Q. a metre scale made of steel is calibrated at 20⁰C to give correct readin. Find d distance b/n 50 cm mark and 51 cm mark if d scale is used at 10⁰C. Coefficient of linear expansion of steel is 1.1 *10^-5/⁰C.

ans is1.00011 cm.

if u get dis ans plz explain me d logic.

i got .999. as it should contract wen temp decreases.

To understand the problem of measuring distances on a steel scale at different temperatures, we need to consider how materials expand or contract with temperature changes. In this case, we are dealing with a steel scale that is calibrated at 20 °C and is being used at a lower temperature of 10 °C. Let's break down the logic step by step.

Understanding Thermal Expansion

Every material has a property known as the coefficient of linear expansion, which indicates how much it expands or contracts per degree change in temperature. For steel, this coefficient is given as 1.1 x 10^-5 / °C. This means that for every degree Celsius increase in temperature, a unit length of steel will expand by 1.1 x 10^-5 times its original length.

Calculating the Change in Length

In this scenario, since the temperature is decreasing from 20 °C to 10 °C, we expect the steel to contract. The change in temperature (ΔT) is:

• ΔT = Final Temperature - Initial Temperature = 10 °C - 20 °C = -10 °C

Now, we can calculate the change in length (ΔL) using the formula:

ΔL = L₀ × α × ΔT

Where:

• L₀ is the original length (1 cm in this case, since we are looking at the distance between the 50 cm and 51 cm marks).
• α is the coefficient of linear expansion (1.1 x 10^-5 / °C).
• ΔT is the change in temperature (-10 °C).

Plugging in the values:

ΔL = 1 cm × (1.1 x 10^-5) × (-10) = -1.1 x 10^-4 cm

Finding the New Length Between the Marks

The original distance between the 50 cm and 51 cm marks is 1 cm. Since the scale contracts, we need to subtract the change in length from the original length:

New Length = Original Length + ΔL

New Length = 1 cm - 1.1 x 10^-4 cm = 0.99989 cm

Understanding the Result

Now, the result we calculated (0.99989 cm) is very close to what you mentioned (0.999 cm). However, the answer you provided (1.00011 cm) seems to indicate a misunderstanding. The contraction of the scale at a lower temperature should indeed yield a length slightly less than 1 cm. The correct interpretation is that the scale contracts, and thus the distance between the 50 cm and 51 cm marks at 10 °C is approximately 0.99989 cm.

Final Thoughts

In summary, when using a steel scale calibrated at a higher temperature in a cooler environment, the scale contracts, leading to a shorter measurement between the marks. This is a fundamental principle of thermal expansion and contraction in materials. If you have any further questions or need clarification on any part of this process, feel free to ask!