Problem:
A string has an unstretched length of 20cm. A 6N weight stretched it to 32cm. If 6N weight is replaced by 5N weight, its new stretched length is?

Solution:

To solve this problem, we need to use Hooke's law, which states that the force required to stretch a spring or a string is proportional to the extension of the spring or string.

Mathematically, we can write Hooke's law as:

F = kx

Where F is the force applied, x is the extension of the spring or string, and k is the spring constant.

In this problem, we are given that:

- The unstretched length of the string is 20cm.
- A 6N weight stretched the string to 32cm.

Therefore, we can calculate the spring constant as follows:

k = F/x
k = 6N/(32cm - 20cm)
k = 6N/12cm
k = 0.5 N/cm

Now, we need to find the new stretched length of the string when a 5N weight is applied. We can use Hooke's law again to solve this problem.

We know that the spring constant is the same for the string, so we can write:

F = kx

Where F is the force applied, x is the extension of the string, and k is the spring constant.

We are given that the force applied is 5N, so we can write:

5N = 0.5 N/cm * x

Solving for x, we get:

x = 10cm

Therefore, the new stretched length of the string when a 5N weight is applied is 20cm + 10cm = 30cm.

Answer:
The new stretched length of the string when a 5N weight is applied is 30cm.

L of string=20cm
change in L when 6N is applied =32-20= 12cm
change due to 1N= 12/6= 2cm
L when 5 N weight is applied =30 cm