Measurement of density - class-VIII

On stretching, volume remains constant. Hence on decreasing the radius by half, the length of the wire becomes quadrupled. 
Mass per unit length $\displaystyle = \frac{20}{4 \times 50 cm}$
$=0.1  g  cm^{-1}$

 Density of water = 1 g $ cm^{-3 }$ = 1000 kg $ m^{-3}$

Density of kerosene $= \displaystyle \frac{mass}{volume}$
$\displaystyle = \frac{5  kg}{5  litres}= \frac{5kg}{5 \times 10^{-3}m^3} = 1000 kg  m^{-3}$

$\displaystyle \frac{d _1}{d _2} = \frac{1}{3} ; \frac{m _1}{m _2} = \frac{3}{4}$
$d = m/v$
$\therefore \displaystyle \frac{V _1}{V _2}  = \frac{m _1}{m _2} \times \frac{d _2}{d _1} = \frac{3}{4} \times \frac{3}{1} = 9:4$

The density of air at NTP is $1.293 kg m^{-3}$.

Given:

density of water in S.I. unit= ${10}^3$ kg/${m}^3$

Density of Brine$=\rho _{b}=1.2g/cc$

The Volume of the room is given as $V = L\times B\times H= 5\times 3\times 4 = 60  { m }^{ 3 }.$
We know, Mass $=$ Density$\times$Volume.
Given that the density $=1.3  kg/m^3$.
So, Mass of the air $=1.3\times60  Kg$.
Hence, mass of the air enclosed in the room is 78 kg.

Let mass of $A$ be $m$

Answer is A.

As we know, Density $d = \dfrac{Mass, M}{ Volume, V}$

the density of water in S.I. unit $\rho=1000 kg/{m}^3$

It is a fact that the density of water is 1000 kg / ${m}^3$

 As on cooling volume of air decreases and its mass remains constant so its density increases as  Density = $\dfrac{mass}{volume}$

The hot air balloon has the hot air filled in it and the density of air is more than the hot air. So, the normal air or atmospheric air has the tendency to remain below than the hot air as the objects having the more density remains below than the objects having less density.

Due to increase in temperature, objects vibrational energy increases resulting in decrease in density of the substance.

 Density = $\dfrac {mass}{volume}$ 

If volume of iron = Volume of aluminium

density of a substance is the characterstic property of material defined as mass per unit volume of the substance hence in CGS system its unit is gm/cc.

Density of a substance is a characterstic property of a substance defined as mass of the substance per unit volume.

The mass of the water and kerosene depends on the volume and the density of the each. Since the water and kerosene have different densities, they will have different mass for the same volumes.

Density of a liquid is a characterstic property of a substance & it is independent of any other external factors like volume etc.

Given,    density = $1.04gm/{ cm }^{ 3 }$

Since we know that density of iron is more than density of sugar hence 

Density = $\dfrac{mass}{volume}$  so $\rho$ = $\dfrac{m}{v}$  not $\dfrac{mg}{v}$ as weight = mg

Density of substance is mass per unit volume hence its unit are $kg/{ m }^{ 3 }$ in S/units & $gm/{ cm }^{ 3 }$ in CGS.

Density of substance is a characterstic property of a substance defined as mass of substance per unit volume.

Density is a physical property of a substance and is defined as the mass per unit volume of a substance. The density of a body is the amount of mass contained per unit volume of the body.

Relative density of a substance is ratio of density of substance to that of water. Hence density of substance can be found by multiplying relative density with 1000$kg/{ m }^{ 3 }$ (density of water).

The mass of the substance is dependent on the density and volume of the substance. The substance having the same volume might have different mass because they have different densities.

The fundamental property of a substance is the property which depends on the body and does not change due to any physical factor. Mass is a fundamental property of a substance. Another such property is the charge.

Density of a substance is defined as the mass per unit volume. Its SI unit is $kgm^{-3}$. For example, the density of water at room temperature and atmospheric pressure is $1\ kgm^{-3}$.

$Answer: C$

Density, mass, volume are the characterstic property of a substance whereas Pressure is an extensive property which depends on area of application of force.

Density of the material is defined as mass per unit volume. Since, volume is temperature dependent, increasing the temperature of the material (with few exceptions) decreases the density of material by increasing its volume. 

Density of the substance is defined as the ratio of mass of substance to its volume.

