What is the change in weight of a hollow cylinder of height 8 cm and radius 3 cm when the air is pumped out of it
1 answer:
This question involves the concepts of density, weight, and volume.
The change in weight of the cylinder will be "2.72 x 10⁻³ N".
The change in weight of the cylinder will be equal to the weight of the air inside the cylinder which is being removed from it:
where,
= change in weight of cylinder = ?
= Weight of air inside cylinder =
= density of air = 1.225 kg/m³
- V = Volume of air inside the cylinder = πr²h = π(0.03 m)²(0.08 m) = 2.26 x 10⁻⁴ m³
- g = acceleration due to gravity = 9.81 m/s²
Therefore,
<u>ΔW = 2.72 x 10⁻³ N</u>
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Answer:
option C
Explanation:
given,
diameter of circular room = 8 m
rotational velocity of the rider = 45 rev/min
=
=4.712 rad/s
here in this case normal force is equal to centripetal force
N = m r ω²
N = m x 4 x 4.712²
N = 88.83m
frictional force = μ N
= 88.83m x μ
now, for the body to not to slide
gravity force is equal to frictional force
m g = 88.83 m x μ
g = 88.83 x μ
9.8 = 88.83 x μ
μ = 0.11
hence, the correct answer is option C