A rod AB moves with a uniform velocity v in a uniform magnetic field as shown in figure
(##HCV_VOL2_C38_E01_033_Q01##)

Options

A. `sqrt(10)m//s`
B. `2sqrt(10)m//s`
C. `sqrt(20)m//s`
D. `sqrt(5)m//s`

Answer

C

Explanation

Two particles ………….
mmotion of `P` with respect to `Q` parabolic

`sqrt(3)V_(0)`
`x = sqrt(3)V_(0)t`
`y = (5)/(2)t^(2)`
`y = (x^(2))/(4)`
`M` is foot perpendicular draw on path of `P` with `Q`.
Let `M` is `(X_(0), (X_(0)^(2))/(4))`
`(-2)/(X_(0)) = ((X_(0)^(2))/(4)-0)/((x_(0)-3))`
`x_(0) = 2`
So `M` is `(2, 1)`
So `QM` is `sqrt(2)` meter
`2= sqrt(3)xx sqrt((10)/(3)t)`
`t = (2)/(sqrt(10))sec`.
`V_(x) = sqrt(3)V_(0) = sqrt(10)m//s`
`V_(y) = 5t implies At t = 2s`.
`V_(y) = sqrt(10) m//s`
speed `= sqrt(10+10) = sqrt(20) m//s`