A brass rod and a steel rod are both measured at $0^{\circ}$ C. Their lengths are found to be 150 cm and 150.2 cm. At what common temperature will their lengths be equal?
($\alpha_{steel} = 12 \times 10^{-6}/^{\circ} C, \alpha_{brass} = 18 \times 10^{-6}/^{\circ} C$)
A. 111.4$^{\circ} $C
B. 167.2$^{\circ}$C
C. 222.8$^{\circ}$C
D. 238.3$^{\circ}$C
Answer
654.6k+ views
Hint: Both the rods upon heating will expand linearly. With the use of formula for linear thermal expansion, we will get two equations with two unknowns. Substitute for one if the variables (length) to find the temperature.
Formula used:
If at an initial temperature, the length of a rod is known as l, then upon heating, the length of the rod becomes:
$l' = l[1 + \alpha \Delta T]$
Where $\Delta T$ is the difference between final and initial temperatures and $\alpha$ is the coefficient of linear thermal expansion.
Complete answer:
We are given:
Initial temperature = 0$^{\circ}$ C;
Initial length for steel, $l_{steel}$ = 150.2 cm;
Initial length for steel, $l_{brass}$ = 150 cm;
The coefficient for linear thermal expansion of steel $\alpha_{steel} = 12 \times 10^{-6} / ^{\circ}C$;
And brass $\alpha_{brass} = 18 \times 10^{-6} / ^{\circ}C$ .
Keeping these values one by one in the formula we get:
$l'_{steel} = 150.2[1 + 12 \times 10^{-6} \Delta T]$
$l'_{brass} = 150[1 + 18 \times 10^{-6} \Delta T]$
There are exactly two unknowns in these two equations.
Now as the initial temperature was 0$^{\circ}$ C, we could write $\Delta T$ =T - 0 = T$^{\circ}$ C. And after this temperature, the lengths of both the rods become equal. So, we have:
$l'_{steel} = l'_{brass} $ .
Therefore, upon equating these two we can get:
$150.2[1 + 12 \times 10^{-6} T] = 150[1 + 18 \times 10^{-6} T]$
$150.2 - 150 = (150 \times 18 - 150.2 \times 12) \times 10^{-6}$
$\dfrac{0.2 \times 10^6}{897.6} = T$
$T = 222.816^{\circ}$C
At this temperature, given brass and steel rods have equal lengths.
Therefore, the correct answer is option (C).
Note:
Unit conversions to S.I. units like from cm to m or Celsius to Kelvin are not required here as the answer is supposed to be in Celsius. Also, cm will cancel on both sides anyway so there is no requirement for any conversion of lengths to meter. In any other question requiring the same formula, one must check if the units on both sides completely match.
Formula used:
If at an initial temperature, the length of a rod is known as l, then upon heating, the length of the rod becomes:
$l' = l[1 + \alpha \Delta T]$
Where $\Delta T$ is the difference between final and initial temperatures and $\alpha$ is the coefficient of linear thermal expansion.
Complete answer:
We are given:
Initial temperature = 0$^{\circ}$ C;
Initial length for steel, $l_{steel}$ = 150.2 cm;
Initial length for steel, $l_{brass}$ = 150 cm;
The coefficient for linear thermal expansion of steel $\alpha_{steel} = 12 \times 10^{-6} / ^{\circ}C$;
And brass $\alpha_{brass} = 18 \times 10^{-6} / ^{\circ}C$ .
Keeping these values one by one in the formula we get:
$l'_{steel} = 150.2[1 + 12 \times 10^{-6} \Delta T]$
$l'_{brass} = 150[1 + 18 \times 10^{-6} \Delta T]$
There are exactly two unknowns in these two equations.
Now as the initial temperature was 0$^{\circ}$ C, we could write $\Delta T$ =T - 0 = T$^{\circ}$ C. And after this temperature, the lengths of both the rods become equal. So, we have:
$l'_{steel} = l'_{brass} $ .
Therefore, upon equating these two we can get:
$150.2[1 + 12 \times 10^{-6} T] = 150[1 + 18 \times 10^{-6} T]$
$150.2 - 150 = (150 \times 18 - 150.2 \times 12) \times 10^{-6}$
$\dfrac{0.2 \times 10^6}{897.6} = T$
$T = 222.816^{\circ}$C
At this temperature, given brass and steel rods have equal lengths.
Therefore, the correct answer is option (C).
Note:
Unit conversions to S.I. units like from cm to m or Celsius to Kelvin are not required here as the answer is supposed to be in Celsius. Also, cm will cancel on both sides anyway so there is no requirement for any conversion of lengths to meter. In any other question requiring the same formula, one must check if the units on both sides completely match.
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
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

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

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

