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QJEEM24Phc5cd: JEE Main 2024 Physics - Class 12 JEEM24Phc5cd Solution: Wheatstone bridge principle is used to measure the specific…

Wheatstone bridge principle is used to measure the specific resistance (S1)\left(S_1\right) of given wire, having length LL , radius rr . If XX is the resistance of wire, then specific resistance is ; S1=X(πr2L)S_1=X\left(\frac{\pi r^2}{L}\right) . If the length of the wire gets doubled then the value of specific resistance will be :

Multiple Choice Options

  • A S14\frac{S_1}{4}
  • B 2 S12 \mathrm{~S}_1
  • C S12\frac{\mathrm{S}_1}{2}
  • D S1S_1

Step-by-Step Verified Solution

The specific resistance (or resistivity) of a material is a fundamental property that describes how much the material resists the flow of electric current. The resistivity is typically denoted by the symbol ρ\rho (rho), and it can be calculated by using the resistance XX of a uniform specimen of the material, along with its physical dimensions. In the case of a wire, the resistivity formula in terms of its resistance XX , length LL , and cross-sectional area A=πr2A=\pi r^2 is given by:

ρ=X⋅AL\rho = \frac{X \cdot A}{L}

This formula is a reinterpretation of Ohm's law, and it states that the specific resistance is proportional to the area of the cross-section of the wire and inversely proportional to its length.

Given that the specific resistance of the wire S1S_1 is determined using the formula:

S1=X(πr2L)S_1 = X \left(\frac{\pi r^2}{L}\right)

Now, let's see what happens to the specific resistance if the length of the wire is doubled. If we denote the new length as 2L2L , the resistance of the wire with the new length will change because resistance is directly proportional to the length of the wire. However, resistivity (specific resistance) is an intrinsic property of the material and does not depend on its length or shape, only on its temperature. Therefore, even if we change the length of the wire, the specific resistance should remain the same.

If we calculate the new resistance X′X' with the doubled length, it would be:

X′=X(2LL)=2XX' = X \left(\frac{2L}{L}\right) = 2X

Thus, we would use the new resistance X′X' and the new length 2L2L to calculate the specific resistance again:

S1′=X′(πr22L)=2X(πr22L)=X(πr2L)S_1' = X' \left(\frac{\pi r^2}{2L}\right) = 2X \left(\frac{\pi r^2}{2L}\right) = X \left(\frac{\pi r^2}{L}\right)

Since this formula is essentially the same as our original formula for S1S_1 , we can conclude that:

S1′=S1S_1' = S_1

Therefore, the value of the specific resistance S1S_1 will remain the same even if the length of the wire gets doubled. The correct answer to the question is:

Option D: S1S_1

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