1. **Stating the problem:** Calculate the torque when a force of 200 N is applied perpendicular to a wrench of length 25 cm. Then find the torque if the wrench length is doubled to 50 cm. 2. **Recall the torque formula:** Torque $\tau$ is given by: $$\tau = F \times r \times \sin(\theta)$$ where $F$ is the force, $r$ is the lever arm length, and $\theta$ is the angle between the force and lever arm. 3. **Applying given values:** Since force is perpendicular to the wrench, $\theta = 90^\circ$, so $\sin(90^\circ) = 1$. Convert wrench length from cm to meters: $25\text{ cm} = 0.25\text{ m}$. 4. **Calculate torque for the first wrench:** $$\tau_1 = 200 \times 0.25 \times 1 = 50\text{ Nm}$$ 5. **Calculate torque for the doubled wrench length:** New length: $2 \times 0.25 = 0.50\text{ m}$. $$\tau_2 = 200 \times 0.50 \times 1 = 100\text{ Nm}$$ 6. **Interpretation:** Doubling the wrench length doubles the torque applied on the bolt. **Final answers:** - Torque with 25 cm wrench: $50\text{ Nm}$ - Torque with 50 cm wrench: $100\text{ Nm}$