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Question: An iron ring measuring 15.00cm in diameter is to be shrunk on a pulley which is 15.05cm in diameter....

An iron ring measuring 15.00cm in diameter is to be shrunk on a pulley which is 15.05cm in diameter. All measurements refer to the room temperature 20∘C20{}^\circ C. At what temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when its temperature comes down to the room temperature. Coefficient of linear expansion of iron =12×10−6/∘C.=12\times {{10}^{-6}}/{}^\circ C.
A. temperature =278∘C278{}^\circ C, Strain = 33×10−333\times {{10}^{-3}}
B. temperature=298∘C298{}^\circ C, strain = 3.33×10−33.33\times {{10}^{-3}}
C. temperature=298∘C298{}^\circ C, strain= 3.33×10−53.33\times {{10}^{-5}}
D. temperature=278∘C278{}^\circ C, strain=3.33×10−53.33\times {{10}^{-5}}

Explanation

Solution

The change in the length of a material due to heat or any external force shows the elasticity of the material. Temperature is one of the external factors on which the dimension of an object depends. The coefficient of linear expansion of a material represents the ability of the material to change its length concerning the changes in the external physical-conditions.
According to the given data
Measuring of iron ring (l1)({{l}_{1}})is 15cm
Measuring of the pulley (l2)({{l}_{2}})is 15.5cm
Referring or initial temperature(T1)({{T}_{1}}) is 20∘C20{}^\circ C

Formula used:
l2=l1(1+αΔT) strain=(l2−l1)l1 \begin{aligned} & {{l}_{2}}={{l}_{1}}\left( 1+\alpha \Delta T \right) \\\ & strain=\dfrac{({{l}_{2}}-{{l}_{1}})}{{{l}_{1}}} \\\ \end{aligned}

Complete step by step answer:
When a material is heated to a certain temperature such that there is a change in the dimension (length) of the object. The new length of the material is dependent on the difference of temperature at two states, the coefficient of linear expansion, and the initial length of the material.
Mathematical the relationship is given by,
l2=l1(1+αΔT){{l}_{2}}={{l}_{1}}(1+\alpha \Delta T)
Where, l2{{l}_{2}} = final length of the material (ring)
l1{{l}_{1}} = initial length of the ring
α = coefficient of linear expansion of material (iron)
ΔT\Delta T= the difference in temperature at the initial and final length of the ring
By putting the values from the given data:
15.5=15(1+12×10−6ΔT)15.5=15(1+12\times {{10}^{-6}}\Delta T)
By rearranging, the change in the temperature is given by:
ΔT=(l2−l1)αl1 ΔT=0.512×10−6×15 ΔT=278∘C \begin{aligned} & \Delta T=\dfrac{({{l}_{2}}-{{l}_{1}})}{\alpha {{l}_{1}}} \\\ & \Delta T=\dfrac{0.5}{12\times {{10}^{-6}}\times 15} \\\ & \Delta T=278{}^\circ C \\\ \end{aligned}
The change in the temperature is:
ΔT=T2−T1\Delta T={{T}_{2}}-{{T}_{1}}
By taking the reference or initial temperature into consideration from the given data:
T2−20=278 T2=298∘C \begin{aligned} & {{T}_{2}}-20=278 \\\ & {{T}_{2}}=298{}^\circ C \\\ \end{aligned}
The strain experienced by the ring is expressed as:
strain=(l2−l1)l1strain=\dfrac{({{l}_{2}}-{{l}_{1}})}{{{l}_{1}}}
So as per the given information:
strain=(15.5−15)15 =3.33×10−3 \begin{aligned} & strain=\dfrac{(15.5-15)}{15} \\\ & =3.33\times {{10}^{-3}} \\\ \end{aligned}

So, the correct answer is “Option B”.

Note: The strain is the ratio of the changed length and the original length of the material. It occurs when an external force is applied to the object. For the value of the final temperature at which the required length can be achieved we have to add the change in the temperature with the reference temperature.