Density of a fluid - class-XI

Newton's law of cooling states that the rate of heat loss is proportional to temperature difference, or time t is proportional to (mc / A). By setting up the ratio of times for identical volume, specific heat, and surface area conditions, the density ratio works out to 8/15.

Density of water at $4^{\circ}C$ in S.I. unit is $1000 kgm^{-3}$.

As 1 gm per cc  = 1000 kg per mc

The time to attain terminal velocity is related to the relaxation time of the sphere in the fluid. Based on the equation of motion for a sphere in a viscous fluid, the characteristic time constant is 2/9 * (rho * r^2 / eta). Given the acceleration condition, the result is 4/9 * (rho * r^2 / eta).

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

$1\ kg/m^3 = 10^3g/10^6cm^3=10^{-3}g/cm^3$

When a liquid cylinder of density rho rotates at angular speed omega, the pressure variation follows the hydrostatic equation dp/dr = rho omega^2 r. Integrating from the center (r = 0) to the radius R gives the pressure increase as Delta P = rho omega^2 R^2 / 2, making option A correct.

Since masses are same,(weight$=mg$) is same for all three.

Using g = G * M / R^2 and M = density * (4/3) * pi * R^3, we get g = G * density * (4/3) * pi * R. Solving for density = 3g / (4 * pi * G * R). Plugging in g=9.8, G=6.67e-11, R=6.4e6, we get approx 5.48e3 kg/m^3.

$1$ $\dfrac{kg}{m^3}  =1\times 1000 $ $g \times \dfrac{1}{10^6 (cm)^3}  = \dfrac{1}{1000}$  $\dfrac{g}{cm^3}$

Density is defined as the mass per unit volume of a substance. Density = mass (kg or g)/volume ($\displaystyle { { m }^{ 3 } }or\quad { { cm }^{ 3 } }$). In MKS system i.e. the SI system the unit of density is ${ kg }/{ { m }^{ 3 } }$ and in CGS system it is ${ g }/{ { cm }^{ 3 } }$. To convert MKS unit to CGS unit the number must be divided by 1000. Therefore, ${ 62.3 }/{ 1000 }=0.0623{ g }/{ { cm }^{ 3 } }$

The coefficient of real expansion gamma_R = (V2 - V1) / (V1 * deltaT). Using the height change as a proxy for volume change: gamma_R = (80.4 - 80) / (80 * 100) = 0.4 / 8000 = 0.00005 = 5e-5 per degree Celsius.

The equilibrium of a floating plank depends on the balance of buoyancy and gravity. For a plank of specific gravity 0.5, the angle theta with the horizontal is determined by the geometry of the submerged portion. The correct angle is pi/6.

The loss in weight in water is 24 - 21 = 3 gm, which is the buoyant force. Since the volume is constant, the buoyant force in another liquid is proportional to its specific gravity. Thus, the loss in the new liquid is 3 * 1.1 = 3.3 gm, and the new weight is 24 - 3.3 = 20.7 gm.

Surface Tension of Water $(T) = 0.00072 N/cm $

A smooth rubber ball has less surface friction than a tennis ball, which has a fuzzy texture. When spun in water, the tennis ball experiences more drag due to its surface texture, making the smooth ball move more easily.

$\begin{array}{l} \dfrac { { { V^{ 2 } } } }{ 2 } +gh+\dfrac { p }{ d } ={ { constant } } \ d=density \ R=pressure \ \dfrac { { V _{ t }^{ 2 } } }{ 2 } +\dfrac { { { P _{ t } } } }{ d } =\dfrac { { V _{ B }^{ 2 } } }{ 2 } +\frac { { { P _{ B } } } }{ d }  \ { P _{ B } }-{ P _{ t } }=\dfrac { d }{ 2 } \left( { V _{ t }^{ 2 }-V _{ B }^{ 2 } } \right) =1018, , N/{ M^{ 2 } } \ where, , { V _{ t } }+{ V _{ B } }=speeds, , over, , top, , and, , bottom \ K.\in =\frac { 1 }{ 2 } \times 100\times { \left( { 11150 } \right) ^{ 2 } }m/s \ =6.2\times { 10^{ 9 } }, , joules \ \in scope, , velocity=11.15, , KM/s=11150, , m/s \end{array}$

If pressure increases, density generally increases in response. The pressure reduces the volume taken up while the mass remains the same. That follows from the definition of density as mass per unit volume. This effect varies greatly between materials however, gases compress readily, such that density increases nearly in proportion to pressure, liquids and solids much less so.

density of aluminium = $2.7g/{ cm }^{ 3 }$

Molar mass of mixture $=\cfrac{total mass}{total mole}$

Density rho = mass / volume. The relative error in density is the sum of the relative errors in mass and volume: (delta_rho / rho) = (delta_m / m) + (delta_V / V). Here, (0.05 / 5.00) + (0.05 / 1.00) = 0.01 + 0.05 = 0.06, which is 6%.

Explanation: In MKS system i.e. the SI system the unit of density is $\displaystyle { kg }/{ { m }^{ 3 } }$ and in CGS system it is $\displaystyle { g }/{ { cm }^{ 3 } }$. Density = mass/volume = $\displaystyle 20/2.5=8{ g }/{ { cm }^{ 3 } }$ in CGS system. In MKS system density = $\displaystyle 8\times 1000=8000{ kg }/{ { m }^{ 3 } }$

Let volume of gold in ornament $V _1$ and that of copper $=V _2$ and density of gold $=\rho g$ and that of copper$=\rho _c$