The coefficient of volumetric expansion of mercury is 18 × 10-50 C -1. The thermometer bulb has a volume of 10-6 ⋅ m 3 and cross-section of the stem is 0.002 cm 2. Assuming that bulb is filled with mercury at 0° C, the increase in length of the mercury column at 100° C will be

Question

The coefficient of volumetric expansion of mercury is 18 × 10-50 C -1. The thermometer bulb has a volume of 10-6 ⋅ m 3 and cross-section of the stem is 0.002 cm 2. Assuming that bulb is filled with mercury at 0° C, the increase in length of the mercury column at 100° C will be (A) 9 cm (B) 9 mm (C) 18 cm (D) 18 mm

Tardigrade Admin · Accepted Answer

The increase in the length of mercury column is Δ L Hg =(&gamma; Hg V Δ T/A) =((18 × 10-5 ° C -1)(10-6 m 3)(100° C )/(0.002 × 10-4 m 2)) =9 × 10-2 m =9 cm

Q. The coefficient of volumetric expansion of mercury is 18×10−50C−1. The thermometer bulb has a volume of 10−6⋅m3 and cross-section of the stem is 0.002cm2. Assuming that bulb is filled with mercury at 0∘C, the increase in length of the mercury column at 100∘C will be

9 cm

9 mm

18 cm

18 mm

The increase in the length of mercury column is ΔLHg​=AγHg​VΔT​ =(0.002×10−4m2)(18×10−5∘C−1)(10−6m3)(100∘C)​ =9×10−2m=9cm

The increase in the length of mercury column is
$Δ L_{H g} = \frac{γ _{H g} V Δ T}{A}$
$= \frac{( 18 \times 1 0 ^{- 5} ^{\circ} C ^{- 1} ) ( 1 0 ^{- 6} m ^{3} ) ( 10 0 ^{\circ} C )}{( 0.002 \times 1 0 ^{- 4} m ^{2} )}$
$= 9 \times 1 0^{- 2} m = 9 c m$