Tag: measurement of density

Questions Related to measurement of density

A block of ice floats on a liquid of density $1.2$ in a beaker then level of liquid when ice completely melt.

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. Remains same

  2. Rises

  3. Lowers

  4. Either (b) or (c)

Show answer
Correct Option: B
Explanation:

Rises.

The level of the liquid in the beaker will rise. This is because the density of the water formed by the melting of ice is less than that of the density of liquid in the beaker 

An orifice of radius $r$ is present at the bottom of a container $r$. The maximum height to which the container can be filled with a liquid of density $\rho$ for which liquid will not come out of the orifice is?

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $\dfrac {T}{r\rho g}$

  2. $\dfrac {3T}{r\rho g}$

  3. $\dfrac {2T}{r\rho g}$

  4. $\dfrac {4T}{r\rho g}$

Show answer
Correct Option: A

A rectangular block 5m x 4m x 2m lies on a table with its largest surface in contact with the table. The work done to keep it so that the block rests on the smallest surface is, if its density is 600 k $m ^ { - 3 }$

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $352800 \mathrm { J }$

  2. zero

  3. $376000 \mathrm { J }$

  4. $24,0000 \mathrm { J }$

Show answer
Correct Option: A
Explanation:

The volume of Rectangular block $ = 5m \times 4m \times 2m$

$ = 40\,\,{m^3}$
Mass = Volume $ \times$ Density
$\begin{array}{l} =40\, \, { m^{ 3 } }\times 600\, \, k{ m^{ -3 } } \ =24,000=2.4\times { 10^{ 4 } }\, \, hg \end{array}$
Potential energy of block when it is lying on its largest surface $= mgh$
$ = 2.4 \times {10^4} \times 9.8 \times 1\,\,joule$
Potential energy of the block when it is lying on 
$ = mgh = 2.4 \times {10^4} \times 9.8 \times 2.5$
Work done = difference in potential energy
$\begin{array}{l} =\left( { 2.4\times { { 10 }^{ 4 } }\times 9.8\times 2.5-2.4\times { { 10 }^{ 4 } }\times 9.8\times 1 } \right) joule \ =2.4\times { 10^{ 4 } }\times 9.8\times 1.5\, \, joule \ =352800\, \, joule \end{array}$
Option A

When two liquid of same volume but different densities $\rho _{1}$ and $\rho _{2}$ are mixed together, then the density of the mixture is

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $\dfrac {p _{1}+p _{2}}{2}$

  2. $p _{1}+p _{2}$

  3. $\dfrac {2p _{1}p _{2}}{p _{1}+p _{2}}$

  4. $2p _{1}+2p _{2}$

Show answer
Correct Option: A
Explanation:

$Density \,\,of \,\,mixture = \dfrac{Total\,\, mass }{ Total \,\,volume}$

Let $1^{st}$ liquid have mass $M _1$, density $p _1$ and Volume $V$

and
$2^{st}$ liquid have mass $M _2$, density $p _2$ and Volume $V$
So,Density of mixture  $= \dfrac{M _1 +M _2}{2V}$$=$$\dfrac {p _1V+p _2V}{2V}$

                                 $\rho _m=\dfrac{p _1+p _2}{2}$

If the length of the cylinder is measured as 25 mm, the diameter is 3.09cm and mass of the cylinder is measured as 50.0 gm find the density of the cylinder in proper significant figures.(in $gm/cm^3$)

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 2.700 $gm/cm^3$

  2. 2.7 $gm/cm^3$

  3. 0.27 $gm/cm^3$

  4. 2.70$gm/cm^3$

Show answer
Correct Option: B
Explanation:

$p\, =\, \displaystyle \frac{m}{\pi (d^2\, /\, 4)h}$
$p\, =\, \displaystyle \frac{(50.0) gm}{3.14\, \times\, (3.09/2)^2\, \times\, (25\, \times\, 10^{-1}) cm^3}\, =\, 2.7\, gm/cm^3$
(in two S.F.)

