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Archimedes' Principle

A body weighs...

Question

A body weighs $W_{1}$ in a liquid of density $d_{1}$ and $W_{2}$ in a liquid of density $d_{2}$ . What is the weight in a liquid of density $d_{3}$ ?

$W_{3}$ = $\frac{W_{1} (d_{3} - d_{1}) - W_{2} (d_{3} - d_{2})}{d_{2} - d_{1}}$

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$W_{3}$ = $\frac{W_{2} (d_{3} - d_{1}) - W_{1} (d_{3} - d_{2})}{d_{2} - d_{1}}$

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$W_{3}$ = $\frac{W_{2} (d_{1} - d_{2}) - W_{1} (d_{3} - d_{2})}{d_{2} - d_{1}}$

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$W_{3}$ = $\frac{W_{2} (d_{3} - d_{1}) - W_{1} (d_{3} - d_{2})}{d_{1} - d_{2}}$

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Solution

The correct option is B
$W_{3}$ = $\frac{W_{2} (d_{3} - d_{1}) - W_{1} (d_{3} - d_{2})}{d_{2} - d_{1}}$

$W_{1}$ = mg $[1 - \frac{d_{1}}{D}]$ = W $[1 - \frac{d_{1}}{D}]$ .........(a)

Similarly, $W_{2}$ = mg $[1 - \frac{d_{2}}{D}]$ = W $[1 - \frac{d_{2}}{D}]$ .........(b)

and $W_{3}$ = mg $[1 - \frac{d_{3}}{D}]$ = W $[1 - \frac{d_{3}}{D}]$ .........(c)

Using (a) , (b) & (c) we obtain

$W_{3}$ = $\frac{W_{2} (d_{3} - d_{1}) - W_{1} (d_{3} - d_{2})}{d_{2} - d_{1}}$

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A body weighs W1 in a liquid of density d1 and W2 in a liquid of density d2. What is the weight in a liquid of density d3?

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A body weighs $W_{1}$ in a liquid of density $d_{1}$ and $W_{2}$ in a liquid of density $d_{2}$ . What is the weight in a liquid of density $d_{3}$ ?