The volume of the bulb of a hydrometer is 7.5 cm3 and the cross sectional area of stem is 0.3 cm2. The mass of the hydrometer is 10.5 g.
1. Find the relative density of liquid in which it floats with its bulb only immersed.
2. Find the length of the stem from the top of the bulb (i.e., the graduation mark) when immersed in a liquid of relative density (a) 1.3, (b) 1.2
Hence show that the graduations on the stem get closer with increase in relative density downwards.
Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM
Given, mass of hydrometer = 10.5 g, volume of bulb of hydrometer = 7.5 cm3, area of cross section of stem = 0.3 cm2.
1. To float in a liquid with bulb only immersed, the condition is :
Weight of hydrometer = Weight of liquid displaced by the immersed part (i.e.,bulb) of hydrometer
or 10.5 = 7.5 (where = density of liquid)
Hence R.D of liquid = 1.4
Thus graduation for R.D = 1.4 will be marked just at the top of the bulb.
2. Let the hydrometer be floating with length l1 from the top of the bulb when immersed in a liquid of relative density 1.3. Then for floatation,
Weight of hydrometer = Weight of liquid displaced by the bulb and length l1 of stem of hydrometer
or 10.5 = [7.5 + (l1 x 0.3)] x 1.3
on solving l1 = 1.923 cm
Thus the graduation for R.D = 1.3 will be at a distance 1.923 cm from the top of the bulb graduated for R.D equal to 1.4.
b. Let the hydrometer be floating with length l2 from the top of the bulb when immersed in a liquid of relative density 1.2. Then for floatation,
Weight of hydrometer = Weight of liquid displaced by the bulb and length l2 of stem of hydrometer.
or 10.5 = [7.5 + (l2 X 0.3)] X 1.2, On solving l2 = 4.166 cm
i.e., the graduation for R.D = 1.2 will be at distance 4.166 cm from the top of the bulb.
Thus, separation between the marks 1.2 and 1.3 is (4.166 - 1.923) cm= 2.243 cm, while the separation between the marks 1.3 and 1.4 is 1.923 cm. Hence the graduations on stem get closer downwards with the increase in relative density.
Answered by | 4th Jun, 2014, 03:23: PM
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