A) \[{{x}^{2}}+{{y}^{2}}+x+2y+4=0\]
B) \[{{x}^{2}}+{{y}^{2}}+x-2y+4=0\]
C) \[{{x}^{2}}+{{y}^{2}}-x-2y+4=0\]
D) \[{{x}^{2}}+{{y}^{2}}-x+2y+4=0\]
Correct Answer: C
Solution :
Given, \[\Rightarrow \] \[[\because |adj\,(adj\,A)|=\,A{{|}^{{{(n-1)}^{2}}}}]\] and \[=100\,\left( \frac{\Delta \phi }{\Delta t} \right)\] Let the required circle be \[=5\times {{10}^{-3}}\times 100=0.5\,\,V\] Now, by condition of orthoganality,\[R=v\sqrt{\frac{2h}{g}}\] ?(i) \[u=v,\] \[u=v+at\] ?(ii) and \[0=v-at\] ?(iii) \[\therefore \] \[-a=\frac{0-v}{t}=-\frac{v}{t}\] and \[f=\mu R=\mu mg\] \[a=\mu g\] \[\therefore \] [from Eq. (i)] \[t=\frac{v}{a}=\frac{v}{g\mu }\] \[\eta =\frac{{{P}_{0}}}{{{P}_{i}}}\] \[\therefore \] Required equation of circle \[\frac{1}{2}\,m{{v}^{2}}=16\,\,J\] \[v=4\,m{{s}^{-1}}\]You need to login to perform this action.
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