Question

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 90.8 cm and diameter 2.75 cm from a storage room to a machinist. Calculate the weight of the rod, w. The acceleration due to gravity, g = 9.81 m/s2 .

Answer 1

Weight of the rod, W = 41.202 N

Step-by-step explanation:

It is given that,

Density of the cylindrical iron rod,
`d=7800\ kg/m^3`

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,
`g=9.81\ m/s^2`

We know that the mass per unit volume is called density of a substance. Its relation is given by :

`d=(m)/(V)`

`m=d* \pi r^2 l`

`m=7800* \pi (0.01375)^2 * 0.908`

m = 4.20 kg

Weight of the rod,

W = mg

`W=4.2\ kg* 9.81\ m/s^2`

W = 41.202 N

So, the weight of the rod is 41.202. Hence, this is the required solution.