Two identical lead bells are immersed into a measuring jar containing water, the level of water raises from $26$ cm $^{3}$ to 38 cm $^{3}$ . If one of the above lead ball and another metallic ball is immersed simultaneously than the level of water in measuring jar rise from $28$ cm $^{3}$ to 36 cm $^{3}$ . Find the volume of the lead ball and metallic ball.

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Solution

$2 (v o l u m e o f l e a d b a l l) = (38 - 26) {c m}^{3}$
$2 \times V_{l e a d} = 12 {c m}^{3}$
Volume of lead ball = $6 {c m}^{3}$
and Volume of lead ball + metallic ball = 36-28
$V_{l e a d} + V_{m e t a l} = 8 {c m}^{3}$
$V_{m e t a l} = 8 - 6 = 2 {c m}^{3}$ .

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2(volumeofleadball)=(38−26)cm32×Vlead=12cm3Volume of lead ball = 6cm3and Volume of lead ball + metallic ball = 36-28 Vlead+Vmetal=8cm3 Vmetal=8−6=2cm3.

$2 (v o l u m e o f l e a d b a l l) = (38 - 26) {c m}^{3}$
$2 \times V_{l e a d} = 12 {c m}^{3}$
Volume of lead ball = $6 {c m}^{3}$
and Volume of lead ball + metallic ball = 36-28
$V_{l e a d} + V_{m e t a l} = 8 {c m}^{3}$
$V_{m e t a l} = 8 - 6 = 2 {c m}^{3}$ .