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A spring balance has a scale that reads from 0 to 50 kg. the length of the scale is 20 cm. a block of mass m is suspended from this balance, when displaced annd released, it oscillates with a period 0.5 s. the value of m is (Take `g=10ms^(-2)`)A. 8 kgB. 12 kgC. 16 kgD. 20 kg |
Answer» Correct Answer - C The 20 cm length of the scaler reads upto 50 kg. `thereforeF=mg=(50kg)(10ms^(-2))=500N` and `x=20cm=0.2m` `therefore`Spring constant, `k=(F)/(x)=(500N)/(0.2m)=2500Nm^(-1)` As `T=2pisqrt((m)/(k))` Squaring both sides, we get `T^(2)=(4pi^(2)m)/(k)` `m=(T^(2)k)/(4pi^(2))=((0.5)^(2)xx(2500Nm^(-1)))/(4xx(3.14)^(2))=16kg` |
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