\,kg$ and that of the mantle is $4 \times 10^{24}\,kg$, find the average density of the planet.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ec80dce4-eb49-4aed-8485-bc670f6eefb04117229037196228982.png"/> 

First find the density of each layer assuming that they are spherically layered, following which the mean of all layer densities will yield the appropriate result.  

 $ \text{ For the inner core:}\\ \text{ Mass:}\, m_1 = 10^{23}kg\\ \text{ Volume:}\, V_1= \dfrac{4}{3}\pi (1216\times 10^3)^3\\ \text{ Density:}\, \rho_1 = \dfrac{m_1}{V_1}\\ \Rightarrow\rho_1= \dfrac{10^{23}}{\left(\dfrac{4}{3}\pi (1216\times 10^3)^3\right)} \\ \Rightarrow\rho_1= 1.33\times 10^4kgm^{-3}\\ \text{ For the outer core:}\\ \text{ Mass:}\,m_2 = 1.8\times 10^{24}\,kg\\ \text{ Volume:}\,V_2= \dfrac{4}{3}\pi (2270\times 10^3)^3\\ \text{ Density:}\, \rho_2 = \dfrac{m_2}{V_2}\\ \Rightarrow\rho_2= \dfrac{1.8 \times 10^{24}}{\left(\dfrac{4}{3}\pi (2270\times 10^3)^3\right)} \\ \Rightarrow\rho_2= 3.68 \times 10^4\,kgm^{-3}\\ \text{ For the mantle(s):}\\ \text{ Mass:}\,m_3 = 4\times 10^{24}kg.\\ \text{ Volume:}\, V_3= \dfrac{4}{3}\pi (2885\times 10^3)^3\\ \text{ Density:}\,\rho_3 = \dfrac{m_3}{V_3} \\ \Rightarrow\rho_3 = \dfrac{4\times 10^{24}}{\left(\dfrac{4}{3}\pi (2885\times 10^3)^3\right)} \\ \Rightarrow\rho_3= 3.98 \times 10^4\,kgm^{-3}\\ \text{ Therefore, the average density of the earth-like planet will be:}\\ \rho_{avg} = \dfrac{\rho_1+\rho_2+\rho_3}{3} \\ \Rightarrow\rho_{avg}  = \dfrac{(1.33 + 3.68 + 3.98)\times 10^{4}}{3} \\ \Rightarrow\rho_{avg}  = 2.99 \times 10^{4}kgm^{-3} \\ \Rightarrow\rho_{avg}  = 2.99 \times 10^{4} \times 10^{-3} \\ \therefore\rho_{avg}  = 29.9\,gcm^{-3}$