## Question

A spring balance reads 10 kg when a bucket of water is suspended from it. What will be the reading of the balance when an iron piece of mass 7.2 kg suspended by a string is immersed with half its volume inside the water in the bucket? Relative density of iron is 7.2.

### Solution

0.5 kg

.

Volume of the iron piece is

= 10^{–3} m^{3}

Now, the volume of the iron piece under water is , which is the volume of water displaced. Hence weight of water displaced

= volume of water displaced dinsity of water *g*

newton = weight of 0.5 kg.

This is the upthrust exerted by water on the iron piece. Hence the iron piece will exert an equal and opposite force on water in the downward direction, which will increase the reading of the balance by 0.5 kg.

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