Question

A wire ring of diameter 20cm is gently
lowered on a liquid surface so that its
plane is parallel to the surface and then
pulled up. The extra force applied is
0.084N. The surface tension of the
liquid is
(A) 0.066Nm-1
(B) 0.66Nm-1
(C) 6.6Nm-1
(D) 66Nm-1​

Answer 1

The surface tension of the liquid is 0.066 nm^-1

Option (A) is correct.

Explanation:

We are given that:

• The diameter of the ring = 20 cm
• An extra force applied on the ring = 0.084 N
• To Find: The surface tension of the liquid = ?

Solution:  

The formula of surface tension is given by.

Surface tension x 2 x π x r x 2 = Force exerted

Now radius = d / 2 = 20 / 2 = 10 cm

S x 2 x 22/7 x 10 x 2 = 0.084 N

S x 2 x 3.14 x 10 x 2 = 0.084 N

S x 6.28 x 10 x 2 = 0.084 N

S x  125.6 = 0.084 N

S = 0.084 / 125.6

S = 0.00066 Ncm^-1

S = 0.00066 x 100 Nm^-1

S = 0.066 Nm^-1

Answer 2

Explanation:

A wire ring of diameter 20 cm is gently  lowered on a liquid surface so that its  plane is parallel to the surface and then  pulled up. The extra force applied is  0.084 N. The surface tension of the  liquid is

(A) 0.066Nm-1

(B) 0.66Nm-1

(C) 6.6Nm-1

(D) 66Nm-1​

• Given According to question a wire ring of diameter 20 cm is gently lowered on a liquid surface. We need to find the surface tension of the liquid.
• surface tension will be inside and outside the ring.
• Therefore total surface tension is given by (2 π R x S) x 2 (inside and outside)
•                            2 x 22 / 7 x 10 x S x 2 = 0.084    (r = d/2 = 20 / 2 = 10)
•                                   44/7 x 20 x S = 0.084
•                                     88 S = 0.588
•                          Or S = 0.00668 Nm^-1

Reference link will be

https://brainly.in/question/16130242