To solve this problem, we will begin by understanding the formula for specific resistance \( S_1 \) in terms of the physical dimensions of the wire and demonstrate how the length of the wire affects specific resistance.
The specific resistance \( S_1 \) of a wire is given by:
\(S_1 = X\left(\frac{\pi r^2}{L}\right)\)
Here:
The formula originates from the basic relation of resistance in terms of resistivity, length, and area of cross-section:
\(R = \rho \left(\frac{L}{A}\right)\)
where:
Now, let's analyze what happens if the length of the wire is doubled. The new length, \( L' \), is:
\(L' = 2L\)
Substituting for \( L' \) in the expression for specific resistance, we get a new specific resistance \( S_1' \):
\(S_1' = X\left(\frac{\pi r^2}{L'}\right)\)
Rewriting this with \( L' = 2L \):
\(S_1' = X\left(\frac{\pi r^2}{2L}\right)\)
However, we need to reconsider as specific resistance \( S_1 \) should be independent of length since \( S_1 \) strictly corresponds to the resistivity of the material, not dependent on geometrical changes once established initially.
The specific resistance, or resistivity, essentially remains the same regardless of doubling the length because it is a material property. Thus, \( S_1' = S_1 \).
Therefore, the correct answer is that the value of the specific resistance remains \(S_1\), which matches the given correct option:
\(S_1\)
The specific resistance (or resistivity) of a material is defined as: \(Sl = X (\frac{πr^2}{L})\) ,
where X is the resistance, r is the radius, and L is the length of the wire. Specific resistance is a material property and does not change with changes in dimensions such as length or radius. Doubling the length of the wire affects the resistance X, but the specific resistance Sl remains unchanged.
The Correct answer is: \( S_1 \)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,




What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)