CBSE Class 11 Physics Chapter 9: Mechanical Properties of Solids NCERT Solutions

We are given the following information:

• Length of the steel wire, $L\_1\; =\; 4.7\; \backslash text\{\; m\}$
• Area of cross-section of the steel wire, $A\_1\; =\; 3.0\; \backslash times\; 10^\{-5\}\; \backslash ,\backslash mathrm\{m\}^2$
• Length of the copper wire, $L\_2\; =\; 3.5\; \backslash text\{\; m\}$
• Area of cross-section of the copper wire, $A\_2\; =\; 4.0\; \backslash times\; 10^\{-5\}\; \backslash ,\backslash mathrm\{m\}^2$
• The extension in both wires is the same, so $\backslash Delta\; L\_1\; =\; \backslash Delta\; L\_2\; =\; \backslash Delta\; L$.
• The load applied is the same for both wires, meaning the force $F$ is the same.

The formula for Young's modulus ($Y$) is given by:

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

For the steel wire, Young's modulus ($Y\_1$) is:

$$Y\_1\; =\; \backslash frac\{F\}\{A\_1\}\; \backslash times\; \backslash frac\{L\_1\}\{\backslash Delta\; L\}\; =\; \backslash frac\{F\}\{3.0\; \backslash times\; 10^\{-5\}\; \backslash ,\backslash mathrm\{m\}^2\}\; \backslash times\; \backslash frac\{4.7\; \backslash text\{\; m\}\}\{\backslash Delta\; L\}\; \backslash quad\; \backslash cdots\; (i)$$

For the copper wire, Young's modulus ($Y\_2$) is:

$$Y\_2\; =\; \backslash frac\{F\}\{A\_2\}\; \backslash times\; \backslash frac\{L\_2\}\{\backslash Delta\; L\}\; =\; \backslash frac\{F\}\{4.0\; \backslash times\; 10^\{-5\}\; \backslash ,\backslash mathrm\{m\}^2\}\; \backslash times\; \backslash frac\{3.5\; \backslash text\{\; m\}\}\{\backslash Delta\; L\}\; \backslash quad\; \backslash cdots\; (ii)$$

To find the ratio of the Young's modulus of steel to that of copper, we divide equation (i) by equation (ii):

$$\backslash frac\{Y\_1\}\{Y\_2\}\; =\; \backslash frac\{\backslash frac\{F\}\{3.0\; \backslash times\; 10^\{-5\}\}\; \backslash times\; \backslash frac\{4.7\}\{\backslash Delta\; L\}\}\{\backslash frac\{F\}\{4.0\; \backslash times\; 10^\{-5\}\}\; \backslash times\; \backslash frac\{3.5\}\{\backslash Delta\; L\}\}$$

We can cancel out the common terms $F$ and $\backslash Delta\; L$:

$$\backslash frac\{Y\_1\}\{Y\_2\}\; =\; \backslash frac\{4.0\; \backslash times\; 10^\{-5\}\}\{3.0\; \backslash times\; 10^\{-5\}\}\; \backslash times\; \backslash frac\{4.7\}\{3.5\}$$

$$\backslash frac\{Y\_1\}\{Y\_2\}\; =\; \backslash frac\{4.0\}\{3.0\}\; \backslash times\; \backslash frac\{4.7\}\{3.5\}\; =\; \backslash frac\{4\; \backslash times\; 4.7\}\{3\; \backslash times\; 3.5\}\; =\; \backslash frac\{18.8\}\{10.5\}$$

$$\backslash frac\{Y\_1\}\{Y\_2\}\; \backslash approx\; 1.79$$

Therefore, the ratio of the Young's modulus of steel to that of copper is approximately 1.79:1.

The provided figure shows a stress-strain curve for a material. We need to determine its Young's modulus and approximate yield strength from this graph.

(a) Young's Modulus:

Young's modulus ($Y$) is defined as the ratio of stress to strain within the elastic limit, where the curve is linear. From the graph, we can observe a point in the linear region:

• When the strain is $0.002$, the corresponding stress is $150\; \backslash times\; 10^6\; \backslash text\{\; N/m\}^2$.

Using the formula $Y\; =\; \backslash frac\{\backslash text\{Stress\}\}\{\backslash text\{Strain\}\}$:

$$Y\; =\; \backslash frac\{150\; \backslash times\; 10^6\; \backslash text\{\; N/m\}^2\}\{0.002\}$$

$$Y\; =\; \backslash frac\{150\; \backslash times\; 10^6\}\{2\; \backslash times\; 10^\{-3\}\}\; \backslash text\{\; N/m\}^2$$

$$Y\; =\; 75\; \backslash times\; 10^9\; \backslash text\{\; N/m\}^2\; =\; 7.5\; \backslash times\; 10^\{10\}\; \backslash text\{\; N/m\}^2$$

Thus, the Young's modulus for this material is $7.5\; \backslash times\; 10^\{10\}\; \backslash text\{\; N/m\}^2$.

(b) Approximate Yield Strength:

The yield strength is the maximum stress that a material can withstand before it begins to deform plastically (i.e., without returning to its original shape upon removal of the stress). This point is typically where the stress-strain curve deviates significantly from linearity, or it can be the point of maximum stress before fracture if no distinct yield point is shown.

Observing the graph, the stress reaches approximately $300\; \backslash times\; 10^6\; \backslash text\{\; N/m\}^2$ before the curve starts to flatten out considerably, indicating the onset of significant plastic deformation or the ultimate tensile strength.

Therefore, the approximate yield strength of this material is $300\; \backslash times\; 10^6\; \backslash text\{\; N/m\}^2$, which can also be written as $3\; \backslash times\; 10^8\; \backslash text\{\; N/m\}^2$.