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A steel ball of diameter 3 mm falls through glycerine and covers a distance of 25 cm in 10S. The specific gravity of steel and glycerine are 7.8 and 1.26 respectively. The viscosity of glycerine is about $pa^{-s}$
$ \eta = { 2 r^2 \over 9 } { (\rho - \sigma) g \over v_1 } $
A steel ball of diameter 3.2 mm falls under gravity through an oil of density $920 kg m^{-3}$ and viscosity $1.64 Ns m^{-2}$. The density of steel. maybe taken as $7820 kg m^{-3} $. What is the terminal velocity of the ball ?
$ v_t = { 2r^2 \over 9 } { (\rho - \sigma) g \over \eta } $
64 equaldrops of water are falling through air witha steadyvelocity $V_O$, if the drops coalesce, what is their new velocity
$ \upsilon = k r^2 $ When drops coalesce , their volume is constant $R^3 = 64 r^3 or R =4r $ $ { \upsilon \over \upsilon_o } = { R^2 \over r^2 } $
The C.G.S. unit of coefficient of viscosity is poise andthe SI unit is Pa-S. What is the relation between the two ?
The terminal velocity of fall of lead shots in the cylindrical vessel is influenced by the nearness of the walls of the vessel. Hence the velocity in formula is divided by a factor$ \left( 1 + {2.4 r \over R} \right) $ Where r is the radius of the lead shots and R the radius of the vessel. This is called ..............
The formula for the terminal velocity of fall of lead shots in the cylindrical vessel, influenced by the nearness of the walls of the vessel and divided by the factor $$ rac{1}{ig(1 + rac{2.4r}{R}ig)},$$ is known as the Ladnberg correction.
A bubble of air 2 mm in diameter rises in a liquid of viscosity 0.075 SI units and density $1350 kg m^-3$. The terminal velocity of the bubble is about________ m/s
$ \upsilon_t = { 2 \over 9 } \left( r^2 ( \rho - \sigma ) g \over \eta \right) $
A liquid takes 5 minute to cool from $ 80 ^\circ C to 50 ^\circ C$ . The temperature of the surrounding is $ 20 ^\circ C$ . What is the time it will take to cool from $ 60 ^\circ C to 30 ^\circ C$ ?
Using the equation $ { \theta_1 - \theta _2 \over t } = K \left( {\theta_1 + \theta_2 \over 2} - \theta_0 \right) \theta_0$ = where = temperature of surrounding
Two spheres of the same material have radii 1 m and 4 m and temperatures 2000 k and 4000 k respectively. If the energy radiated by the spheres are $E_1 $ and $E_2 $ respectively then find ratio of $ { E_1 \over E_2 } $
$ { E_1 \over E_2 } = \left( { R_1 \over R_2 } \right)^2 \left( { T_1 \over T_4 } \right)^4$
A body cools in 5 minute from $ 60 ^\circ C to 40 ^\circ C $ .The temperature of the surroundings is $ 10 ^\circ C$ . What is its temperature after the next 5 minute ?
According to newton's law approximately, $ { \theta_1 - \theta _2 \over t } = K \left( {\theta_1 + \theta_2 \over 2} - \theta_0 \right) $ $ 160 - 4 \theta = 20 + \theta \Rightarrow \theta = 28 C $
What is the units of emissive power in stefan's law ?
Emissive power is defined as the radiant energy emitted per sec per unit area of the surface. Hence $ [E} = [ P/A] \Rightarrow unit of E = Wm^{-2 } $
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