Question

Two springs A and B fixed at the top and are stretched by 8 cm and 16 cm respectively, when loads of 20 N and 10 N are suspended at the lower ends. The ratio of the spring constants of the springs A and B is:

• A. \( 1 : 1 \)
• B. \( 2 : 1 \)
• C. \( 3 : 1 \)
• D. \( 4 : 1 \)

Hint:

In problems involving the spring constant, remember that the ratio of the spring constants can be found using Hooke's Law, which relates force, spring constant, and displacement. In this case, compare the force per unit displacement for each spring.

Answer 1

The Correct Option is D

The force on a spring is given by Hooke's Law: \[ F = k \cdot x \] where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the extension of the spring. For spring A: \[ F_A = k_A \cdot x_A \quad \text{where} \quad F_A = 20 \, \text{N}, \, x_A = 8 \, \text{cm} \] For spring B: \[ F_B = k_B \cdot x_B \quad \text{where} \quad F_B = 10 \, \text{N}, \, x_B = 16 \, \text{cm} \] Now, using Hooke’s Law for both springs: \[ k_A = \frac{F_A}{x_A} = \frac{20}{8} = 2.5 \, \text{N/cm} \] \[ k_B = \frac{F_B}{x_B} = \frac{10}{16} = 0.625 \, \text{N/cm} \] Thus, the ratio of the spring constants \( \frac{k_A}{k_B} \) is: \[ \frac{k_A}{k_B} = \frac{2.5}{0.625} = 4 : 1 \] Hence, the correct answer is option (4) \( 4 : 1 \).

Answer 2

Step 1: Use Hooke’s Law.
Hooke’s law relates the force \( F \) applied to a spring with the extension \( x \) as: \[ F = kx \Rightarrow k = \frac{F}{x} \] Where \( k \) is the spring constant.

Step 2: Calculate spring constants individually.
For spring A:
\[ k_A = \frac{F_A}{x_A} = \frac{20}{8} = 2.5 \, \text{N/cm} \]
For spring B:
\[ k_B = \frac{F_B}{x_B} = \frac{10}{16} = 0.625 \, \text{N/cm} \]
Step 3: Find the ratio.
\[ \frac{k_A}{k_B} = \frac{2.5}{0.625} = 4 \Rightarrow k_A : k_B = 4 : 1 \]
Final Answer: \( \boxed{4 : 1} \)