Question

A same force is acting on two wires made of same material. One wire has length $l$ and diameter $d$. The other wire has length $\left(\frac{l}{2}\right)$ and diameter $2d$. If the extensions in the two wires are $l_1$ and $l_2$ such that $l_1 + l_2 = 1$cm, then the values of $l_1$ and $l_2$ are

• A. 0.80 cm, 0.20 cm
• B. 0.89 cm, 0.11 cm
• C. 0.90 cm, 0.10 cm
• D. 0.95 cm, 0.05 cm

Answer 1

Answer: (B)

0.89 cm, 0.11 cm

Answer 2

The extension $\Delta L$ of a wire under a force $F$ is given by the formula derived from Young's Modulus:

$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$

Rearranging for $\Delta L$, we get:

$\Delta L = \frac{FL}{AY}$

where $L$ is the original length, $A$ is the cross-sectional area, and $Y$ is Young's modulus of the material.

The cross-sectional area of a wire with diameter $d$ is $A = \pi (d/2)^2 = \frac{\pi d^2}{4}$.

So, the extension can be written as:

$\Delta L = \frac{FL}{(\pi d^2/4)Y} = \frac{4FL}{\pi d^2 Y}$

For the first wire:

Length $L_1 = l$ Diameter $d_1 = d$ Extension $\Delta L_1 = l_1$

Since the material is the same, Young's modulus $Y_1 = Y$. The force is the same, $F_1 = F$.

$l_1 = \frac{4Fl}{\pi d^2 Y}$

For the second wire:

Length $L_2 = l/2$ Diameter $d_2 = 2d$ Extension $\Delta L_2 = l_2$

Material is the same, $Y_2 = Y$. Force is the same, $F_2 = F$.

$l_2 = \frac{4F(l/2)}{\pi (2d)^2 Y} = \frac{4Fl/2}{\pi (4d^2) Y} = \frac{2Fl}{4\pi d^2 Y} = \frac{Fl}{2\pi d^2 Y}$

Now, let's find the ratio of the extensions $l_1$ and $l_2$:

$\frac{l_1}{l_2} = \frac{\frac{4Fl}{\pi d^2 Y}}{\frac{Fl}{2\pi d^2 Y}}$

$\frac{l_1}{l_2} = \frac{4Fl}{\pi d^2 Y} \times \frac{2\pi d^2 Y}{Fl}$

$\frac{l_1}{l_2} = \frac{4 \times 2}{1} = 8$

So, $l_1 = 8l_2$.

We are given that the sum of the extensions is 1 cm:

$l_1 + l_2 = 1$ cm

Substitute $l_1 = 8l_2$ into the sum equation:

$8l_2 + l_2 = 1$ $9l_2 = 1$ $l_2 = \frac{1}{9}$ cm

Now find $l_1$ using $l_1 = 8l_2$:

$l_1 = 8 \times \frac{1}{9} = \frac{8}{9}$ cm

The exact values are $l_1 = \frac{8}{9}$ cm and $l_2 = \frac{1}{9}$ cm.

Let's convert these to decimal values:

$l_1 = \frac{8}{9} \approx 0.888...$ cm $l_2 = \frac{1}{9} \approx 0.111...$ cm

Comparing these values with the given options, option (b) provides values 0.89 cm and 0.11 cm. These are the values of $8/9$ and $1/9$ rounded to two decimal places. $0.89 + 0.11 = 1.00$ cm (Matches the sum) $0.89 / 0.11 \approx 8.09$ (Close to the calculated ratio 8, the difference is due to rounding).

Thus, the values of $l_1$ and $l_2$ are approximately 0.89 cm and 0.11 cm.