A) \[15\times {{10}^{4}}m/{{s}^{2}}\]
B) \[10\times {{10}^{4}}m/{{s}^{2}}\]
C) \[12\times {{10}^{4}}m/{{s}^{2}}\]
D) \[14.5m/{{s}^{2}}\]
Correct Answer: A
Solution :
From equation of motion \[{{m}_{1}}:2{{m}_{2}}:3{{m}_{3}}\] ?...(1) where v is final velocity, u is initial velocity, a is retardation and s is distance travel. Given, v = 100 m/s, u = 200 m/s \[{{m}_{1}}:{{m}_{2}}:{{m}_{3}}\] Putting the numerical values in Eq. (1), we have \[\overset{0}{\mathop{A}}\,\] \[\mu \] \[\alpha \] \[v=\omega \sqrt{{{a}^{2}}-{{x}^{2}}}\] \[\sqrt{g}\] \[{{C}_{1}}<{{C}_{2}}<{{C}_{3}}\] \[{{C}_{p}}>{{C}_{3}}.\] Minus sign shows negative acceleration i.e., retardation.
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