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\[{{s}_{2}},\text{ }{{s}_{4}}\]and\[{{s}_{6}}\]respectively Now, \[{{s}_{2}}=\frac{1}{2}g{{(2)}^{2}}=2g\] \[{{s}_{4}}=\frac{1}{2}g{{(4)}^{2}}=8g\] \[{{s}_{6}}=\frac{1}{2}g{{(6)}^{2}}=18g\] Hence, the distance travelled in first two seconds \[{{({{s}_{i}})}_{2}}={{S}_{2}}-{{S}_{0}}=2g\] \[{{({{S}_{m}})}_{2}}={{S}_{4}}-{{S}_{2}}=8g-2g=6g\] \[{{({{S}_{f}})}_{2}}={{S}_{6}}-{{S}_{4}}\] \[=18g-8g\] \[=10g\] Now the ratio becomes \[=2g:6g:10g\] \[=1:3:5\]
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