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) \[14.5\text{ }m/{{s}^{2}}\]
Correct Answer: A
Solution :
From equation of motion \[{{\nu }^{2}}={{u}^{2}}+2as\] where v is final velocity, u is initial velocity, a is retardation and s is distance travel. Given, \[\operatorname{v} = 100 m/s,\,\,u =200 m/s\] \[s=10\,cm\,\,=\,\,10\times {{10}^{-}}^{2}m\] Putting the numerical values in Eq. (a), we have \[\therefore \,\,\,\,\,~{{\left( 100 \right)}^{2}}={{\left( 200 \right)}^{2}}+2\times a\times 10\times 1{{0}^{-2}}\] \[\Rightarrow \,\,\,\, 30000= 20a\,\times 1{{0}^{-}}^{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,a=-\frac{30000}{20\,\,\times \,\,{{10}^{-2}}}\] \[\,=-15\times {{10}^{4}}\,m/{{s}^{2}}\] Minus sign shows negative acceleration, i.e., retardation.
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