Question

What is the weight of a bowling ball with a 5 in. radius if we know that one cubic inch weighs 1/100th of a pound?

Answer 1

Answer:

The weight of a bowling ball is 5.24 pounds.

Option (B) is correct.

Step-by-step explanation:

Formula

$Volume\ of\ a\ sphere = \frac{4}{3}\pi\ r^{3}$

Where r is the radius of a sphere.

As given

The radius of the ball is 5 in.

As the shape of the ball is spherical .

Thus

$Volume\ of\ a\ ball = \frac{4}{3}\pi\ 5^{3}$

$\pi = \frac{22}{7}$

Thus

$Volume\ of\ a\ ball = \frac{4\times 22\times 5\times\ 5\times 5}{3\times 7}$

$Volume\ of\ a\ ball = \frac{11000}{21}$

Volume of a ball = 523.8 in³ (Approx)

As

$1\ in^{3} = \frac{1}{100}\ pound$

Thus

Convert  523.8 in³ into pounds.

$523.8\ in^{3} = \frac{523.8}{100}\ pound$

$523.8\ in^{3} = 5.24\ pound\ (Approx)$

Therefore the weight of a bowling ball is 5.24 pounds.

Therefore Option (B) is correct.

Answer 2

Answer:

Option B is correct.

Weight of  a bowling ball is 5.24 Ib

Step-by-step explanation:

Assume: The shape of the bowling ball is perfectly spherical.  

Given:  

Radius of a bowling ball= 5 inches (r)  .

One cubic inch weighs $\frac{1}{100}$th of a pound.

Density of a bowling ball = $\frac{1}{100} Ibs/in^3$

Volume of sphere is given by:

$V = \frac{4}{3} \pi r^3$ where V is the volume and r is the radius of the sphere.

Substitute the value of r =5 and $\pi = 3.14$ in above we get;

$V = \frac{4}{3} \cdot 3.14 \cdot 5^3 =\frac{4}{3} \cdot 3.14 \cdot 125$

Simplify:

$V = 523.3333... in^3$

To find the weight of a bowling ball:

$Weight = Volume \times Density$

Then;

$Weight = 523.33333.. \times \frac{1}{100} =\frac{523.3333..}{100} = 5.2333...$

Therefore, the weight of a bowling ball ≈ 5.24 Ib