The initial level of water in a measuring cylinder was $150 m l$ . When an iron piece of irregular shape is completely immersed, the water level changed to $190 m l$ . Find the volume of iron.

4 \times 10^{- 6} m^{3}

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40 \times 10^{- 6} m^{3}

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400 \times 10^{- 6} m^{3}

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4000 \times 10^{- 6} m^{3}

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Solution

The correct option is B $40 \times 10^{- 6} m^{3}$
Given:
Initial level of water, $V_{1} = 150 m l$
Final level of water, $V_{2} = 190 m l$

The increase in volume is equal to the volume of iron piece. Hence, volume of iron piece is:
$V = V_{2} - V_{1}$
$V = 190 - 150 = 40 m l$

From unit conversions, $1 m l = 10^{- 6} m^{3}$

$⟹ V = 40 \times 10^{- 6} m^{3}$

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The initial level of water in a measuring cylinder was 150 ml. When an iron piece of irregular shape is completely immersed, the water level changed to 190 ml. Find the volume of iron.

The correct option is B 40×10−6 m3Given: Initial level of water, V1=150 ml Final level of water, V2=190 ml The increase in volume is equal to the volume of iron piece. Hence, volume of iron piece is: V=V2−V1 V=190−150=40 ml From unit conversions, 1 ml=10−6 m3 ⟹V=40×10−6 m3

The initial level of water in a measuring cylinder was $150 m l$ . When an iron piece of irregular shape is completely immersed, the water level changed to $190 m l$ . Find the volume of iron.