Answer:
Weight of the rod, W = 41.202 N
Explanation:
It is given that,
Density of the cylindrical iron rod, [tex]d=7800\ kg/m^3[/tex]
Length of the cylindrical iron rod, l = 90.8 cm = 0.908 m
Diameter of the rod, d = 2.75 cm = 0.0275 m
Radius, r = 0.01375 m
Acceleration due to gravity, [tex]g=9.81\ m/s^2[/tex]
We know that the mass per unit volume is called density of a substance. Its relation is given by :
[tex]d=\dfrac{m}{V}[/tex]
[tex]m=d\times \pi r^2 l[/tex]
[tex]m=7800\times \pi (0.01375)^2 \times 0.908[/tex]
m = 4.20 kg
Weight of the rod,
W = mg
[tex]W=4.2\ kg\times 9.81\ m/s^2[/tex]
W = 41.202 N
So, the weight of the rod is 41.202. Hence, this is the required solution.
