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| Hydrometer |
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| Hydrometer is an instrument used for determining the density of a liquid. It usually consists of a glass float with a long thin stem which is graduated. The glass float is a large hollow bulb which increases the buoyancy so that the hydrometer floats. The narrow stem increases the sensitivity of the hydrometer. The bottom of the hydrometer is made heavier by loading it with lead shots so that it floats vertically. |
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| Hydrometer |
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| It works on the principle of flotation. Is sinks more in a lighter liquid so as to displace more volume of the lighter liquid whose weight is equal to the weight of the hydrometer. Hence, it will sink less in a denser liquid so that it has to displace less volume of the denser liquid whose weight will be equal to the hydrometer. Thus, the lowest density is marked on the top of the stem of a hydrometer and the maximum density is marked at the bottom of the graduated stem. |
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| It is special type of hydrometer used for testing the purity of milk or to check richness of milk. It has a range of relative density 1.105 to 1.045. |
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| A battery hydrometer is shown in figure below. It is used for measuring the relative density of accumulator acid. It is kept inside a glass tube fitted with a rubber bulb at the upper end. The lower end of the glass tube is connected with a narrow tube which is made of acid resistant material. When in use, this end is submerged in the acid in the accumulator. |
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| A hydrometer used for testing the acid in a battery. |
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| When the rubber bulb is squeezed, the air escapes from the narrow tube at the other end of the glass tube. When the pressure is removed as there is a partial vacuum in the glass tube, the greater atmospheric pressure outside pushes acid up into the glass tube. Now the hydrometer floats in the acid. The density of the acid can then be read on the floating hydrometer. |
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| The acid in a fully charged cell should have a relative density of 1.25 to 1.30. A reading of less than 1.18 indicates that recharging is necessary. |
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| A body weighs 600 g in air and 400 g in water. Calculate (i) Upthrust on the body, (ii) Volume of the body, and (iii) Relative density of the solid. |
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| Weight of the substance in air = 600 g |
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| Weight of the substance in water = 400 g |
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| Lost weight of the substance in water = 600 - 400 = 200 g |
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| Upthrust acting on the body = Loss of weight of the substance in water |
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| = 200 g |
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| Weight of water displaced = 200 g |
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| Volume of water displaced = 200 cm3 |
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| Volume of the substance = 200 cm3 |
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| A piece of aluminium of volume 200 cm3 and density 2.7 gcm-3 is completely immersed in kerosene. Determine upthrust exerted on the piece of aluminium and also determine how much will it weigh in kerosene. (density of kerosene = 0.8 g cm-3) |
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| Weight of aluminium piece = V x d1 = 200 x 2.7 = 540 g |
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| Upthrust = Weight of the liquid displace = V x d2 = 200 x 0.8 = 160 gf |
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| Weight of aluminium in the kerosene = 540 - 160 = 380 g |
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| A slab of ice of volume 800 cm3 and of density 0.9 gcm-3 floats in water of density 1.1 gcm-3. What fraction of ice is above salt water? |
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| Weight of ice = V1 x d = 800 x 0.9 = 720 gf. |
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| Let x be the volume of ice immersed in salt water, |
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| Weight of ice = Weight of water displaced |
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| Volume of ice above water = 800 - 654.5 = 145.5 cm3. |
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| A solid weighs 600 g in air, 450 g in water and 480 g in a liquid. Find (i) volume of the solid, (ii) R.D. of the solid, and (iii) R.D. of the liquid. |
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| Weight of the solid in air = 600 g |
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| Weight of the solid in water = 450 g |
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| Loss of weight of the solid in water = 600 - 450 = 150 g |
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| Mass of water displaced = 150 g |
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| (i) Volume of water displaced = 150 cm3 |
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(ii) R.D. of the solid  |
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(iii) R.D. of the liquid  |
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| Loss of weight of the solid in liquid = Weight of solid in air - weight of solid in liquid |
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| = 600 - 480 = 120 g |
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| (i) A solid X weighing 400 g in air is suspended in water as shown. Using the above information, calculate the R.D. of the solid X. |
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| (ii) The solid X is then completely immersed in a liquid as shown. Calculate the R.D. of the liquid. |
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| Weight of the solid in air = 400 g |
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| Using principle of moments |
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| X x 20 = 400 x 15 |
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| Weight of the solid in water = 300 g |
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| Weight of the solid in liquid, |
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| X x 20 = 560 x 10 |
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| Loss of weight of the solid in water = Weight of the sold in air - weight of solid in water |
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| = 400 g - 300 g |
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| = 100 g |
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| Loss of weight of the solid in liquid = 400 - 280 = 120 g |
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| = 1.2 |
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| A hollow cylinder closed at one end weighs 85 g which floats vertically in water when 35 g of lead shots are added into it. If the depth of immersion is 10 cm, calculate the |
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| (i) upthrust acting on the cylinder, |
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| (ii) area of cross section of the cylinder, |
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| (iii) depth of immersion in a liquid of R.D. equal to 12. |
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| (i) Upthrust = Weight of the floating body = 85 + 35 = 120 g |
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| (ii) Area of cross-section of the cylinder = a |
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Mass of water displaced = a x 10 x 1  |
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| \ 10a = 120 |
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| Therefore, a = 12 cm2 |
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| (iii) Let the cylinder float in the liquid by submerging upto depth of h cm. |
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| V x d (of liquid) = 120 wt. of the solid |
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| V = h x a = h x 12 |
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| \ h x 12 x 1.2 = 120 |
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| = 8.3 cm |
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| The cylinder will float by submerging 8.3 cm in the liquid of R.D. = 1.2 |
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| A wooden block in the form of a cube of side 10 cm is floating in water with 4 cm above the surface of water. If the density of water is 1 gcm-3, find the density of wood. |
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| Now volume of the wooden block |
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| = 10 x 10 x 10 = 1000 cm3 |
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| When the block floats, the volume of water displaced by the submerged part |
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| = 10 x 10 x 6 = 600 cm3 |
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| Mass of this water |
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| = V x d = 600 x 1 = 600 g |
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| As the block is floating, |
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| Weight of the displaced water = Weight of the block |
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Hence, density of the block  |
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| = 0.6 gcm-3 |
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