
Two rods A and B of identical dimensions are at temperature \[{30^ \circ }C\]. If A is heated up to \[{180^ \circ }C\] and B up to \[{T^ \circ }C\], then new lengths are the same. If the ratio of coefficients of linear expansions of A and B is 4:3, then the value of T is?
A. \[{270^ \circ }\]
B. \[{230^ \circ }\]
C. \[{250^ \circ }\]
D. \[{200^ \circ }\]
Answer
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Hint: In this question, the coefficient of expansion for both the rods is given, and both were heated from temperature \[{30^ \circ }C\] up to \[{180^ \circ }C\]and \[{T^ \circ }C\]respectively, so by using the change in length due to temperature formula, we will find the temperature \[{T^ \circ }C\].
Complete step by step answer:
The coefficients of linear expansions of rod A \[ = 4\]
Coefficients of linear expansions of rod B \[ = 3\]
The initial temperature of both the rods is \[{30^ \circ }C\]
It is said that both the rods are initially at the same length and then they are heated up to different temperature and due to rise in temperature both the rods expand up to the same length,
Now we know that the change in length due to the rise in temperature is \[\Delta l = l\alpha \Delta T - - (i)\],
Since both the rod expands to the same length; hence we can write \[\Delta {l_1} = \Delta {l_2} - - (ii)\]
Now it is said that the rod A was initially at \[{30^ \circ }C\] and it was heated up to \[{180^ \circ }C\], rod B was also initially at \[{30^ \circ }C\]and it was heated up to \[{T^ \circ }C\]; hence we can write equation (ii) as
$
\Delta {l_1} = \Delta {l_2} \\
l{\alpha _1}\Delta {T_1} = l{\alpha _2}\Delta {T_2} \\
\dfrac{{{\alpha _1}}}{{{\alpha _2}}} = \dfrac{{\Delta {T_2}}}{{\Delta {T_1}}} \\
$
Hence by substituting the values, we get
$
\dfrac{{{\alpha _1}}}{{{\alpha _2}}} = \dfrac{{\Delta {T_2}}}{{\Delta {T_1}}} \\
\dfrac{4}{3} = \dfrac{{T - 30}}{{180 - 30}} \\
\implies \dfrac{4}{3} = \dfrac{{T - 30}}{{150}} \\
\implies 600 = 3T - 90 \\
\implies 3T = 690 \\
\therefore T = {230^ \circ } \\
$
Hence the value of \[T = {230^ \circ }\].
So, the correct answer is “Option B”.
Note:
Change in length due to the rise in temperature is given by the formula \[\Delta l = l\alpha \Delta T\], where \[l\] is the initial length \[\alpha \] is the coefficient of thermal expansion, and \[\Delta T\] is the change in temperature.
Complete step by step answer:
The coefficients of linear expansions of rod A \[ = 4\]
Coefficients of linear expansions of rod B \[ = 3\]
The initial temperature of both the rods is \[{30^ \circ }C\]
It is said that both the rods are initially at the same length and then they are heated up to different temperature and due to rise in temperature both the rods expand up to the same length,
Now we know that the change in length due to the rise in temperature is \[\Delta l = l\alpha \Delta T - - (i)\],
Since both the rod expands to the same length; hence we can write \[\Delta {l_1} = \Delta {l_2} - - (ii)\]
Now it is said that the rod A was initially at \[{30^ \circ }C\] and it was heated up to \[{180^ \circ }C\], rod B was also initially at \[{30^ \circ }C\]and it was heated up to \[{T^ \circ }C\]; hence we can write equation (ii) as
$
\Delta {l_1} = \Delta {l_2} \\
l{\alpha _1}\Delta {T_1} = l{\alpha _2}\Delta {T_2} \\
\dfrac{{{\alpha _1}}}{{{\alpha _2}}} = \dfrac{{\Delta {T_2}}}{{\Delta {T_1}}} \\
$
Hence by substituting the values, we get
$
\dfrac{{{\alpha _1}}}{{{\alpha _2}}} = \dfrac{{\Delta {T_2}}}{{\Delta {T_1}}} \\
\dfrac{4}{3} = \dfrac{{T - 30}}{{180 - 30}} \\
\implies \dfrac{4}{3} = \dfrac{{T - 30}}{{150}} \\
\implies 600 = 3T - 90 \\
\implies 3T = 690 \\
\therefore T = {230^ \circ } \\
$
Hence the value of \[T = {230^ \circ }\].
So, the correct answer is “Option B”.
Note:
Change in length due to the rise in temperature is given by the formula \[\Delta l = l\alpha \Delta T\], where \[l\] is the initial length \[\alpha \] is the coefficient of thermal expansion, and \[\Delta T\] is the change in temperature.
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