We are given a solid cylinder with diameter $1.2$ m and height $4$ m. Its mass is $29$ kg. We need to find the density of the object in kg/m$^3$, rounded to the nearest hundredth.

1. Problem Description

We are given a solid cylinder with diameter $1.2$ m and height $4$ m. Its mass is $29$ kg. We need to find the density of the object in kg/m$^3$, rounded to the nearest hundredth.

2. Solution Steps

First, we need to find the radius of the cylinder. The diameter is $1.2$ m, so the radius $r$ is half of that:

$r = \frac{1.2}{2} = 0.6$ m

Next, we need to find the volume $V$ of the cylinder. The formula for the volume of a cylinder is:

$V = \pi r^2 h$

where $r$ is the radius and $h$ is the height.

Substituting the values $r=0.6$ m and $h=4$ m, we get:

$V = \pi (0.6)^2 (4)$

$V = \pi (0.36)(4)$

$V = 1.44\pi$ m$^3$

Now, we can find the density $\rho$ using the formula:

$\rho = \frac{m}{V}$

where $m$ is the mass and $V$ is the volume.

The mass is given as $m = 29$ kg. Therefore:

$\rho = \frac{29}{1.44\pi}$

Using a calculator with the $\pi$ button, we calculate:

$\rho = \frac{29}{1.44\pi} \approx \frac{29}{4.523893421} \approx 6.410452$

Rounding to the nearest hundredth, we get:

$\rho \approx 6.41$ kg/m$^3$

3. Final Answer

Density = 6.41 kg/m$^3$