
Two rods of different materials having coefficients of linear expansion ${\alpha _1},{\alpha _2}$ and Young's moduli \[{Y_1}\] and \[{Y_2}\] respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. The rods are heated such that they undergo the same increases in temperature. There is no bending of rods. If\[{\alpha _1}:{\alpha _2} = 3:2\]. The thermal stress developed in the two rods are equally provided \[{Y_1}:{Y_2}\] is equal to-
A. \[2:3\]
B. \[1:1\]
C. \[3:2\]
D. \[4:9\]
Answer
497.4k+ views
Hint: We can see from the question that the increase in temperature is the same for both the rods and as they are fixed between two walls, their initial lengths will also be equal. We use the formula which relates the Young’s Modulus to the stress and strain of the rods. From this method, we can obtain the required ratio.
Complete step by step solution:
Let us consider the length of the rods to be l, the respective changes in the lengths due to heating will be as given below
For the first rod we can take$\Delta {l_1} = l{\alpha _1}\Delta t$, and for the second rod we get $\Delta {l_2} = l{\alpha _2}\Delta t$
Now the strain can be given due to the change in its initial length as
For the first rod, we get \[\dfrac{{\Delta {l_1}}}{l} = {\alpha _1}\Delta t\] and for the second rod \[\dfrac{{\Delta {l_2}}}{l} = {\alpha _2}\Delta t\]
Now we know that the stress is given by the product of the young’s modulus and strain, and substituting for strain we get the equation
\[{Y_1}{\alpha _1}\Delta t = {Y_2}{\alpha _2}\Delta t\]
Given that \[{\alpha _1}:{\alpha _2} = 3:2\] we get the final ratio as,
$\dfrac{{{Y_1}}}{{{Y_2}}} = \dfrac{3}{2}$
Note:
The rise in length of any side of a body is influenced by the original length\[{l_0}\], the temperature rise \[t,\] , and the coefficient of linear expansion\[\alpha \]. The coefficient of linear expansion depends on the temperature of the object. It varies from substance to substance and it can be defined as the rise in length per unit length when the temperature is raised by \[1^\circ C\].
Complete step by step solution:
Let us consider the length of the rods to be l, the respective changes in the lengths due to heating will be as given below
For the first rod we can take$\Delta {l_1} = l{\alpha _1}\Delta t$, and for the second rod we get $\Delta {l_2} = l{\alpha _2}\Delta t$
Now the strain can be given due to the change in its initial length as
For the first rod, we get \[\dfrac{{\Delta {l_1}}}{l} = {\alpha _1}\Delta t\] and for the second rod \[\dfrac{{\Delta {l_2}}}{l} = {\alpha _2}\Delta t\]
Now we know that the stress is given by the product of the young’s modulus and strain, and substituting for strain we get the equation
\[{Y_1}{\alpha _1}\Delta t = {Y_2}{\alpha _2}\Delta t\]
Given that \[{\alpha _1}:{\alpha _2} = 3:2\] we get the final ratio as,
$\dfrac{{{Y_1}}}{{{Y_2}}} = \dfrac{3}{2}$
Note:
The rise in length of any side of a body is influenced by the original length\[{l_0}\], the temperature rise \[t,\] , and the coefficient of linear expansion\[\alpha \]. The coefficient of linear expansion depends on the temperature of the object. It varies from substance to substance and it can be defined as the rise in length per unit length when the temperature is raised by \[1^\circ C\].
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