We know that, density of silver is $10.5 \times 10^3 \dfrac{kg}{m^3}$ and density of water is $1.00 \times 10^3 \dfrac{kg}{m^3}$.

Let the mass of the substance be $m$ and its volume be $V$. Then the density of the substance is defined as the ratio of mass to the volume of the body.

SI unit of density $=\dfrac { kg }{ { m }^{ 3 } } =\dfrac { 1000 }{ 10^6 } \dfrac { g }{ cc } $

A vessel contains oil density $=0.8gm/cm^3$

Power = Pressure × (Volume/Time). Pressure = ρgh = (13.6×10³ kg/m³)(10 m/s²)(0.150 m) = 20,400 Pa. Flow rate = 5 L/min = 5×10⁻³ m³/60 s = 8.33×10⁻⁵ m³/s. Power = (20,400)(8.33×10⁻⁵) = 1.70 W. Option D is correct.

Since copper and iron have different density, hence both balls will have different volumes for same mass.

If passenger drinks water then the weight of boat get increase and the boat will displace the same amount of water in the tank resulting no net change in the water level.

It is very well known that mass per unit volume is defined as density. 

Density of water is $1000kg/m^3$.

Density of oil=0.80 g/cc=800 kg/m$^{3}$

Volume of ice= Mass/Density= $M/P$

Given Pressure in lungs${P} _{lungs}=750$ mm of Hg

Maximum density of water is at $4^0C$

The true mass M_true is found by equating the forces: M_true * g - V * d * g = M * g - V * d2 * g, where V is volume. Substituting V = M_true / d1 leads to the correct buoyant force balance equation.

Since at both times, container is same . which means that volume remains constant. By applying ideal gas equation.

For the ball to be suspended: electric force = weight - buoyant force. qE = ρ_ball V g - ρ_oil V g. q = (ρ_ball - ρ_oil)Vg/E. Volume of sphere = (4/3)πr³ = (4/3)π(0.5 cm)³. After unit conversions and calculation, q ≈ 1.1 μC. Option A is correct.

When block is immersed in liquid, tension decreases by buoyant force. T' = T - ρ_liquid V g. Fundamental frequency f ∝ √T. Beats occur when frequencies differ from tuning fork. For equal beats, the tension ratio gives: (1 - ρ_liquid/20)/1 = 1/4. Solving gives ρ_liquid = 3.8 g/cm³. Option A is correct.

$m = 20.00g$

$V _s=5cc$ As given

The $C.G.S.$ unit of kg is Gm$.$ and that of meter is cm$.$ When we convert $13600,kg{m^{ - 3}}$ then $13600 \times 1000\left( {100 \times 100 \times 100} \right) = 13.6,gc{m^{ - 3}}$

$According, to, question..................... \ q=\int  _{ 0 }^{ R }{ pdv }=\int  _{ 0 }^{ R }{ po }\left( { 1-\dfrac { r }{ R }  } \right) , ., 4\pi { r^{ 2 } }drr \ \Rightarrow q=po.4\pi , \left[ { \int  _{ 0 }^{ R }{ { r^{ 2 } }dr-\int  _{ 0 }^{ R }{ \dfrac { { { r^{ 3 } } } }{ R } dr } } } \right]  \ \Rightarrow q=po.4\pi \left( { \dfrac { { { R^{ 3 } } } }{ 3 } -\dfrac { { { R^{ 3 } } } }{ 4 }  } \right) =po\left( { 4\pi  } \right) \dfrac { { { R^{ 3 } } } }{ { 12 } }  \ , , , , , , \therefore , , , q=\dfrac { { po.\pi { R^{ 3 } } } }{ 3 }  \ E=\dfrac { 1 }{ { 4\pi { \in _{ 0 } } } } \times , \dfrac { q }{ { { r^{ 2 } } } } =\dfrac { 1 }{ { 4\pi { \in _{ 0 } } } } , \times \dfrac { { po, .\pi { R^{ 3 } } } }{ { 3{ r^{ 2 } } } } =\dfrac { { po, .{ R^{ 3 } } } }{ { 12{ \in _{ 0 } }{ r^{ 2 } } } }  \ , , , \therefore , , , , E=\dfrac { { po, .{ R^{ 3 } } } }{ { 12{ \in _{ 0 } }{ r^{ 2 } } } }  \ there, for, the, correct, option, is, B.$

For floating body: weight = buoyant force. Ah₁ρ_body g = Ah₂ρ_liquid g, where h₁ is total height and h₂ is submerged height. Given ρ_liquid = 3ρ_body, we get h₁ = 3h₂. Exposed height = h₁ - h₂ = 2h₂ = 2/3 h₁. Option A is correct.

