Answer:
B. 6 cm
Explanation:
First, we calculate the spring constant of a single spring:
[tex]k = \frac{F}{\Delta x}\\[/tex]
where,
k = spring constant of single spring = ?
F = Force Applied = 10 N
Δx = extension = 4 cm = 0.04 m
Therefore,
[tex]k = \frac{10\ N}{0.04\ m}\\k = 250\ N/m\\[/tex]
Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:
[tex]k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\[/tex]
For a load of 30 N, applying Hooke's Law:
[tex]\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\[/tex]
Hence, the correct option is:
B. 6 cm
