Unsolved Problems BL Theraja Vol-1-Tutorial Problems No. 1.1

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 mm2 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 (m2)

According to the given data, we have,
L= 100 m,        A=0.1 mm2,     ρ= 50 x 10-8 Ω m        
R = 50×10-8 ………………………………. (from equation 1)
= 50×10-8x102x107
R = 50×10-8x109
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,

problem1

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 (m2)

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 (I1) carried 80 % more current than the latter (I2)
That means,

If, I2 = 1 A
Then, I1 = 1.8 A

Similarly for the length,
If, L1 = 1
Then, L2 = 1.47

Now,

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

I1 = V1/ R1 and I1 = V2/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 Rx, Ry, and Rz 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 mm2 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 mm2,        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 1013.

Read Next :

Please follow and like us:

1 thought on “Unsolved Problems BL Theraja Vol-1-Tutorial Problems No. 1.1”

Leave a Comment