Here is the solution to Question 5 on elasticity: Question 5: A light elastic string has natural length $0.8$ m and modulus of elasticity $100$ N. a) Find the tension in the string when its length is $1.2$ m. Step 1: Identify the given values. Natural length, $L_0 = 0.8$ m Modulus of elasticity, $\lambda = 100$ N Extended length, $L = 1.2$ m Step 2: Calculate the extension ($e$). The extension is the difference between the extended length and the natural length. $$e = L - L_0$$ $$e = 1.2 \text{ m} - 0.8 \text{ m}$$ $$e = 0.4 \text{ m}$$ Step 3: Apply Hooke's Law to find the tension ($T$). Hooke's Law states that $T = \frac{\lambda e}{L_0}$. $$T = \frac{(100 \text{ N})(0.4 \text{ m})}{0.8 \text{ m}}$$ $$T = \frac{40}{0.8} \text{ N}$$ $$T = 50 \text{ N}$$ The tension in the string is $\boxed{\text{50 N}}$. b) Find the length of the string when the tension is $150$ N. Step 1: Identify the given values. Natural length, $L_0 = 0.8$ m Modulus of elasticity, $\lambda = 100$ N Tension, $T = 150$ N Step 2: Apply Hooke's Law to find the extension ($e$). Rearrange Hooke's Law ($T = \frac{\lambda e}{L_0}$) to solve for $e$: $$e = \frac{T L_0}{\lambda}$$ $$e = \frac{(150 \text{ N})(0.8 \text{ m})}{100 \text{ N}}$$ $$e = \frac{120}{100} \text{ m}$$ $$e = 1.2 \text{ m}$$ Step 3: Calculate the extended length ($L$). The extended length is the natural length plus the extension. $$L = L_0 + e$$ $$L = 0.8 \text{ m} + 1.2 \text{ m}$$ $$L = 2.0 \text{ m}$$ The length of the string when the tension is $150$ N is $\boxed{\text{2.0 m}}$.