A cable stretched by an amount d when it supports a crate of mass M. The cable is replaced by another cable of the same material having the same length and twice the diameter. If the same crate is supported by the thicker cable, by how much will the cable stretch ?

Options

A. `(d)/(4)`
B. `(d)/(2)`
C. d
D. 2d

Answer

To solve the problem step by step, we will use the concepts of Young's modulus and the relationship between stress, strain, and the properties of the material.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a cable that stretches by an amount \( d \) when a crate of mass \( M \) is hung from it.
- We replace this cable with another cable made of the same material, having the same length but twice the diameter.

2. **Identify the Relevant Formulas**:
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Where:
- \( F \) is the force applied (weight of the crate, \( mg \)),
- \( A \) is the cross-sectional area of the cable,
- \( \Delta L \) is the change in length (stretch),
- \( L \) is the original length.

3. **Calculate the Stress and Strain for the First Cable**:
- For the first cable:
- The force \( F = Mg \).
- The cross-sectional area \( A_1 = \frac{\pi d^2}{4} \) (where \( d \) is the diameter of the first cable).
- The strain is given by \( \text{Strain} = \frac{d}{L} \).
- The stress can be expressed as:
\[ \text{Stress} = \frac{F}{A_1} = \frac{Mg}{\frac{\pi d^2}{4}} = \frac{4Mg}{\pi d^2} \]
- Therefore, we can write Young's modulus for the first cable:
\[ Y = \frac{\frac{4Mg}{\pi d^2}}{\frac{d}{L}} \implies Y = \frac{4MgL}{\pi d^3} \]

4. **Calculate the Stress and Strain for the Second Cable**:
- For the second cable (with diameter \( 2d \)):
- The cross-sectional area \( A_2 = \frac{\pi (2d)^2}{4} = \frac{\pi (4d^2)}{4} = \pi d^2 \).
- The stress in the second cable is:
\[ \text{Stress} = \frac{F}{A_2} = \frac{Mg}{\pi d^2} \]
- The strain for the second cable can be expressed as:
\[ \text{Strain} = \frac{\Delta L_2}{L} \]
- Using Young's modulus for the second cable:
\[ Y = \frac{\frac{Mg}{\pi d^2}}{\frac{\Delta L_2}{L}} \implies Y = \frac{MgL}{\pi d^2 \Delta L_2} \]

5. **Equating Young's Modulus**:
- Since both cables are made of the same material, Young's modulus \( Y \) is the same for both cables:
\[ \frac{4MgL}{\pi d^3} = \frac{MgL}{\pi d^2 \Delta L_2} \]
- Canceling \( MgL \) from both sides:
\[ \frac{4}{d^3} = \frac{1}{d^2 \Delta L_2} \]
- Rearranging gives:
\[ \Delta L_2 = \frac{d^2}{4} \]

6. **Final Stretch Calculation**:
- Since the original stretch \( d \) was for the first cable, the stretch for the second cable is:
\[ \Delta L_2 = \frac{d}{4} \]

### Conclusion:
The new stretch in the thicker cable is \( \frac{d}{4} \).