A) All equal, being equal to \[\sqrt{2/g}\] second
B) In the ratio of the square roots of the integers 1, 2, 3.....
C) (c) In the ratio of the difference in the square roots of the integers i.e. \[\sqrt{1},\,(\sqrt{2}-\sqrt{1}),\,(\sqrt{3}-\sqrt{2}),\,(\sqrt{4}-\sqrt{3})\]...
D) In the ratio of the reciprocal of the square roots of the integers i.e.,. \[\frac{1}{\sqrt{1}},\,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{4}}\]
Correct Answer: C
Solution :
\[h=ut+\frac{1}{2}g{{t}^{2}}\Rightarrow 1=0\times {{t}_{1}}+\frac{1}{2}gt_{1}^{2}\Rightarrow {{t}_{1}}=\sqrt{2/g}\] Velocity after travelling 1m distance \[{{v}^{2}}={{u}^{2}}+2gh\Rightarrow {{v}^{2}}={{(0)}^{2}}+2g\times 1\Rightarrow v=\sqrt{2g}\] For second 1 meter distance \[1=\sqrt{2g}\times {{t}_{2}}+\frac{1}{2}gt_{2}^{2}\Rightarrow gt_{2}^{2}+2\sqrt{2g}{{t}_{2}}-2=0\] \[{{t}_{2}}=\frac{-2\sqrt{2g}\pm \sqrt{8g+8g}}{2g}=\frac{-\sqrt{2}\pm 2}{\sqrt{g}}\] Taking +ve sign \[{{t}_{2}}=(2-\sqrt{2})/\sqrt{g}\] \[\therefore \]\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\sqrt{2/g}}{(2-\sqrt{2})/\sqrt{g}}=\frac{1}{\sqrt{2}-1}\] and so on.
You need to login to perform this action.
You will be redirected in
3 sec
