A) \[1:4:9\]
B) \[5:3:1\]
C) \[1:3:5\]
D) \[1:2:3\]
Correct Answer: C
Solution :
From equation of motion \[S=ut+\frac{1}{2}g{{t}^{2}}\] Since, the body is falling from rest \[S=\frac{1}{2}g{{t}^{2}}\] Suppose the distance travelled in \[t=2S,\,t=4S,\,\,t=6S\] And \[{{S}_{2}},{{S}_{4}},\] and \[{{S}_{6}}\]respectively Now, \[{{S}_{2}}=\frac{1}{2}g\,{{(2)}^{2}}=2g\] \[{{S}_{4}}=\frac{1}{2}g\,{{(4)}^{2}}=8g\] \[{{S}_{6}}=\frac{1}{2}g\,{{(6)}^{2}}=18g\] Therefore, the distance travelled in first two seconds \[{{({{S}_{i}})}_{2}}={{S}_{2}}-{{S}_{0}}=2g.\] \[{{({{S}_{m}})}_{2}}={{S}_{4}}-{{S}_{2}}\] \[{{({{S}_{f}})}_{2}}={{S}_{6}}-{{S}_{4}}=18g-8g=10g\] Hence, the ratio becomes \[=2g:6g:10g=1:3:5\]
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