A gas molecule of mass m at the surface of the earth has kinetic energy equivalent to 0oc.0oc. If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. (kbkb is boltzmann constant)
Answers
Since you refer to "sink" I will assume that they are falling in water. There is no simple answer to this question because the drag force on a sinking object depends on its geometry. For example, suppose one object is a big flat sheet and the other is a long thin needle suspended below the sheet which behaves like a parachute; clearly, this will sink with a much smaller speed than if both objects were long thin needles. The most appropriate approximation of the magnitude of the drag force f in water is a quadratic function of the velocity v, f=½ρwCDAv2, where ρw=103 kg/m3 is the density of water, A is the area presented to the onrushing water as the object falls, and CD is a constant which depends on the geometry of the object, called the drag coefficient. For a sphere of radius R the drag coefficient is given by CD=0.47. The upward force f is only one force on the object as it sinks; it also has a weight down W=mg and a buoyant force up B=Vρwg where V=4πR3/3 is the volume of the sphere. If the sphere is solid and has a density ρ=m/V, we can write B=mg(ρw/ρ). Also note that we can write R as R=3√[3m/(4πρ)]. Finally, we can write the total force on a sphere: F=[mg(ρw-ρ)/ρ]+½ρwCDAv2. Note that when a speed is reached where F=0, the sphere will fall with a constant velocity called the terminal velocity, vt=√{[mg(ρ-ρw)/ρ]/[½ρwCDA]}.
So, let's do a specific example to be able to answer your question for a particular situation. Suppose we have two 1 kg balls of solid iron for which ρ=7.87x103 kg/m3. Then the radius of each ball will be R=0.0312 m. Doing the arithmetic, F=-8.56+0.72v2. So each of two balls of mass 1 kg, tied together will accelerate until their speeds would be vt=√(8.56/0.72)=3.45 m/s. On the other hand, if we have one sphere of mass 2 kg, its radius will be 0.0393 m. So, F=-17.1+1.14v2. So one ball of mass 2 kg will accelerate until vt=√(17.1/1.14)=3.87 m/s.
I hope this will help you
Answer is D
Average translational kinetic energy of molecules=
KE=
2
3
k
B
T=
2
819
k
B
asT=273K
Now. KE= PE
Or,
2
819
k
B
=Mgh
h=
2Mg
819k
B