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A balance similar to a Roman steel yard is constructed suchthat the length of the scale on the beam of the balance is 50 cm and the least count on thescale is 1 mm. The zero mark of the scale is at a distance of 15 cmfrom the point of suspension of thebalance. The hook that isused to attach the loads and the centre of gravity of thebeam are at distance 8 cm and 5 cm, respectively,from its fulcrum. If the weight of the rider of the balance is 50 gf, find the weight of thebeam, the maximum load that can be measured using the balance and its least count. |
Answer» Solution :The information given in the question can be depicted as shown in the figure.  G is the POSITION of centre ofgravity. H is theposition of the hook, O is the fulcrum, A and B are zero cm mark and 50 cm mark of the scale on the beam Given OG=5 cm, OH = 8 cm andOA = 15 cm Mass of the rider R = 50 gf. Let x g be themass of the beam Then x (OG) = R(OA) `X=(R(OA))/(OG)=(50xx15)/(5)=150g` THUS, the weight of thebeamis 150 gf Let L be the maximum load that is attached to the balance. Then `L(OH)+ x(OG)=R(OB)` `L=(R(OB)-x(OG))/(OH)=(50(65)-150(5))/(8)` `=312.5 g` The distance moved by the rider on the scale from its zero mark is directlyproportional to the load attached. Thus, 50 cm (=500 mm) corresponds to 312.5 g. Thus, the least count of the balance = 0.625 g |
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