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A piece of steel has a weight `w` in air, `w_(1)` when completely immersed in water and `w_(2)` when completely immersed in an unknown liquid. The relative density (specific gravity) of liquid isA. `(w-w_(1))/(w-w_(2))`B. `(w-w_(2))/(w-w_(1))`C. `(w_(1)-w_(2))/(w-w_(1))`D. `(w_(1)-w_(2))/(w-w_(2))` |
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Answer» Correct Answer - B Relative density of steel=`("weight in air")/("change in weight in water")` `=(w)/(w-w_(1))` Now change in weight in the given liquid = upthrust in this liquid or `w-w_(2)={(w//g)/((w//w-w_(1))rho_(w))}d_(2)xxrho_(w)xxg` `therefore d_(2)=(w-w_(2))/(w-w_(1))`= relative density of given liquid. |
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