Breaking stress = Force / Area = (Mass * g) / Area = (Volume * density * g) / Area. Since Volume = Area * height, stress = density * g * height. Solving for density: 10^6 = density * 10 * 5, so density = 2 * 10^4 kg/m^3.

Rises.

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

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

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

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

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, $\rho=\displaystyle \frac{Mass}{Volume}=\frac{m}{V}=\frac{36\times 10^{-3}}{24\times 10^{-6}}$
$=1500kg m^{-3}$

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

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

Their exists less no. of molecules at the high temperature and high no. of molecules at the low temperature i.e. in the cold conditions.

Ventilators are provided in the rooms at the top of the roofs because if the air inside the room gets hot, the hot air rises up and flows through these ventilators and thus cool air remains at bottom. It brings cool and fresh air in the room. Thus ventilators maintain conventional currents to keep the air fresh in the room.

Let V is the volume of the vessel.

Now, let $p _{1}$ and $p _{2}$ be the partial pressure, then using gas law: 

$p _{1}V = \dfrac{m _1}{M _1}RT\\$

$p _{2}V = \dfrac{m _2}{M _2}RT,\ p _{0}  = p _{1} +  p _{2}\\$

$p _{0} = \left(\dfrac{m _1}{M _1} + \dfrac{m _2}{M _2}\right)\dfrac{RT}{V}\\$

$V = \left(\dfrac{m _1}{M _1} + \dfrac{m _2}{M _2}\right)\dfrac{RT}{p _{0}}\\$

$\because \rho _{mix}=\dfrac{(m _{1} + m _{2})}{V}\\$

$rho _{mix}=\dfrac {(m _1 + m _2)M _1 M _2} {(m _1M _2 + m _2M _1)} \times \dfrac{p _0}{RT}\\$

Substituting values,

$\rho _{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$

$= 1.446 \ per \ litre$

Option A is correct.

$\begin{array}{l} d=density, , of, , cylin{ { de } }r \ A=area, , of, , cross-sectional, , of, , cylinder \ U\sin  g, , law, , of, , floation, \ weight, , of, , cylinder=up, thrust, , by, , two, , liquids \ L\times A\times d\times g \ =n\rho \times \left( { pL\times A } \right) g+\rho \left( { L-pL } \right) Ag \ d=np\rho +\rho \left( { 1-p } \right) =\left( { np+1-p } \right) \rho  \ d=\left{ { 1+\left( { n-1 } \right) p } \right} \rho  \end{array}$

The density of a body is given by:

$\begin{array}{l} dM=s\left( { 4\pi { x^{ 2 } }dx } \right)  \ ={ \rho _{ 0 } }\left( { 1+\dfrac { x }{ R }  } \right) \left( { 4\pi { x^{ 2 } }dx } \right)  \ dI=\dfrac { 2 }{ 3 } dM{ x^{ 2 } } \ \dfrac { 2 }{ 3 } { \rho _{ o } }\int _{ 0 }^{ R }{ \left( { 4\pi { x^{ 4 } }dx+\dfrac { { 4\pi  } }{ R } { x^{ 5 } }dx } \right)  }  \ =\dfrac { { 8\pi { \rho _{ 0 } } } }{ 3 } \left[ { \dfrac { { { x^{ 5 } } } }{ 5 } +\dfrac { { { R^{ 5 } } } }{ 6 }  } \right]  \ =\dfrac { 8 }{ 3 } \pi { \rho _{ o } }\times \dfrac { { 11{ R^{ 5 } } } }{ { 30 } }  \ =\dfrac { { 44 } }{ { 45 } } \pi { \rho _{ 0 } }{ R^{ 5 } } \end{array}$

Volume of wind flowing per second $= Av$
Mass of wind flowing per second $= Av\rho$
Mass of air passing in time $t\ s = Av\rho t$.

Suppose distance between the parallel plates is $D$ and applied voltage $V _{(t)} = V _02\pi vt$.