Question

a needle 5cm long can just rest on the surface of the water without wetting. what is it's weight? surface tension of the water is 0.07N/m​

Answer 1

The weight of the needle is 0.007 N.

Explanation:

Surface tension is the force acting on the surface of any fluid or film due to the attractive forces on the surface of them. So it is measured as force acting per unit effective length on the surface. The effective length is the distance or the length which is in contact with the surface.

For example, in the present case, the needle can be in contact with the water body from both sides of the needle. So effective length of the needle will be 2 times the length of the needle. Then the surface tension on the water due to the needle will be

$Surface Tension T = \frac{Force}{effective length} = \frac{Force}{2*length of needle}$

As the length of the needle is given as 5 cm = 0.05 m and the surface tension of water is given as 0.07 N/m, then the weight of the needle will be equal to the force applied by the needle.

$Force = Surface Tension*2*length of the needle$

$Force = 0.07*2*0.05 = 0.007 N$

So, the weight of the needle is 0.007 N.

Answer 2

The weight of the needle = 0.007 N

Explanation:

Step 1 :

Given,

Length of the needle = 5 cm = 5 × $10^{(-2)}$ m

Surface tension of the water =0.07 N/m​

We need to find the weight of the needle.

Step 2 :

We have

weight = (surface tension) × 2l

where l is the length of the object

The total contact length between the needle and water is 2 times the length of the needle.

Therefore

weight of the needle = 0.07 × 2 × 5 × $10^{(-2)}$  = 0.007 N

Step 3 :

Answer :

The weight of the needle = 0.007 N