The greatest length of the wire made of material of breaking stress $ 8 \times 10^8 N/m^2 $ and density $ 8 \times 10^3kg m^3 $ that can be suspended from a rigid support without breaking is $ ( g= 10m/s^2) $ 

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $1 km$

  2. $0.1 km$

  3. $10 km$

  4. $20$

Show answer
Correct Option: C
Explanation:

$ stress= length \times desity \times g $
$ length = \dfrac {stress}{g \times density} $
$= \dfrac { 8 \times 10^8}{10 \times 8 \times 10^3 } $


$=10km$

A tank  $5 { m }$  high is half filled with water and then is filled to the top with oil of density  $0.85 g / c m ^ { 3 } .$  The pressure at the bottom of the tank, due to these liquids is

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $1.85{ g }/{ cm }^{ { 2 } }$

  2. $89.25{ g }/{ cm }^{ { 2 } }$

  3. $462.5{ g }/{ cm }^{ { 2 } }$

  4. $500{ g }/{ cm }^{ { 2 } }$

Show answer
Correct Option: C
Explanation:
Given,

Total height of tank $=5\,m$

Density of water, $\rho _1=1\,g/cm^3$

Density of oil, $\rho _2=0.85\,g/cm^3$

We know that,

Half tank is filled with water and half tank is filled with oil.

So,  $h _1=h _2=\dfrac 52=2.5\,m=250\,cm$

Pressure at the bottom of the tank is

$P=h _1\rho _1g+h _2\rho _2g$

$P=g(250\times 1+250\times 0.85)$

$P=462.5\,g/cm^2$

Suppose that a student finds that $24 \ mL$ of a certain liquid weighs $36 \ g$. What is the density of this liquid in SI unit?

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $1500  \ kg m^{-3}$

  2. $1200  \ kg m^{-3}$

  3. $1550 \  kg m^{-3}$

  4. $1300  \ kg m^{-3}$

Show answer
Correct Option: A
Explanation:

Given, mass of liquid, $m=36g=36\times 10^{-3}kg$;
density of liquid, $\rho=?$
Volume of liquid, $V=24mL=24\times 10^{-3}L$
$=24\times 10^{-3}\times 10^{-3}m^3$
$=24\times 10^{-6}m^3$
Always remember, $1L=10^{-3}m^3=10^3 cm^3$,
Also, $1mL=1cm^3$(also called $1 \ cc$)
Density, $\rho=\displaystyle \frac{Mass}{Volume}=\frac{m}{V}=\frac{36\times 10^{-3}}{24\times 10^{-6}}$
$=1500kg m^{-3}$

Brine has a density of 1.2 g/cc. 40 cc of it is mixed with 30 cc of water. The density of the solution is:

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 2.11 g/cc

  2. 1.11 g/cc

  3. 12.2 g/cc

  4. 20.4 g/cc

Show answer
Correct Option: B
Explanation:

${ \rho  } _{ mixture }=\dfrac { { \rho  } _{ 1 }{ V } _{ 1 }+{ \rho  } _{ 2 }{ V } _{ 2 } }{ { V } _{ 1 }+{ V } _{ 2 } } $


               $=\dfrac { 1.2\times 40+1\times 30 }{ 70 } =\dfrac { 78 }{ 70 } $


${ \rho  } _{ mixture }=1.11gm/cc$

The densities of three liquids are D, 2D and 3D. What will be the density of the resulting mixture if equal volumes of the three liquids are mixed? 

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 6D

  2. 1.4D

  3. 2D

  4. 3D

Show answer
Correct Option: C
Explanation:

let $V$ be the volume of each liquid then the total volume of the mixture becomes $3V$. 

$\rho = \dfrac{total\ mass}{total\ volume}$

Therefore, the mass of the liquids can be written as:
$m _1=D\times V=DV$
$m _1=2D\times V=2DV$
$m _1=3D\times V=3DV$

the total mass of the liquids is
$M= DV+2DV+3DV=6DV$

Therefore, the density of the mixture is:
$\rho=\dfrac{6DV}{3V}$

$\rho = 2D$