Final answer:
The spring constant of a suspended body (0.1 kg) stretching a spring by 10 cm is approximately 10 N/m, rounded from the calculated value of 9.8 N/m using Hooke's Law. Therefore correct answer is option B.
Step-by-step explanation:
The question asks us to find the spring constant (also known as the force constant) of a spring from which a body of mass 100 grams (0.1 kg) is suspended, causing it to stretch by 10 cm (0.1 m). To calculate the spring constant, we can use Hooke's Law, which is given by the equation F = kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
Firstly, we need to calculate the force (F) which is also the weight of the body in this case. Weight can be calculated by multiplying the mass (m) by the acceleration due to gravity (g), which is approximately 9.8 m/s². Thus, F = mg = 0.1 kg × 9.8 m/s² = 0.98 N. Now, applying Hooke's Law F = kx, we have 0.98 N = k × 0.1 m, which gives us k = 0.98 N / 0.1 m = 9.8 N/m.
However, since none of the multiple-choice options are exactly 9.8 N/m, and considering that the spring constant should be an integer value in this context, we round to the closest option, which is 10 N/m (Option B).