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Oliga [24]
3 years ago
13

Two capillary tubes, made of the same substance, are lowered into a water bath. The radius of the larger tube is twice that of t

he smaller tube. If water rises a height 8.8 cm above the surface of the water bath in the smaller tube, how high (in cm) will it rise in the larger tube?
Physics
1 answer:
Alinara [238K]3 years ago
8 0

Answer: 4.4 cm.

Explanation:

Rise of water in the smaller tube= h1=8.8 centimeter(cm)

Radius of the smaller tube= r

Rise of water in the larger tube= h2 (in centimeter).

The radius of the larger tube is twice that of the smaller tube means that;

The radius in the larger tube is 2r ( 2 multiply by the radius r, of the smaller tube)

Using Jurin's law;

height or rise of liquid is inversely proportional to its radius, r.

That is; hr= constant.

Therefore, we have;

h1 × r1 = h2 × r2.

Rise in smaller tube × radius of the smaller tube = height of the larger tube × radius of the larger tube.

8.8 cm × r = h2 × 2r

= (8.8cm)r = (h2) 2r

Divide both sides by 2r, we then have;

8.8cm r/ 2r = h2

h2= 4.4cm

Therefore, the height or rise in large tube is half of that of the smaller tube.

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