Question

a silver cylindrical rod has a length od .5m and radius of .4m.find the density of the rod if its mass is 2640kg

Answer 1

v of cylinder=πr^2h

=3.14×(4^2)×5

=251.33 m3

density=mass/volume

=2640/251.33

=10.5

Answer 2

Answer:

The density of silver cylindrical rod is equal to 10.51 Kg/m³

Explanation:

Given, the length of the rod = 5m

The radius of the rod, r = 4m

We can consider  the rod as cylinder. So the length of the rod can be treated as height of the cylinder.

We know the volume of the cylinder, $V = \pi r^2 h$

where h = 5m and r = 4m,

Substitute the values in the formula of volume.

$V = (3.14)(4)^2(5)$

$V =251.2 \;m^3$

Given, the mass of rod = 2640 Kg

Now we know that density of the rod can be calculated as:

Density of rod = mass/ volume

$Density =\frac{2640 \; Kg}{251.2\;m^3}$

Density of the rod = 10.51 Kg/m³

Therefore, the density of rod is equal to 10.51 Kg/m³.

To know more references links:-

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