First slide

Pressure in a fluid-mechanical properties of fluids

Question

An ornament weighing 36 g in air weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it.
Specific gravity of gold is 19.3 and that of copper is 8.9.

Difficult

A

2.2 g
B

4.4 g
C

1.1 g
D

3.6 g

Solution

$\begin{array}{rcl} \begin{array}{l} Given that m_{real} = 36 g, m_{app} = 34 g . \\ Density of gold ρ_{Au} = 19.3 g / cc \\ Density of copper ρ_{Cu} = 8.9 g / cc \\ We know that loss of weight = weight of displaced \end{array} \\ \begin{array}{l} water = 36 - 34 = 2 g = Buoyant force = B \\ Here, m_{real} = m_{Au} + m_{Cu} = 36 g . . . . . . . . . . (i) \\ Let v {be the volume of the ornament in cm}^{3} . Then \\ \begin{array}{ll} B = v \times ρ_{w} \times g = 2 \times g \\ \Rightarrow (\frac{m_{Au}}{ρ_{Au}} + \frac{m_{Cu}}{ρ_{Cu}}) ρ_{w} \times g = 2 \times g \\ \Rightarrow & m_{Au} ρ_{Cu} + m_{Cu} ρ_{Au} = 2 ρ_{A u} ρ_{Cu} \\ \Rightarrow & 8.9 m_{Au} + 19.3 m_{Cu} = 2 \times 19.3 \times 8.9 = 343.54 . . . . . . . . . (i i) \end{array} \\ From Eqs. (i) and (ii), 8.9 m_{Au} + 19.3 m_{Cu} = 343.54 \\ \Rightarrow 8.9 (m_{Au} + m_{Cu}) + 10.4 m_{Cu} = 343.54 \\ \begin{array}{l} \Rightarrow 8.9 \times 36 + 10.4 m_{Cu} = 343.54 \\ \Rightarrow m_{Cu} = 2.225 g \\ So the amount of copper in the ornament is 2.2 g . \end{array} \end{array} \end{array}$

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