Solveeit Logo
Login

Question

Question: The density of a substance at \(0\,{}^ \circ C\) is \(10\,gc{m^{ - 3}}\) and at \(100\,{}^ \circ C\)...

The density of a substance at 0 ∘C0\,{}^ \circ C is 10 gcm−310\,gc{m^{ - 3}} and at 100 ∘C100\,{}^ \circ C its density is 9.7 gcm−39.7\,gc{m^{ - 3}}. The coefficient of linear expansion of the substance is:
(A) 102{10^2}
(B) 10−2{10^{ - 2}}
(C) 10−3{10^{ - 3}}
(D) 10−4{10^{ - 4}}

Explanation

Solution

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,
ρ0=ρ(1+3α×ΔT){\rho _0} = \rho \left( {1 + 3\alpha \times \Delta T} \right)
Where, ρ\rho is the density of the substance in maximum temperature, ρ0{\rho _0} is the density of the substance at zero temperature, α\alpha is the coefficient of linear expansion of the substance and ΔT\Delta T is the difference in the temperatures.

Complete step by step answer:
Given that,
The density of a substance at 0 ∘C0\,{}^ \circ C is, ρ0=10 gcm−3{\rho _0} = 10\,gc{m^{ - 3}}
The density of a substance at 100 ∘C100\,{}^ \circ C is, ρ=9.7 gcm−3\rho = 9.7\,gc{m^{ - 3}}
Now,
The thermal expansion of the substance is determined by,
ρ0=ρ(1+3α×ΔT) ................(1){\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,
⇒10=9.7(1+3α×(100−0))\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,
⇒10=9.7(1+3α×100)\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,
⇒10=9.7(1+300α)\Rightarrow 10 = 9.7\left( {1 + 300\alpha } \right)
By multiplying the terms in the above equation, then the above equation is written as,
⇒10=9.7+2910α\Rightarrow 10 = 9.7 + 2910\alpha
By rearranging the terms in the above equation, then the above equation is written as,
⇒2910α=10−9.7\Rightarrow 2910\alpha = 10 - 9.7
By subtracting the terms in the above equation, then the above equation is written as,
⇒2910α=0.3\Rightarrow 2910\alpha = 0.3
By rearranging the terms in the above equation, then the above equation is written as,
⇒α=0.32910\Rightarrow \alpha = \dfrac{{0.3}}{{2910}}
By dividing the terms in the above equation, then the above equation is written as,
⇒α=1×10−4\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.