Answer to Question #202614 in Mechanics | Relativity for karen

Question

your friend is concerned about a gold ring she bought on a recent trip. The ring was expensive, and she would like to know whether it is really made of gold or of something else. You decide to help her, using your knowledge of physics. You weigh the ring and find that it has a weight of 0.158 N. using a string, you suspend the ring from the scale and, with the ring submerged in water, weight it again to find a new reading of 0.150 N. Is the ring pure gold?

Answer 1

Explanations & Calculations

• When the ring is weighed in air, the scale reads the actual weight of it: $\small mg$
• When weighed submerged, the scale reads something which is lower than the actual weight. That is due to the upthrust imposed on the ring.
• The difference between the two readings is equal to that.

• Take the density of the ring to be $\small d$, the volume of the ring to be $\small V$.

• Then, from the reading taken in the air,

$\qquad\qquad \begin{aligned} \small mg&=\small 0.158\\ \small (Vd)g&=\small 0.158\cdots(\because\,m=Vd)\\ \small V&=\small \frac{0.158}{dg}\cdots(1) \end{aligned}$

• From the readings taken submerged,

$\qquad\qquad \begin{aligned} \small U&=\small 0.158-0.150=0.008\,N\\ \small V\rho_wg&=\small0.008\\ \small V&=\small \frac{0.008}{\rho_wg}\cdots(2) \end{aligned}$

• Then, by (1) = (2) we can calculate the density of gold & compare it with the standard value.

$\qquad\qquad \begin{aligned} \small \frac{0.158}{dg}&=\small \frac{0.008}{\rho_wg}\\ \small d&=\small \frac{0.158\times1000}{0.008}\\ &=\small 19750\,kgm^{-3} \end{aligned}$

• This is more than the density of Karat 24 gold which is $\small 19300 kgm^{-3}$
• The ring is almost pure gold.