Question

An alloy is prepared by mixing equal volumes of two metals. The specific gravity of alloy is 4. But when equal masses of two same metals are mixed together, the specific gravity of the alloy is 3. The specific gravity of each metal is:
A. 2, 4
B. 6, 4
C. 6, 2
D. 4, 8

Answer 1

The density of the material is the ratio of mass and its volume. Use the equation of specific gravity of the alloy of two metals of different specific gravity. Solve the simultaneous equations if any.

Complete step by step answer:
We have given, on mixing of equal volume of two metals, the specific gravity is 4.
The specific gravity of an alloy on mixing of two metals of volume ${V_1}$ and ${V_2}$ respectively is,
${\rho _{alloy}} = \dfrac{{{\rho _1}{V_1} + {\rho _2}{V_2}}}{{{V_1} + {V_2}}}$
Here, ${\rho _1}$ is the specific gravity of the first metal and ${\rho _2}$ is the specific gravity of the second metal.
Since the volume of the two metals is the same. Substitute ${V_1} = {V_2} = V$ in the above equation. Also, the specific gravity of the mixture is 4.
$4 = \dfrac{{{\rho _1}V + {\rho _2}V}}{{V + V}}$
$\Rightarrow 4 = \dfrac{{{\rho _1} + {\rho _2}}}{2}$
$\Rightarrow {\rho _1} + {\rho _2} = 8$ …… (1)
We also have given, on mixing the same amount of mass of two metals, the specific gravity of the resultant alloy is 3.
Therefore, the specific gravity of the alloy is,
${\rho _{alloy}} = \dfrac{{{\rho _1}{V_1} + {\rho _2}{V_2}}}{{{V_1} + {V_2}}}$
Since the density of the metal is the ratio of mass and volume, we can write the above equation as follows,
${\rho _{alloy}} = \dfrac{{{m_1} + {m_2}}}{{\dfrac{{{m_1}}}{{{\rho _1}}} + \dfrac{{{m_2}}}{{{\rho _2}}}}}$
Since the mass of the two metals is equal, substitute ${m_1} = {m_2} = m$ and 3 for ${\rho _{alloy}}$ in the above equation.
$3 = \dfrac{{m + m}}{{\dfrac{m}{{{\rho _1}}} + \dfrac{m}{{{\rho _2}}}}}$
$\Rightarrow 3 = \dfrac{{2m}}{{m\left( {\dfrac{1}{{{\rho _1}}} + \dfrac{1}{{{\rho _2}}}} \right)}}$
$\Rightarrow 3 = \dfrac{{2{\rho _1}{\rho _2}}}{{{\rho _1} + {\rho _2}}}$
But, ${\rho _1} + {\rho _2} = 8$. Therefore, the above equation becomes,
$3 = \dfrac{{2{\rho _1}{\rho _2}}}{8}$
$\Rightarrow {\rho _1}{\rho _2} = 12$ …… (2)
From equation (1),
${\rho _1} = 8 - {\rho _2}$
Substitute ${\rho _1} = 8 - {\rho _2}$ in equation (2).
$\left( {8 - {\rho _2}} \right){\rho _2} = 12$
$\Rightarrow 8{\rho _2} - \rho _2^2 = 12$
$\Rightarrow \rho _2^2 - 8{\rho _2} + 12 = 0$
Solving the above equation, we get the value of specific gravity of the second metal ${\rho _2} = 6$.
Substitute ${\rho _2} = 6$ in equation (2).
${\rho _1}\left( 6 \right) = 12$
$\Rightarrow {\rho _1} = 2$
Therefore, the specific gravity of each metal is 6 and 2.

So, the correct answer is “Option C”.

Note:
To solve the second-degree linear equation $a{x^2} + bx + c = 0$, use the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. This will give two values of x. In the given question, ${\rho _2}$ has two values. Pick any one of them and substitute it in the former equation.