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Question: Statement 1: The temperature dependence of the resistance is usually given as \( R = {R_0}\left( {1 ...

Statement 1: The temperature dependence of the resistance is usually given as R=R0(1+αΔT)R = {R_0}\left( {1 + \alpha \Delta T} \right). The resistance of a wire changes from 100Ω100\Omega to 150Ω150\Omega when its temperature is increased from 27C{27^ \circ }C to 227C{227^ \circ }C. This implies that α=2.5×103/C\alpha = 2.5 \times {10^{ - 3}}{/^ \circ }C .
Statement 2: R=R0(1+αΔT)R = {R_0}\left( {1 + \alpha \Delta T} \right) is valid only when the change in the temperature ΔT\Delta T is small and ΔR=(RR0)<<R0\Delta R = \left( {R - {R_0}} \right) < < {R_0} .
(A) Statement 1 is True, Statement 2 is False.
(B) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1.
(C) Statement 1 is True, Statement 2 is True; Statement 2 is not the correct explanation for Statement 1.
(D) Statement 1 is False, Statement 2 is True.

Explanation

Solution

Hint As temperature increases of a wire, it’s resistance also increases. We are having a relation given for change in resistance with respect to temperature. So we can get a solution using this equation.

Complete step by step answer
The temperature dependence of the resistance in the question is given as
R=R0(1+αΔT)\Rightarrow R = {R_0}\left( {1 + \alpha \Delta T} \right)
Now, according to the question, we have R0=100Ω{R_0} = 100\Omega, R=150ΩR = 150\Omega, T0=27C{T_0} = {27^ \circ }C, and T=227CT = {227^ \circ }C .
So, ΔT=TT0\Delta T = T - {T_0}
ΔT=227C27C=200C\Rightarrow \Delta T = {227^ \circ }C - {27^ \circ }C = {200^ \circ }C
Putting these values in (1) we get
150=100(1+200α)\Rightarrow 150 = 100\left( {1 + 200\alpha } \right)
200α=0.5\Rightarrow 200\alpha = 0.5
On solving we get
α=2.5×104/C\Rightarrow \alpha = 2.5 \times {10^{ - 4}}{/^ \circ }C
This value matches with the value given in the Statement 1.
But while deriving the expression for the variation of the resistance with the temperature, it is assumed that the change in temperature is very small. But in this case, the change in temperature is
ΔT=200C\Rightarrow \Delta T = {200^ \circ }C
This is a quite large value.
Also, while the derivation of the equation (1) is carried out, it is assumed that the change in resistance is very small compared to the original value, that is
(RR0)<<R0\Rightarrow \left( {R - {R_0}} \right) < < {R_0}
But in this case the change in resistance
(RR0)=150100\Rightarrow \left( {R - {R_0}} \right) = 150 - 100
(RR0)=50Ω\Rightarrow \left( {R - {R_0}} \right) = 50\Omega
Which is comparable to the original value of resistance. So the above equation (1) cannot be applied to this case. Thus the value of α\alpha which is obtained above is incorrect.
Thus the Statement 1 is False.
Also the Statement 2 is True at the same time due to the reasons already stated above.
Hence the correct answer is option D.

Note
Do not blindly jump to the conclusion that Statement 1 is correct after getting the value of α\alpha same as that given in the Statement. That value is intentionally given to be the same. The question basically wants to judge the knowledge of the concept mentioned in the Statement 2.