A cylindrical metal radius 7cm and height 15cm has its centre drilled with a hole of 5cm in cross-section. If 152cm(cube) of d metal weight 125kg, what is the weight of d hallowed cylinder?
How do I solve this please?
figure out the volume of the inner drilled out hole (PI*5^2*15 cm^3)
then mass of metal out of hole
= density*volume
density= 125kg/152 kg/cm^3 Wow that is a dense metal
mass metal out ofhole=density*volumehole
To solve this problem, we need to find the volume and weight of the hollowed cylinder using the given information.
1. Start by finding the volume of the original solid cylinder. The volume of a cylinder is given by the formula:
Volume = π * radius^2 * height
The radius of the original cylinder is given as 7 cm, and the height is given as 15 cm. Substitute these values into the formula to find the volume of the original solid cylinder.
2. Next, calculate the volume of the drilled hole. The hole is cylindrical in shape as well, so we can use the same formula as before to find its volume. The radius of the hole is given as 5 cm. Use this value to find the volume of the drilled hole.
3. Now, subtract the volume of the hole from the volume of the original cylinder. This will give us the volume of the hollowed cylinder. Note that we are assuming that the drilled hole goes straight through the center of the cylinder.
4. The volume of the hollowed cylinder is given as 152 cm^3. Set up an equation using this volume and the volume calculated in the previous step.
5. Now, we need to find the weight of the hollowed cylinder. We are given that 152 cm^3 of the original solid cylinder weighs 125 kg. Divide this weight by the volume of the solid cylinder to find the weight per unit volume (kg/cm^3).
6. Finally, multiply the weight per unit volume by the volume of the hollowed cylinder calculated earlier. This will give you the weight of the hollowed cylinder.
Follow these steps to solve the problem and find the weight of the hollowed cylinder from the given information.