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Error in Metal Scale Reading Due To Expansion and Contraction

Error in Metal Scale Reading Due To Expansion and Contraction :- In this article we are going to discuss about error in metal scale reading due to change in temperature.

Suppose we have a metal scale. As we know, with rise and fall in temperature, the length of metal scale changes and hence the distance between the graduations on the scale also changes. So with the change in temperature there arises error in the reading of a metal scale.

Error in Metal Scale Reading due to Rise in temperature :-

Let a metal scale gives correct reading at a temperature T₀ºC and the temperature at which we are measuring the length of an object is T₁ºC(T₁> T₀). With the rise in temperature the distance between any two scale divisions increases in the ratio 1: (1 + αΔT).

Here ΔT = T₁ – T₀and α = coefficient of linear expansion of the material of the scale.

If the scale is measuring the length of an object as L cm, then its actual reading must be L(1 + αΔT)cm and hence a correction of +LαΔT should be applied.

In above figure the actual length of the object is 8 cm, but as the temperature rises to T₁ºC , the length of the scale increases and the scale reads less i.e., 7.5 cm.

Here

Error = M.V. (measured value) – T.V. (True value) = 7.5cm – 8cm = -0.5 cm

Correction = +0.5 cm

Error in Metal Scale Reading due to Fall in temperature :-

As the temperature falls, the length of the metal scale reduces and the scale reads more.

Let the temperature be T₂ºC(T₂ < T₀).

With the fall in temperature the distance between any two scale divisions decreases in the ratio 1: (1 – αΔT).

Here ΔT = T₀ – T₂and α = coefficient of linear expansion of the material of the scale.

If the scale is measuring the length of an object as L cm, then its actual reading must be L(1 – αΔT)cm and hence a correction of -LαΔT should be applied.

In above figure the actual length of the object is 8 cm, but as the temperature falls to T₂ºC , the length of the scale decreases and the scale reads more i.e., 8.5 cm.

Here

Error = M.V. (measured value) – T.V. (True value) = 8.5cm – 8cm = +0.5 cm

Correction = -0.5 cm

Example 1.

A meter scale is made of steel and measures correct length at 16ºC. What will be the percentage error if the scale is used

On a summer day when temperature is 46ºC and
On a winter day when the temperature is 6ºC ?

Coefficient of linear expansion of steel is 11×10^-6ºC^-1.

Solution :

On a summer day, the scale will read less than the actual length. Hence, the percentage error will be :-

percentage error = –(Δl/l) ×100

⇒ percentage error = -(αΔT) ×100

⇒ percentage error = -( 11×10^-6) ×(46-16)×100

⇒ percentage error = -0.033%

On a winter day, the scale will read more than the actual length. Hence, the percentage error will be :-

percentage error = +(Δl/l) ×100

⇒ percentage error = +(αΔT) ×100

⇒ percentage error = +( 11×10^-6) ×(16-6)×100

⇒ percentage error = +0.011%

Example 2.

A bar measured with a vernier calliper is found to be 180 mm long. The temperature during the measurement is 10 ºC. The measurement error will be if the scale of the vernier calliper has been graduated at a temperature of 20 ºC : (α = 1.1 × 10^-5 °C^-1. Assume that the length of the bar does not change.)

(A) 1.98 × 10^–1 mm (B) 1.98 × 10^–2 mm (C) 1.98 × 10^–3 mm (D) 1.98 × 10^–4 mm

Solution :

As the temperature falls, the length of the metal scale reduces and the scale reads more, hence

Error = M.V. (measured value) – T.V. (True value) = αLΔT

∴ Error = 1.1 × 10^-5× 180mm × 10 = 1.98 × 10^–2 mm

Barometer Errors and Temperature Correction

A mercury barometer is a standard tool for measuring atmospheric pressure, but its readings are affected by temperature. To obtain a precise and comparable reading, two main corrections must be applied to the observed value.

(1). Scale Expansion : The brass or steel scale used to measure the height also expands when heated. This makes the markings on the scale farther apart, causing the observed reading to be slightly less than the true length of the liquid column.

(2). Mercury Expansion : As temperature increases, the mercury in the column expands, causing its density to decrease. This means the height of the mercury column will be greater for the same atmospheric pressure.

Finding the True Reading at 0°C

Let the brass scale of barometer is calibrated at temperature T_cal and at any temperature T_t let the observed reading (height of mercury column) is h_observed . As with rise in temperature, the observed reading is less than the true reading, so true height of mercury column at temperature T_t will be $\displaystyle {{h}_{observed}}[1+{{\alpha }_{scale}}\left( {{T}_{t}}-{{T}_{cal}} \right)]$ .

Now this is the apparent height of mercury column at temperature T_t , but we have to find the true height of mercury column at reference temperature of 0°C. Using the formula of pressure exerted by a liquid column,

$\displaystyle {{\left( P \right)}_{at\text{ }0{}^\circ C}}={{\left( P \right)}_{at\text{ }t{}^\circ C}}$

$\displaystyle \Rightarrow {{h}_{0}}{{\rho }_{0}}g={{h}_{t}}{{\rho }_{t}}g$

$\displaystyle \Rightarrow {{h}_{0}}=\frac{{{h}_{t}}}{{{\rho }_{0}}}{{\rho }_{t}}=\frac{{{h}_{observed}}[1+{{\alpha }_{scale}}\left( {{T}_{t}}-{{T}_{cal}} \right)]}{{{\rho }_{0}}}\times \frac{{{\rho }_{0}}}{[1+{{\gamma }_{Hg}}\left( {{T}_{t}}-{{T}_{0}} \right)]}$

$\displaystyle \Rightarrow {{h}_{true}}={{h}_{0}}={{h}_{observed}}\left[ \frac{1+{{\alpha }_{scale}}\left( {{T}_{t}}-{{T}_{cal}} \right)}{1+{{\gamma }_{Hg}}\left( {{T}_{t}}-{{T}_{0}} \right)} \right]$

The above formula can be used to find the true reading of barometer/true value of atmospheric pressure at 0°C.

