Thermal Expansion

Let

p_0%20%5C%2C%20%5C%2C%20and%20%5C%2C%20%5C%2C%20p_T

be densities of glycerin at

0%5E%7B%5Ccirc%7DC

and T

%5E%7B%5Ccirc%7D

C respectively. Then

p_T%20%3D%20p_0%20(1-%20%5Cgamma%20%5CDelta)

where

%5Cgamma

is the coefficient of volume expansion of
glycerin and

%5CDelta%20T

is rise in temperature.

%5Cfrac%7Bp%20_T%7D%7B%20p_0%7D%20%3D%201-%20%5Cgamma%20%5CDelta%20T%20%5C%2C%20%5C%2C%20or%20%5C%2C%20%5C%2C%20%5Cgamma%20%5CDelta%20T%20%3D%201-%5Cfrac%7Bp%20_T%7D%7B%20p_0%7D

Thus ,

%5Cfrac%7Bp_0%20-%20p_T%7D%7Bp_0%7D%20%3D%20%5Cgamma%20%5CDelta%20T

Here,

%5Cgamma%20%3D5%20%5Ctimes%2010%5E%7B-4%7D%20K%5E%7B-1%7D%20%5C%2C%20%5C%2C%20and%20%5C%2C%20%5C%2C%20%5CDelta%20T%20%3D%2040%5E%5Ccirc%20C%20%3D%2040%20K

%5Ctherefore%20%5C%2C%20%5C%2C

The fractional change in the density of glycerin

%3D%5Cfrac%7Bp_0%20-%20p_T%7D%7Bp_0%7D%20%3D%20%5Cgamma%20%5CDelta%20T%20%3D%20(5%20%5Ctimes%2010%5E%7B-4%7D%20K%5E%7B-1%7D)%20(40%20K)%20%3D0.020

So, the correct answer is (D): 0.020.

To solve this problem, we need to determine the length of the aluminum rod such that its increase in length with temperature change is independent, or the same as that of the copper rod. This involves using the concept of linear expansion.

The formula for linear expansion is given by:

%5CDelta%20L%20%3D%20L%20%5Ccdot%20%5Calpha%20%5Ccdot%20%5CDelta%20T

where:

$%5CDelta%20L$ is the change in length,
$L$ is the original length of the material,
$%5Calpha$ is the coefficient of linear expansion,
$%5CDelta%20T$ is the change in temperature.

For the lengths to be independent of the change in temperature, the increase in length $%5CDelta%20L$ due to temperature should be equal for both rods. Hence, we set the equations of change in length for copper and aluminum rods equal:

%5CDelta%20L_%7BCu%7D%20%3D%20%5CDelta%20L_%7BAl%7D

Since the temperature change $%5CDelta%20T$ will cancel out, we have:

L_%7BCu%7D%20%5Ccdot%20%5Calpha_%7BCu%7D%20%3D%20L_%7BAl%7D%20%5Ccdot%20%5Calpha_%7BAl%7D

We know:

$L_%7BCu%7D%20%3D%2088%20%5C%2C%20cm$
$%5Calpha_%7BCu%7D%20%3D%201.7%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%20K%5E%7B-1%7D$
$%5Calpha_%7BAl%7D%20%3D%202.2%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%20K%5E%7B-1%7D$

Substitute these values into the equation:

88%20%5Ccdot%201.7%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20L_%7BAl%7D%20%5Ccdot%202.2%20%5Ctimes%2010%5E%7B-5%7D

Solve for $L_%7BAl%7D$ :

L_%7BAl%7D%20%3D%20%5Cfrac%7B88%20%5Ccdot%201.7%20%5Ctimes%2010%5E%7B-5%7D%7D%7B2.2%20%5Ctimes%2010%5E%7B-5%7D%7D

L_%7BAl%7D%20%3D%20%5Cfrac%7B88%20%5Ccdot%201.7%7D%7B2.2%7D

L_%7BAl%7D%20%3D%20%5Cfrac%7B149.6%7D%7B2.2%7D

L_%7BAl%7D%20%5Capprox%2068%20%5C%2C%20cm

Therefore, the length of the aluminum rod is approximately 68 cm.

High-Yield Trend

Chapter Questions
4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Thermal Expansion

High-Yield Trend

Chapter Questions 4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Chapter Questions
4 MCQs