An alloy is prepared by mixing equal volume 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 alloy is $3$ . The specific gravity of each metal is:

2, 4

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6, 4

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6, 2

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4, 8

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Solution

The correct option is C $6, 2$
Case I $V_{1} = V_{2} = V$ Case II $m_{1} = m_{2} = m$ .
$ρ_{a l l o y} = \frac{ρ_{1} V_{1} + ρ_{2} V_{2}}{V_{1} + V_{2}} ρ_{a l l o y} = \frac{m_{1} + m_{2}}{\frac{m_{1}}{ρ_{1}} + \frac{m_{2}}{ρ_{2}}}$
$4 = (ρ_{1} + ρ_{2}) / 23 = (2 ρ_{1} ρ_{2}) / (ρ_{1} + ρ_{2})$

$ρ_{1} + ρ_{2} = 8$

$3 = \frac{ρ_{1} ρ_{2}}{4}$

$ρ_{1} . ρ_{2} = 12$
So $ρ_{1} = 6 & ρ_{2} = 2 o r ρ_{1} = 2 & ρ_{2} = 6$

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An alloy is prepared by mixing equal volume 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 alloy is 3. The specific gravity of each metal is:

The correct option is C 6,2Case I V1=V2=V Case II m1=m2=m.ρalloy=ρ1V1+ρ2V2V1+V2ρalloy=m1+m2m1ρ1+m2ρ2 4=(ρ1+ρ2)/23=(2ρ1ρ2)/(ρ1+ρ2)ρ1+ρ2=8 3=ρ1ρ24 ρ1.ρ2=12So ρ1=6&ρ2=2orρ1=2&ρ2=6

An alloy is prepared by mixing equal volume 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 alloy is $3$ . The specific gravity of each metal is: