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An object of volume 40 cm and in the form of a cylinder is tied to the hook of a spring balance records the weight to be 125 gf. The object lowered in a liquid of density 0.8 GCM[tex]^{–3} [)tex], such that half of it is immersed in the liquid. What is the new reading on the spring balance ? |
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Answer» ct of VOLUME 40 cm and in the form of a CYLINDER is tied to the hook of a spring balance records the weight to be 125 gf. The OBJECT lowered in a LIQUID of density 0.8 GCM, such that half of it is IMMERSED in the liquid. What is the new reading on the spring balance ?New reading of the spring balance is 109 gf.Given:-Initial reading on the spring balance = 125 gfVolume of the object = 40 cmSo, volume of the object immersed in liquid = 40/2 = 20 cmVolume of the liquid displaced (V) = 20 cmDensity of the liquid (d) = 0.8 gmMass of liquid displaced (m) = V × d = (20 cm) × (0.8 gm)So, weight of liquid displaced (m) = 16 g.Upthrust = Weight of liquid displaced = 16 gf.Hence, new reading of the spring balance= (125 – 16) = 109 gf. |
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