A) 200
B) 50
C) 53
D) 100
Correct Answer: B
Solution :
Let k be the spring constant, then time period (T) of spring-mass system is given by \[T=2\pi \sqrt{\frac{m}{k}}\]. where m is mass. \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\] Given, \[{{m}_{1}}=200g,\,\,{{T}_{1}}=1\,s,\,\,{{T}_{2}}-0.5\,s\] \[\Rightarrow \] \[{{m}_{2}}={{m}_{1}}{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2}}\] \[\therefore \] \[{{m}_{2}}=200{{\left( \frac{0.5}{1} \right)}^{2}}\] \[=0.25\times 200=50\,g\]You need to login to perform this action.
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