Question

A metallic V shaped rod OAC is rotated with respect to one of its end in uniform magnetic field (as shown in figure), such that axis of rotation is parallel to the direction of magnetic field. Length of each arm of rod is $L$ and angle between the arms is ${60}^0$. P is the mid-point of section AC. Magnitude of magnetic field is $B$. Then choose the correct relations.

• A. $V_A-V_0=\frac{\omega B{L}^2}{2}$
• B. $V_A-V_C=\frac{\omega B{L}^2}{2}$
• C. $V_C-V_P=\frac{\omega B{L}^2}{8}$
• D. $V_A-V_P=\frac{\omega B{L}^2}{8}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $V_A-V_0=\frac{\omega B{L}^2}{2}$
C $V_C-V_P=\frac{\omega B{L}^2}{8}$
D $V_A-V_P=\frac{\omega B{L}^2}{8}$


OA = OC = $L$
OP = $\frac{\sqrt{3}L}{2}$
$V_A-V_O=\frac{\omega B{L}^2}{2}$
$V_A-V_C=0$
$V_C-V_O=\frac{\omega B{L}^2}{2}$
$V_P-V_O=\frac{\omega B}{2}\left(O{P}^2\right)=\frac{\omega B}{2}\left(\frac{3L^2}{4}\right)=\frac{3\omega B{L}^2}{8}$
$V_C-V_P=(V_C-V_O)-(V_P-V_O)$
$=\frac{\omega B{L}^2}{2}-\frac{3\omega B{L}^2}{8}$
$=\frac{\omega B{L}^2}{8}$
$V_A-V_P=(V_A-V_O)-(V_P-V_O)$
$=\frac{\omega B{L}^2}{2}-\frac{3\omega B{L}^2}{8}$
$=\frac{\omega B{L}^2}{8}$