A) \[\frac{10}{3}\]
B) \[\frac{5}{2}\]
C) \[\frac{1}{2}\]
D) \[\frac{2}{5}\]
Correct Answer: B
Solution :
The coordinates (x, y, z) of masses 1g, 2g, 3g and 4g are \[({{x}_{1}}=0,\,{{y}_{1}}=0,\,{{z}_{1}}=0)\,({{x}_{2}}=0,\,{{y}_{2}}=0,\,{{z}_{2}}=0)\] \[({{x}_{3}}=0,\,{{y}_{3}}=0,\,{{z}_{3}}=0)\,({{x}_{4}}=\alpha ,\,{{y}_{4}}=2\alpha ,\,{{z}_{4}}=3\alpha )\] \[{{x}_{CM}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}+{{m}_{3}}{{x}_{3}}+{{x}_{4}}{{x}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\] \[{{x}_{CM}}=\frac{4\alpha }{1+2+3+4}=\frac{4\alpha }{10}\] \[1=\frac{4\alpha }{10}\Rightarrow \alpha =\frac{5}{2}\] \[{{Y}_{CM}}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}+{{m}_{4}}{{y}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\] \[2=\frac{4\times 2\alpha }{10}\] \[\alpha =\frac{20}{8}=\frac{5}{2}\] \[{{z}_{CM}}=\frac{{{m}_{1}}{{z}_{1}}+{{m}_{2}}{{z}_{2}}+{{m}_{3}}{{z}_{3}}+{{m}_{4}}{{z}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\] \[3=\frac{4\times 3\alpha }{10}\] \[\alpha =\frac{5}{2}\]
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