If a cylinder of diameter \[1.0cm\]at \[{30^o}C\]is to be slid into a hole of diameter \[0.9997cm\]in a steel plate at the same temperature, the minimum required rise in the temperature of the plate is: (coefficient of linear expansion of steel\[ = 12 \times {10^{ - 6}}{/^o}C\]
(A) \[{25^o}C\]
(B) \[{35^o}C\]
(C ) \[{45^o}C\]
(D) \[{55^o}C\]
Answer
588.9k+ views
Hint:We need to find the rise in temperature that can cause the hole in the steel plate to have the same diameter as that of the cylinder.
From the given diameters we first need to calculate the increase in diameter that is required in the hole. Applying the formula for thermal expansion we can then calculate the rise in temperature from the values of thermal coefficient and the rise in diameter given.
Complete Step-By-Step Solution:
As we can infer from the question, we need to find the increase in diameter of the hole such that the cylinder may fit in.
It is clear that for the cylinder to slide inside the hole of the steel plate, diameter of the cylinder as well as diameter of the hole in the steel plate must be equal.
Therefore, after expansion, the diameter of hole must be \[ = 1cm\]
Thus, the increase in diameter of the hole is\[ = 1cm - \]Initial diameter of the hole
So, Change in Diameter is \[ = 1cm - 0.9997cm = 0.003cm\]
Now, let us calculate the required rise in temperature.
We know, the formula of thermal expansion:
\[\Delta d = {d_o}\alpha \Delta t\]
Where,
\[\Delta d = \]Change in diameter of the hole \[{d_o} = \] Initial diameter of the hole
\[\alpha = \] Coefficient of linear expansion of steel
\[\Delta t = \] Rise in temperature
Putting the values as given in the question:
\[0.003 = 0.9997 \times 12 \times {10^{ - 6}} \times \Delta t\]
Rearranging the equation, we obtain:
\[\Delta t = \dfrac{{0.003}}{{0.9997 \times 12 \times {{10}^{ - 6}}}}C\]
Hence, we obtain:
\[\Delta t = {25^o}C\]
This is the required answer, hence option (A) is correct.
Note:Thermal expansion is a tendency of matter to change its state, area, volume. Thermal expansion occurs when due to application of heat, the molecules present states moving vigorously and taking up more space. This change in dimensions occurs over a narrow interval of temperature that is why it is referred to as linear coefficient of expansion.
From the given diameters we first need to calculate the increase in diameter that is required in the hole. Applying the formula for thermal expansion we can then calculate the rise in temperature from the values of thermal coefficient and the rise in diameter given.
Complete Step-By-Step Solution:
As we can infer from the question, we need to find the increase in diameter of the hole such that the cylinder may fit in.
It is clear that for the cylinder to slide inside the hole of the steel plate, diameter of the cylinder as well as diameter of the hole in the steel plate must be equal.
Therefore, after expansion, the diameter of hole must be \[ = 1cm\]
Thus, the increase in diameter of the hole is\[ = 1cm - \]Initial diameter of the hole
So, Change in Diameter is \[ = 1cm - 0.9997cm = 0.003cm\]
Now, let us calculate the required rise in temperature.
We know, the formula of thermal expansion:
\[\Delta d = {d_o}\alpha \Delta t\]
Where,
\[\Delta d = \]Change in diameter of the hole \[{d_o} = \] Initial diameter of the hole
\[\alpha = \] Coefficient of linear expansion of steel
\[\Delta t = \] Rise in temperature
Putting the values as given in the question:
\[0.003 = 0.9997 \times 12 \times {10^{ - 6}} \times \Delta t\]
Rearranging the equation, we obtain:
\[\Delta t = \dfrac{{0.003}}{{0.9997 \times 12 \times {{10}^{ - 6}}}}C\]
Hence, we obtain:
\[\Delta t = {25^o}C\]
This is the required answer, hence option (A) is correct.
Note:Thermal expansion is a tendency of matter to change its state, area, volume. Thermal expansion occurs when due to application of heat, the molecules present states moving vigorously and taking up more space. This change in dimensions occurs over a narrow interval of temperature that is why it is referred to as linear coefficient of expansion.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Which among the following are examples of coming together class 11 social science CBSE

Discuss the various forms of bacteria class 11 biology CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a labelled diagram showing structure of LS of class 11 biology CBSE

