A) \[\frac{2hv}{3}\]
B) \[30\mu A\]
C) \[90\mu A\]
D) \[4mA\]
Correct Answer: C
Solution :
Let the distribution of current in the various arms as shown in figure. Let R is the effective resistance between X and Y then \[90\mu A\] ....(i) Applying Kirchhoffs second law in a second loop XABCDYX, we get \[4mA\] \[2mA\] ?..(ii) Applying Kirchhoffs law in a closed loop ABEFA, we get \[3.6mA\] \[1.8mA\] \[62V,\,\,2\Omega \] ??.(iii) Applying Kirchhoffs law in closed loop BCDEB \[63V,\,\,1\Omega \] \[61V,\,\,1\Omega \] \[64V,\,\,2\Omega \] \[15min\] ??.(iv) From Eqs. (iii), and (iv), we get \[20\text{ }min\] \[7.5min\] From Eq. (iii), we get \[25min\] \[\frac{\omega M}{M+m}\] \[\frac{\omega (M-2M)}{M+2m}\] \[\frac{\omega (M+2M)}{M}\] \[\frac{\omega M}{M+2m}\] \[{{a}_{c}}={{k}^{2}}{{r}^{2}}{{t}^{2}},\] \[2mk{{r}^{2}}t\] From Eq. (i) \[mk{{r}^{2}}{{t}^{2}}\] \[m{{k}^{2}}{{r}^{2}}t\] ..?(iv) From Eq. (ii) \[m{{k}^{2}}r{{t}^{2}}\] ??(vi) From Eqs. (v) and (vi), we get \[1\text{ }km/h\] \[\text{3 }km/h\]You need to login to perform this action.
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