Question

A metal metre scale that is accurate up to 0.5 mm is made at a temperature of 25$^\circ$C. The range of temperatures within which it can be used is (Coefficient of linear expansion of the metal = $10^{-5}$ /$^\circ$C)

• A. +25$^\circ$C to +75$^\circ$C
• B. +25$^\circ$C to +50$^\circ$C
• C. -25$^\circ$C to +75$^\circ$C
• D. 0$^\circ$C to +50$^\circ$C

Hint:

The formula for linear expansion, $\Delta L = \alpha L_0 \Delta T$, gives the change in length. When dealing with "accuracy" or "error", this implies using the absolute value of the change, $|\Delta L|$, as the scale can be either too long (at higher temperatures) or too short (at lower temperatures).

Answer 1

The Correct Option is C

Step 1: Identify the given parameters.
The scale is a "metre scale", so its original length at the calibration temperature is $L_0 = 1$ m $= 1000$ mm. The calibration temperature is $T_0 = 25^\circ$C. The maximum allowed error in measurement is the change in length, $|\Delta L| \le 0.5$ mm. The coefficient of linear expansion is $\alpha = 10^{-5}$ /$^\circ$C.
Step 2: Use the formula for linear thermal expansion.
The change in length $\Delta L$ due to a change in temperature $\Delta T = T - T_0$ is given by: \[ \Delta L = \alpha L_0 \Delta T. \] We are interested in the magnitude of the temperature change, so we use: \[ |\Delta L| = \alpha L_0 |T - T_0|. \]
Step 3: Set up the inequality for the allowed temperature range.
We are given that the error must be at most 0.5 mm. \[ \alpha L_0 |T - 25| \le 0.5. \] Substitute the given values: \[ (10^{-5}) (1000) |T - 25| \le 0.5. \] \[ 10^{-2} |T - 25| \le 0.5. \]
Step 4: Solve the inequality for T.
\[ |T - 25| \le \frac{0.5}{10^{-2}} = 0.5 \times 100 = 50. \] This inequality can be written as: \[ -50 \le T - 25 \le 50. \] Add 25 to all parts of the inequality: \[ -50 + 25 \le T \le 50 + 25. \] \[ -25 \le T \le 75. \] The range of temperatures is from -25$^\circ$C to +75$^\circ$C.