The reading of a barometer on a steel scale is 75 cm, while the room temperature is 30?C. The linear and volumetric expensions for steel and mercury are 1.2× 10-5oC-1 and 1.8× 10-4oC-1 (γ ), respectively. The atmospheric pressure can be obtained as

Question

The reading of a barometer on a steel scale is 75 cm, while the room temperature is 30?C. The linear and volumetric expensions for steel and mercury are 1.2× 10-5oC-1 and 1.8× 10-4oC-1 (&gamma; ), respectively. The atmospheric pressure can be obtained as (A) 74.89 cm (B) 67.62 cm (C) 88.67 cm (D) 42.70 cm

Tardigrade Admin · Accepted Answer

From the formula for the linear expansion, the length at 30Â°C is given as lθ =75[1+α (30)] ...(i) At this temperature suppose the length of mercury column is l0 while its length at 0Â°C is l0, then lθ =l0[ 1+(1/3)&gamma; (30) ] ...(ii) From Eqs. (i) and (ii), we get l0[ 1+(1/3)&gamma; (30) ]=75[1+α (30)] ⇒ l0=7489cm

Q. The reading of a barometer on a steel scale is 75 cm, while the room temperature is 30?C. The linear and volumetric expensions for steel and mercury are 1.2×10−5oC−1 and 1.8×10−4oC−1 (γ), respectively. The atmospheric pressure can be obtained as

74.89 cm

67.62 cm

88.67 cm

42.70 cm

Q. The reading of a barometer on a steel scale is 75 cm, while the room temperature is 30?C. The linear and volumetric expensions for steel and mercury are $1.2 \times 10^{- 5}^{o} C^{- 1}$ and $1.8 \times 10^{- 4}^{o} C^{- 1}$ $(γ),$ respectively. The atmospheric pressure can be obtained as