A spring 20cm is stretched to 25cm by load of 50n. What will be it's length when stretched by 100n. Assuming that the elastic limit is not reached. 40cm,10cm,35cm,40cm,50cm.

Answer 1

We can use Hooke's law to solve this problem:

F = kx

where F is the force applied, x is the displacement from the equilibrium length, and k is the spring constant.

We can first find the spring constant using the initial data:

k = F/x = 50 N / (25 cm - 20 cm) = 10 N/cm

Now we can use this spring constant to find the displacement at 100 N:

x = F/k = 100 N / 10 N/cm = 10 cm

Therefore, the length of the spring when stretched by 100 N will be:

25 cm + 10 cm = 35 cm

So the answer is option C, 35 cm.