A) 1 : 4 : 9
B) 1 : 2 : 4
C) 1 : 3 : 5
D) 1 : 2 : 3
Correct Answer: C
Solution :
Using the relation \[s=ut+\frac{1}{2}g{{t}^{2}}\] As the body is falling from rest,\[u=0\] \[s=\frac{1}{2}g{{t}^{2}}\] Suppose the distance travelled in \[t=2s,\,\,t=4s,\,\,t=6s\] are \[{{s}_{2}},\,\,{{s}_{4}}\] and \[{{s}_{6}}\] respectively. Now, \[{{s}_{2}}=\frac{1}{2}g{{(2)}^{2}}=2\,\,g\] \[{{s}_{4}}=\frac{1}{2}g{{(4)}^{2}}=8\,\,g\] \[{{s}_{6}}=\frac{1}{2}g{{(6)}^{2}}=18\,\,g\] Hence, the distance travelled in First two seconds \[{{({{S}_{i}})}_{2}}={{s}_{2}}-{{s}_{0}}=2\,\,g\] \[{{({{s}_{m}})}_{2}}={{s}_{4}}-{{s}_{2}}\] \[=8\,\,g-2\,\,g\] \[=6\,\,g\] \[{{({{s}_{f}})}_{2}}={{s}_{6}}-{{s}_{4}}\] \[=18\,\,g-8\,\,g\] \[=10\,\,g\] Now, the ratio becomes \[=2g:6g:10s\] \[=1:3:5\]
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