Question

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A piece of metal weighs 'x' Newton in air ,'y' Newton when completey immersed in water and 'z' Newton when completely immeresed in liquid.the relative density of liquid is❓
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Answer 1

Answer:

W-W1÷W-W2

Explanation:

• weight of liquid = L1÷Ls=1–z÷x

Answer 2

Given:

Weight of the metal in the air, $W=x$ $N$

Weight of the metal when immersed in water, $W_w=y$ $N$

Weight of the metal when immersed in a liquid, $W_l=z$ $N$

To Find:

Relative density of the liquid.

Solution:

Weight of the metal in the air $=x N$

Weight of the metal in water= $yN$

∴ The buoyant force produced by water $=(x-y)N$

Using Archimedes' principle weight of the water displaced $=(x-y)N$

Similarly, The buoyant force produced by the liquid $=(x-z)N$

weight of the liquid displaced $=(x-z)N$

Let the volume of the metal be $V$

Since, the same volume of both water and the liquid is displaced by the metal,

The density of water, $d_w=\frac{(x-y)}{gV}$, (Where $g$ is the gravitational acceleration)

The density of the liquid, $d_l=\frac{(x-z)}{gV}$

"The relative density of a substance is the ratio of the density of the substance to the density of water."

∴ Relative density of the liquid $=\frac{(x-y)/gV}{(x-z)/gV}$

                                                   $=\frac{x-y}{x-z}$

Hence, the relative density of the given liquid = $\frac{x-y}{x-z}$.