In this article, we will solve the Unsolved Problems from BL Theraja Vol-1-Tutorial Problems No. 1.1 for better understanding of electrical resistivity.

**Problem-1 **Calculate the resistance of 100 m length of a wire having a uniform cross-sectional area of 0.1 mm^{2} if the wire is made of manganin having a resistivity of 50 × 10−8 Ω-m. If the wire is drawn out to three times its original length, by how many times would you expect its resistance to be increased ?

**Solution:**

R = ρ L / A (1)

Where,

R = resistance of the conductor (ohms, Ω)

ρ = resistivity of the conductor material (ohm metre, Ω-m)

L = length of conductor (m)

A = cross-sectional area of conductor (m^{2})

According to the given data, we have,

L= 100 m, A=0.1 mm^{2}, ρ= 50 x 10^{-8 }Ω m

R = 50×10^{-8} ………………………………. (from equation 1)

= 50×10^{-8}x10^{2}x10^{7}

R = 50×10^{-8}x10^{9}**R = 500 Ω**

Now, according to question if the wire is drawn out to three times its original length

The new length becomes,

L’ = 3L

As we know that Volume can be defined as, (Area of cross section multiplied by Length), and here the volume will remain same and hence this can be expressed as

V= AxL = A’x3L

A= 3A’

A’=A/3

Therefore, from equation 1,

Thus, the resistance will increase by 9 times.

**Problem-2** : **A cube of a material of side 1 cm has a resistance of 0.001 Ω between its opposite faces. If the same volume of the material has a length of 8 cm and a uniform cross-section, what will be the resistance of this length?**

**Solution:**

R = ρ L / A (1)

Where,

R = resistance of the conductor (ohms, Ω)

ρ = resistivity of the conductor material (ohm metre, Ω m)

L = length of conductor (m)

A = cross-sectional area of conductor (m^{2})

According to the given data,

L= 1 cm, R= 0.001 Ω (2)

R = ρ

As Volume is same and we know volume can be calculated by multiplying Area of cross section (A) and Length (L)

Hence,

**R’ = 0.064 Ω**

**Problem-3 : A lead wire and an iron wire are connected in parallel. Their respective specific resistances are in the ratio 49 : 24. The former carries 80 percent more current than the latter and the latter is 47 percent longer than the former. Determine the ratio of their cross-sectional area.**

According to the data given in question,

And now question is saying former (I_{1}) carried 80 % more current than the latter (I_{2})

That means,

If, I_{2 }= 1 A

Then, I_{1} = 1.8 A

Similarly for the length,

If, L_{1} = 1

Then, L_{2} = 1.47

Now,

And as the lead wire and iron wire connected in parallel. By using current division rule,

I_{1} = V_{1}/ R_{1} and I_{1} = V_{2}/R _{2} (2)

By equating above,

**Problem-4 : A rectangular metal strip has the following dimensions : x = 10 cm, y = 0.5 cm, z = 0.2 cm Determine the ratio of resistances R**_{x}, R_{y}, and R_{z} between the respective pairs of opposite faces.

_{x}, R

_{y}, and R

_{z}between the respective pairs of opposite faces.

**Solution:**

As we know the Relation between resistance to its resistivity and dimensions can be calculated by using formula,

R = ρ L / A

According to the question,

**Problem-5 : The resistance of a conductor 1 mm**^{2} in cross-section and 20 m long is 0.346 Ω. Determine the specific resistance of the conducting material.

^{2}in cross-section and 20 m long is 0.346 Ω. Determine the specific resistance of the conducting material.

**Solution:**

R = ρ L / A (1)

Where,

R = resistance of the conductor (ohms, Ω)

ρ = resistivity of the conductor material (ohm metre, Ω m)

L = length of conductor (m)

A = cross-sectional area of conductor (m2)

According to the given data, we have,

L= 20 m, A=1 mm^{2}, R= 0.346Ω

By using equation 1,

R = ρ L / A

ρ = RA / L

Specific resistance of the conducting material is, ρ = 1.73 x 10^{-6} Ω m

**Problem-6 : When a current of 2 A flows for 3 micro-seconds in a copper wire, estimate the number of electrons crossing the cross-section of the wire.**

**Solution:**

Given,

Current through the wire is, I = 2 A

Time taken for this current flow, t = 3 µ s

t = 3×10^{-6} s

Assume,

n = number of electrons that crossed the cross section of wire in time t

e = Charge of the electron

So, the total charge in wire can be calculated as

q = ne (where, q is total charge) (1)

Putting the values in above formula,

Therefore, the number of electrons crossing the cross-section of the wire is 3.75 x 10^{13}.

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