If the calibration temperature of the scale is also 0°C, then the above formula can be simplified as below :

$\displaystyle {{h}_{true}}={{h}_{observed}}\left[ \frac{1+{{\alpha }_{scale}}\left( {{T}_{t}}-{{T}_{0}} \right)}{1+{{\gamma }_{Hg}}\left( {{T}_{t}}-{{T}_{0}} \right)} \right]={{h}_{observed}}\left[ \frac{1+{{\alpha }_{scale}}\Delta T}{1+{{\gamma }_{Hg}}\Delta T} \right]$

$\displaystyle \Rightarrow {{h}_{true}}={{h}_{observed}}(1+{{\alpha }_{scale}}\Delta T){{(1+{{\gamma }_{Hg}}\Delta T)}^{-1}}$

$\displaystyle \Rightarrow {{h}_{true}}={{h}_{observed}}(1+{{\alpha }_{scale}}\Delta T)(1-{{\gamma }_{Hg}}\Delta T)$

$\displaystyle \Rightarrow {{h}_{true}}={{h}_{observed}}\left( 1-{{\gamma }_{Hg}}\Delta T+{{\alpha }_{scale}}\Delta T-{{\alpha }_{scale}}\Delta T\times {{\gamma }_{Hg}}\Delta T \right)$

$\displaystyle \Rightarrow {{h}_{true}}\cong {{h}_{observed}}\left( 1+{{\alpha }_{scale}}\Delta T-{{\gamma }_{Hg}}\Delta T \right)$

Example 3.

The brass scale of a barometer gives correct reading at 0º C. Coefficient of thermal expansion of brass is 0.00002/º C. The barometer reads 75 cm at 27º C. What is the correct atmospheric pressure at 27ºC ?

Solution :

The scale expands with temperature, so a fixed actual height of mercury appears shorter on the expanded scale. Therefore, the true height (i.e., the correct pressure) is :

$\displaystyle {{h}_{true}}={{h}_{observed}}(1+\alpha \Delta T)=75\left[ 1+0.00002\times 27 \right]=75.0405cm$

Aliter

The reading of the barometer at higher temperature is slightly less than the actual value because the brass scale expands. The amount by which the reading is less is approximately equal to the increase in length of the brass scale due to thermal expansion.

error in reading ≈ expansion in brass scale

Δh = h × α × ΔT

Δh = 75 × 0.00002 × 27 = 0.0405 cm

Therefore, the true height (i.e., the correct pressure) :

$\displaystyle {{h}_{true}}=\left( 75+.0405 \right)cm=75.0405cm$

Example 4.

Brass scale of a Barometer gives correct reading at 0 ºC. Coefficient of linear expansion of brass is 18 × 10⁻⁶ ºC^-1 . If the barometer reads 76 cm at 20 ºC , the correct reading is :

(A) 76.426 cm

(B) 75.7 cm

(D) 76.264 cm

[Given γ_Hg = 18 × 10⁻⁵ ºC^-1]

Solution :

With an increase in temperature, the brass scale expands, causing the reading to appear less than the actual height. Therefore, we must add a correction for scale expansion to obtain the true pressure. However, mercury also expands with temperature, which causes the actual column height to increase. This effect partially offsets the scale expansion, so we must subtract a correction for mercury expansion to arrive at the correct reading.

So the correct reading will be :

$\displaystyle {{h}_{true}}={{h}_{observed}}(1+{{\alpha }_{scale}}\Delta T-{{\gamma }_{Hg}}\Delta T)$

You might wonder why the coefficient of volume expansion (γ) of mercury isn’t converted to the coefficient of linear expansion (α = γ/3). The reason is that we assume the expansion of the tube containing mercury is negligible, and the volume expansion of mercury along the x and y axes effectively contributes to an observable change in height along the z-axis.

Hence correct reading :

$\displaystyle {{h}_{true}}=76\left[ 1+\left( 18\times {{10}^{-6}}\times 20 \right)-\left( 18\times {{10}^{-5}}\times 20 \right) \right]$

$\displaystyle \Rightarrow {{h}_{true}}=75.754cm$

So (B) is the correct option.

Example 5.

The coefficient of cubical expansion of mercury is 0.00018/ºC and that of brass 0.00006/ºC. If a barometer having a brass scale were to read 74.5 cm at 30 ºC, find the true barometric height at 0 ºC. The scale is supposed to be correct at 15 ºC.

(A) 74.122 cm

(B) 79.152 cm

(D) 142.39 cm

Solution :

γ_brass= 0.00006/ºC

⇒ α_brass = 0.00002/ºC

Using

$\displaystyle {{h}_{true}}={{h}_{observed}}\left[ 1+{{\alpha }_{scale}}\left( {{T}_{t}}-{{T}_{cal}} \right)-{{\gamma }_{Hg}}\left( {{T}_{t}}-{{T}_{0}} \right) \right]$