Question

If a 20 N force is applied on a spring, it stretches 2 cm. When 100 N force is applied to the spring (within its elastic range) its length change will be:

12 cm
10 cm
11 cm
13 cm

Answer 1

Answer:

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Explanation:

Answer 2

The change in length of the spring in the second case is 10cm. So, option 'B' is correct.

Given:-

Force applied on spring = 20N

Change in length = 2cm

Another Force applied on spring = 100N

To Find:-

The change in length of spring in the second case.

Solution:-

We can easily find out the change in length of spring in the second case by following these simple steps.

As

Force applied on spring (f1) = 20N

Change in length (x1) = 2cm

Another Force applied on spring (f2) = 100N

Change in length in second case (x2) =?

According to the formula of force on spring,

$f = k \times x$

where k is the spring constant.

Here since the spring is the same, then their spring constant must be also the same.

For case 1

$f1 = k \times x1$

$20 = k \times 2$

$k = \frac{20}{2}$

$k = 10$

For case 2

$f2 = k \times x2$

$100 = 10 \times x2$

$x2 = \frac{100}{10}$

$x2 = 10cm$

Hence, The change in length of spring in the second case is 10cm. So, option 'B' is correct.

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