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: \( S_1 \)

Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is cm. 
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.

Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to: