Project ForgeA hydrometer, with stem cross-sectional area of 2.82×10⁻⁵m², is immersed in water, with immersed volume 15×10⁻⁶m³ and stem length above the surface Lw. If water is replaced by a liquid with specific gravity 1.5 and the stem length above the liquid surface is Lf, then the difference, Lf−Lw, is
- A−177 mm
- 177 mm
- C−266 mm
- D266 mm
Explanation
The hydrometer's weight (fixed) equals the weight of liquid displaced in both cases: ρwater×Vimmersed,water=ρliquid×Vimmersed,liquid ⟹ Vimmersed,liquid=15×10⁻⁶/1.5=10×10⁻⁶m³. Since total hydrometer volume=Vimmersed+Astem×(stem length above surface) is constant: 15e-6+2.82e-5×Lw=10e-6+2.82e-5×Lf ⟹ Lf−Lw=5×10⁻⁶/2.82×10⁻⁵≈0.177m=177mm.
Source: GATE CH 2010 (2010)
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