The density of a substance at $0\,{}^ \circ C$ is $10\,gc{m^{ - 3}}$ and at $100\,{}^ \circ C$ its density is $9.7\,gc{m^{ - 3}}$. The coefficient of linear expansion of the substance is:
(A) ${10^2}$
(B) ${10^{ - 2}}$
(C) ${10^{ - 3}}$
(D) ${10^{ - 4}}$
Answer
639k+ views
Hint: The coefficient of the linear expansion of the substance is determined by using the thermal expansion of the substance formula which gives the relation of the temperature of the liquid, density of the liquid and the coefficient of the linear expansion. Then the coefficient of linear expansion of the substance is determined.
Formula Used: The thermal expansion of the substance is determined by,
${\rho _0} = \rho \left( {1 + 3\alpha \times \Delta T} \right)$
Where, $\rho $ is the density of the substance in maximum temperature, ${\rho _0}$ is the density of the substance at zero temperature, $\alpha $ is the coefficient of linear expansion of the substance and $\Delta T$ is the difference in the temperatures.
Complete step by step answer:
Given that,
The density of a substance at $0\,{}^ \circ C$ is, ${\rho _0} = 10\,gc{m^{ - 3}}$
The density of a substance at $100\,{}^ \circ C$ is, $\rho = 9.7\,gc{m^{ - 3}}$
Now,
The thermal expansion of the substance is determined by,
${\rho _0} = \rho \left( {1 + 3\alpha \times \Delta T} \right)\,................\left( 1 \right)$
By substituting the density of the substance in maximum temperature, density of the substance at zero temperature and the difference in the temperatures in the above equation (1), then the equation (1) is written as,
$\Rightarrow 10 = 9.7\left( {1 + 3\alpha \times \left( {100 - 0} \right)} \right)$
By subtracting the terms in the above equation, then the above equation is written as,
$\Rightarrow 10 = 9.7\left( {1 + 3\alpha \times 100} \right)$
By multiplying the terms in the above equation, then the above equation is written as,
$\Rightarrow 10 = 9.7\left( {1 + 300\alpha } \right)$
By multiplying the terms in the above equation, then the above equation is written as,
$\Rightarrow 10 = 9.7 + 2910\alpha $
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow 2910\alpha = 10 - 9.7$
By subtracting the terms in the above equation, then the above equation is written as,
$\Rightarrow 2910\alpha = 0.3$
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow \alpha = \dfrac{{0.3}}{{2910}}$
By dividing the terms in the above equation, then the above equation is written as,
$\Rightarrow \alpha = 1 \times {10^{ - 4}}$
Hence, the option (D) is the correct answer.
Note:The coefficient of the linear expansion is depending on the density substance in maximum temperature and density of the substance at zero temperature and the temperature of the substance. The temperature difference is the difference of the temperature from the maximum temperature to the minimum temperature.
Formula Used: The thermal expansion of the substance is determined by,
${\rho _0} = \rho \left( {1 + 3\alpha \times \Delta T} \right)$
Where, $\rho $ is the density of the substance in maximum temperature, ${\rho _0}$ is the density of the substance at zero temperature, $\alpha $ is the coefficient of linear expansion of the substance and $\Delta T$ is the difference in the temperatures.
Complete step by step answer:
Given that,
The density of a substance at $0\,{}^ \circ C$ is, ${\rho _0} = 10\,gc{m^{ - 3}}$
The density of a substance at $100\,{}^ \circ C$ is, $\rho = 9.7\,gc{m^{ - 3}}$
Now,
The thermal expansion of the substance is determined by,
${\rho _0} = \rho \left( {1 + 3\alpha \times \Delta T} \right)\,................\left( 1 \right)$
By substituting the density of the substance in maximum temperature, density of the substance at zero temperature and the difference in the temperatures in the above equation (1), then the equation (1) is written as,
$\Rightarrow 10 = 9.7\left( {1 + 3\alpha \times \left( {100 - 0} \right)} \right)$
By subtracting the terms in the above equation, then the above equation is written as,
$\Rightarrow 10 = 9.7\left( {1 + 3\alpha \times 100} \right)$
By multiplying the terms in the above equation, then the above equation is written as,
$\Rightarrow 10 = 9.7\left( {1 + 300\alpha } \right)$
By multiplying the terms in the above equation, then the above equation is written as,
$\Rightarrow 10 = 9.7 + 2910\alpha $
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow 2910\alpha = 10 - 9.7$
By subtracting the terms in the above equation, then the above equation is written as,
$\Rightarrow 2910\alpha = 0.3$
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow \alpha = \dfrac{{0.3}}{{2910}}$
By dividing the terms in the above equation, then the above equation is written as,
$\Rightarrow \alpha = 1 \times {10^{ - 4}}$
Hence, the option (D) is the correct answer.
Note:The coefficient of the linear expansion is depending on the density substance in maximum temperature and density of the substance at zero temperature and the temperature of the substance. The temperature difference is the difference of the temperature from the maximum temperature to the minimum temperature.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

What is the maximum height of a projectile class 11 physics CBSE

Derive an expression for maximum height and range of class 11 physics CBSE

Describe various types of epithelial tissues with the class 11 biology CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

State and derive law of conservation of momentum class 11 physics CBSE

