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12 September, 03:09

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

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  1. 12 September, 03:42
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    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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