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Anestetic [448]
3 years ago
11

Seven little spheres of mercury, each with a diameter of 2 mm. When they coalesce to form a single sphere, how big will it be (i

.e. what is its diameter)? How does its surface area compare with the total surface area of the previous Seven little spheres?
Physics
1 answer:
Tanzania [10]3 years ago
6 0

Answer:

<em>The total surface area of the seven little spheres is 1.91 times the total surface area of the bigger sphere.</em>

Explanation:

<u>Volume of a Sphere</u>

The volume of a sphere of radius r is given by:

\displaystyle V=\frac{4}{3}\cdot \pi\cdot r^3

The volume of each little sphere is:

\displaystyle V_l=\frac{4}{3}\cdot \pi\cdot 2^3

V_l=33.51\ mm^3

When the seven little spheres coalesce, they form a single bigger sphere of volume:

V_b=7*V_l=234.57\ mm^3

Knowing the volume, we can find the radius rb by solving the formula for r:

\displaystyle V_b=\frac{4}{3}\cdot \pi\cdot r_b^3

Multiplying by 3:

3V_b=4\cdot \pi\cdot r_b^3

Dividing by 4π:

\displaystyle \frac{3V_b}{4\cdot \pi}= r_b^3

Taking the cubic root:

\displaystyle r_b=\sqrt[3]{\frac{3V_b}{4\cdot \pi}}

Substituting:

\displaystyle r_b=\sqrt[3]{\frac{3*234.57}{4\cdot \pi}}

r_b=3.83\ mm

The surface area of the seven little spheres is:

A_l=7*(4\pi r^2)=7*(4\pi 2^2)=351.86\ mm^2

The surface area of the bigger sphere is:

A_b=4\pi r_b^2=4\pi (3.83)^2=184.33\ mm^2

The ratio between them is:

\displaystyle \frac{351.86\ mm^2}{184.33\ mm^2}=1.91

The total surface area of the seven little spheres is 1.91 times the total surface area of the bigger sphere.

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