A) \[{{V}_{A}}-{{V}_{O}}=\frac{B\omega {{l}^{2}}}{2}\]
B) \[{{V}_{O}}-{{V}_{C}}=\frac{9}{2}B\omega {{l}^{2}}\]
C) \[{{V}_{A}}-{{V}_{C}}=4B\omega {{l}^{2}}\]
D) \[{{V}_{C}}-{{V}_{O}}=\frac{9}{2}B\omega {{l}^{2}}\]
Correct Answer: C
Solution :
The potential difference is given by \[{{V}_{O}}-{{V}_{A}}=\frac{B\omega l}{2}\times l\] \[{{V}_{O}}-{{V}_{C}}=\frac{B\omega (3l)}{2}\times 3l\] Hence, \[{{V}_{A}}-{{V}_{C}}=4\,B\omega {{l}^{2}}\] \[\Rightarrow \] \[{{V}_{A}}>{{V}_{C}}\]
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