I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.
I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:
Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>
</span>
</span>
</span>
<span>
Distance at Perihelion
(</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>
Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.
</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>
1/2 (50%) of that is 43.9845 Earth days
The average of the aphelion and perihelion distances is
1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU
This also happens to be 1/2 of the major axis of the elliptical orbit.
To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as
Where,
P = Tensile Force
L= Length
A = Cross sectional Area
E = Young's modulus
PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then
Therefore the elongaton of the rod in a 200mm gage length is 
PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,
Where,
Poission's ratio
= Lateral strain
= Linear strain
Therefore the change in diameter of the rod is 
Use direct proportionality: If 1.25 m = 72 N, then 0.55 m = y
0.55 m × 72 = 39.6.
39.6÷1,25 gives you your answer which is 31.68 N.
<span>Volume of air in the balloon 1.01 x 10^6 L
Density of air is 1.20 g/l
Mass = Density X Volume
So mass of the air in the Balloon= ( 1.01 x 10^6) X 1.20 = 1.212 x 10^6 g
As the air is heated, the volume of air in the balloon expands to 1.10x 10^6 L
Density= Mass/ voume
So the Density of heated air = 1.212 x 10^6/ 1.10x 10^6 = 1.101 g/l
The answer is 1.101 g/l.</span>
