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PLEASE ANSWER THIS FASTThe apparent weight of a piece of solid in water is 80 gm-wt and that in kerosene oil is 100 gm-wt (density of kerosene =0•8 gm/cm3)A. Calculate the weight of the piece of solid in air B. Calculate its volume C. What will be the apparent weight of the piece of solid in a liquid of density 1•4 gm/cm3?NO SPAMS PLEASE |
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Answer» Answer: (A) 160 gm-wt (B) 80 cc (C) 48 gm-wt Explanation: Let the weight of the piece of solid in air is W Weight of piece of solid in water = Weight of solid in air - bouyant force in water = Weight of solid in air - weight of water displaced by the solid Let the volume of the solid is V Then the weight of water displaced = V × density of water × g = V × 1 × g = VG Thus 80G= W - Vg .............. (1) Similarly for kerosene we can write the equation 100G = W - V × 0.8 × g or, 100g = W - 0.8Vg ........ (2) Multiplying eq (1) by 0.8 and subtracting it from eq (2) 100g - 64g = 0.2W or, 0.2W = 32g or, W = 32g/0.2 = 160g Therefore, the weight of the piece of solid in air is 160 gm-wt (B) Putting the value of W in eq (1) 80g = 160g - Vg or, V = 160 - 80 = 80 cm³ (C) Weight of the liquid displaced = V × density of the liquid × g = 80 × 1.4 × g = 112g Therefore, weight of the piece of solid in the liquid = Weight of the solid - Weight of the liquid displaced = 160g - 112g = 48G Thus, weight of the solid = 48 gm-wt. Hope this answer is helpful. |
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