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A piece of steel has a weight \mathrm{W} in air, \mathrm{W}_1 when completely immersed in water and \mathrm{W_2} when completely immersed in an unknown liquid. The relative density (specific gravity) of liquid is :

Option: 1

\mathrm{\frac{W-W_1}{W-W_2}}


Option: 2

\mathrm{\frac{W-W_2}{W-W_1}}


Option: 3

\mathrm{\frac{W_1-W_2}{W-W_1}}


Option: 4

\mathrm{\frac{W_1-W_2}{W-W_2}}


Answers (1)

best_answer

\mathrm{\begin{aligned} & W-v \times 1 \times g=W_1 \\ & W-v \times \times \times g=W_2 \\ & \Rightarrow W-\left(W-W_1\right) \times x=W_2 \\ & x=\frac{W-W_2}{W-W_1} \end{aligned}}

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Kuldeep Maurya

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