A) 26s
B) 25s
C) 13s
D) 12s
Correct Answer: C
Solution :
Distance travelled in nth sec \[{{s}_{n}}=u+\frac{1}{2}g(2n-1)\] In case of freely falling body u = 0 So, \[{{s}_{n}}=\frac{1}{2}g(2n-1)\] ...(1) Now distance travelled in t sec, is \[s=\frac{1}{2}g\times {{(5)}^{2}}=\frac{25}{2}g\] ??(2) When\[t=5\text{ }sec\], then \[s=\frac{1}{2}g\times {{(5)}^{2}}=\frac{25}{2}g\] ...(2) According to the condition \[{{s}_{n}}=s\], so, .from eqs. (1) and (2) \[\frac{1}{2}g(2n-1)=\frac{g}{2}\times 25\] or \[2n-1=25\] or \[n=13\sec \]You need to login to perform this action.
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