Question

A small silver (10.5 g/cm^3) cylinder of diameter 1.4 cm and a cylinder of lead (11.3 g/cm^3) balance each other when placed on a triple beam balance. If they have the same length, what must be the diameter of the lead cylinder?

Answer 1

Answer:

The diameter of the lead cylinder is 1.35 cm.

Explanation:

Given that,

Density of silver = 10.5 g/cm³

Density of lead = 11.3 g/cm³

Diameter = 1.4 cm

As mass of both is equal.

Let diameter of lead $d_{l}$

We need to calculate the the diameter of the lead cylinder

Using balance equation of density

$V\times \rho_{s}=V\times \rho_{l}$

$\dfrac{\pi\times d_{s}^2\times h}{4}\times\rho_{s}=\dfrac{\pi\times d_{l}^2\times h}{4}\times\rho_{s}$

$d_{l}^2=\dfrac{d_{s}^{2}\times\rho_{s}}{\rho_{l}}$

put the value into the formula

$d_{l}^2=\dfrac{(1.4\times10^{-2})^2\times10.5}{11.3}$

$d_{l}=\sqrt{0.00018212}$

$d_{l}=0.0135\ m$

$d_{l}=1.35\ cm$

Hence, The diameter of the lead cylinder is 1.35 cm.