O-Level Practice Question 2: A helium balloon has a volume of 621 cm3 at ground level where the atmospheric pressure is 101,000 Pa. The balloon rises and expands to 636 cm3. At this point, what is the height of the balloon above ground ? Take the density of air to be 1.25 kg/m3 and constant over the height h.
In doing this problem, realize that the pressure inside the balloon must be equal to the pressure outside (atmospheric pressure at a specific height).
The first step is to find the pressure at height h. Although not explicitly stated, we have to assume that temperature is a constant, which means we can use the following equation,
(P1)(V1) = (P2)(V2)
101,000 Pa (621 cm3) = P2 (636 cm3)
P2 = 98,618 Pa
The difference in pressure between the top and the bottom will be equal to (density)gh. This is the same concept behind saying the pressure under a column of liquid is (density)gh. Go deeper under a lake and pressure increases due to the height of the column of liquid above. Go deeper in the atmosphere and you have the same effect.
101,000 Pa – 98,618 Pa = 2382 Pa = (1.25 kg/m3) (10 m/s2) h
h = 191 m
O-Level Practice Question 1: A rocket with a mass of 1,000 kg takes off from the surface of another planet using a thrust of 30,000 N. An acceleration of 1.5 m/s2 is achieved.
A) What is the resultant force on the rocket?
(1000 kg) (1.5 m/s2)= 1500 N
B) What is the weight of the rocket?
Resultant Force = Thrust – Weight
1500 N = 30,000 N – W
W = 28,500 N
C) What is the acceleration of gravity on the other planet?
Weight = ma
28,500 N = (1000 kg) a
a = 28.5 m/s2