A) \[S=ft\]
B) \[f\]
C) \[f\]
D) \[s=\frac{1}{6}f{{t}^{2}}\]
Correct Answer: C
Solution :
The velocity-time graph for the given situation can be drawn as below. Magnitudes of slope of \[{{G}^{1/2}}{{h}^{1/2}}{{c}^{-3/2}}\] and slope of \[{{G}^{1/2}}{{h}^{1/2}}{{c}^{1/2}}\] \[{{\left[ 2G\frac{({{m}_{1}}-{{m}_{2}}}{r} \right]}^{1/2}}\] \[{{\left[ \frac{2G}{r}({{m}_{1}}+{{m}_{2}}) \right]}^{1/2}}\] \[{{\left[ \frac{r}{2G({{m}_{1}}{{m}_{2}})} \right]}^{1/2}}\] In graph area of \[{{\left[ \frac{2G}{r}{{m}_{1}}{{m}_{2}} \right]}^{1/2}}\] gives distances, \[(\Delta l)\] ?...(i) Area of rectangle ABED gives distance travelled in time t. \[50{{m}^{2}}/{{s}^{2}}\] Distance travelled in timer \[50.5{{m}^{2}}/{{s}^{2}}\] \[51{{m}^{2}}/{{s}^{2}}\] Thus \[521{{m}^{2}}/{{s}^{2}}\] \[\frac{1}{\sqrt{a}}\] \[a\] \[\sqrt{a}\] \[{{a}^{3/2}}\] ???(ii) From Eqs. (i) and (ii), we have \[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\] \[\frac{q}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\] \[\frac{q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\]You need to login to perform this action.
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