Physics, asked by espy404, 1 year ago

At a temperature 20°C the volume of a certain glass flask up to a certain mark on the stem of the flask,
is exactly 100 cm. The flask is filled up to this mark with a liquid at 20°C. The cross section of the stem
is 1 mm- and can be considered constant. How far will the liquid rise or fall in the stem when the
temperature is raised to 40°C? Given, volume expansion coefficient of the liquid is 120 x 10 per 'c,
and linear expansion coefficient of glass is 8 * 10 per degree c.​

Answers

Answered by aditi8272
0

Explanation:

Volume expansion coefficient of glass=3×linear=3α

Change in temperature is 38−18=200C

Increment of volume of flask is ΔV=50cc×3α×200C=0.027cc

Expansion in mercury 50cc×γΔt=50×180×10−6×20=0.180cc

Apparent increment=0.180−0.027=0.153cc

Option C is correct.

Answered by dualadmire
0

Given:

Initial temperature = 20°C

Final temperature = 40°C

Initial height of liquid = 100 cm

Cross section of the stem = 1mm

Coefficient of volume expansion = 120 * 10^-3/°C

Coefficient of linear expansion = 8* 10^-3/°C

To find:

How far will the liquid rise or fall.

Solution:

The formula for linear expansion is:

ΔL = α L ΔT

Where ΔL = change in length

α = linear expansion of coefficient

L = initial length

ΔT = change in temperature

L- 100 = 8*10 * 100 (40 - 20)

L - 100 = 8 * 10^-3 * 100 * (20)

L - 100 = 16

L = 116 cm

Therefore, the liquid will rise up to 116 cm.

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