A block of wood of density $500 k g / m^{3}$ has mass $m k g$ in air. A lead block which has apparent weight of $28 k g$ in water is attached to the block of wood, and both of them are submerged in water. If their combined apparent weight in water is $20 k g$ , find the value of $m$ . Taken density of water $= 1000 k g / m^{3}$ .

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Solution

$S_{1} V_{1} + S_{2} V_{2} - S_{w} (V_{1} + V_{2}) = 20 ⟶ (1)$

$\Rightarrow (S_{1} V_{1} - S_{w} V_{1}) + S_{2} V_{2} - S_{w} V_{2} = 20$

$V_{1} (S_{1} - S_{w}) = 28 ⟶ (2)$

$\Rightarrow 28 + 500 V_{2} - 1000 V_{2} = 20$

$S_{1} V_{1} - S_{w} V_{1} = 28$

$\Rightarrow 28 - 500 V_{2} = 20$

$\Rightarrow {500 V}_{2} = 8$

$\Rightarrow V_{2} = \frac{8}{500} m^{3}$

mass of wooden block $= S_{2} V_{2}$
$= 500 \times \frac{8}{500}$
$= 8 k g$ .

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S1V1+S2V2−Sw(V1+V2)=20⟶(1) ⇒(S1V1−SwV1)+S2V2−SwV2=20 V1(S1−Sw)=28⟶(2)⇒28+500V2−1000V2=20 S1V1−SwV1=28⇒28−500V2=20⇒500V2=8⇒V2=8500m3mass of wooden block =S2V2 =500×8500 =8kg.