A) 8 cm
B) 40 cm
C) 80 cm
D) None of these
Correct Answer: D
Solution :
| (*) The moment 1 kg hits the platform, |
| \[1(v)+0=(1+3){{v}_{1}}\Rightarrow |(\sqrt{2gh})|=4{{v}_{1}}\] |
| \[\Rightarrow \]\[{{v}_{1}}=\frac{\sqrt{2gh}}{4}=\frac{\sqrt{2\times 10\times 100}}{4}=\sqrt{\frac{{{10}^{3}}}{8}}m/s\] |
| (Compression due to masses is negligible.) |
| Using energy conservation principle, |
| \[\frac{1}{2}Mv_{1}^{2}=\frac{1}{2}k{{x}^{2}};\] |
| \[x=\sqrt{\frac{M'}{k}{{v}_{1}}}=\sqrt{\frac{4}{\frac{5}{4}\times {{10}^{6}}}\times \frac{{{10}^{3}}\times {{10}^{4}}}{4}}cm\]\[=2cm\] |
| *None of the given options is correct. |
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