Question

two spheres are cut from a certain uniform rock. one has radius 4.20 cm. the mass of the other is eight times greater. find its radius.

Explanation:

Step1: Recall volume formula for sphere

The volume of a sphere is $V = \frac{4}{3}\pi r^{3}$.

Step2: Relate volumes of two spheres

Since the rock is uniform, mass is proportional to volume. Let $V_1$ be the volume of the first sphere with radius $r_1 = 4.20$ cm and $V_2$ be the volume of the second sphere with radius $r_2$. Given $m_2=8m_1$, so $V_2 = 8V_1$.

Step3: Substitute volume formulas

$\frac{4}{3}\pi r_2^{3}=8\times\frac{4}{3}\pi r_1^{3}$.

Step4: Simplify the equation

Cancel out $\frac{4}{3}\pi$ on both sides, we get $r_2^{3}=8r_1^{3}$. Since $8 = 2^{3}$, then $r_2^{3}=(2r_1)^{3}$, so $r_2 = 2r_1$.

Step5: Calculate $r_2$

Substitute $r_1 = 4.20$ cm into $r_2 = 2r_1$, we have $r_2=2\times4.20 = 8.40$ cm.

Answer:

$8.40$ cm