A) \[{{V}_{A}}-{{V}_{O}}=\frac{B\omega {{l}^{2}}}{2}\]
B) \[{{V}_{O}}-{{V}_{C}}=\frac{3}{2}B\omega {{l}^{2}}\]
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 :
\[{{V}_{O}}-{{V}_{A}}=\frac{B\omega l}{2}\times l\] \[=\frac{B\omega {{l}^{2}}}{2}\] \[{{V}_{O}}-{{V}_{C}}=\frac{B\omega 3l}{2}\times 3l\] \[=\frac{9B\omega {{l}^{2}}}{2}\] Hence, \[{{V}_{A}}-{{V}_{C}}=4B\,\omega {{l}^{2}}\] or \[{{V}_{A}}>{{V}_{C}}\]
You need to login to perform this action.
You will be redirected in
3 sec
