A charge particle of 2μ C accelerated by a potential difference of...

As we know, radius of the circular path is given by,

$r = \dfrac{mv}{qB} = \dfrac{\sqrt{2mK}}{qB} = \dfrac{\sqrt{2 \times m \times qV}}{qB}$

$⇒ {r}^{2} = \dfrac{2mV}{q{B}^{2}}$

$⇒ m = \dfrac{{r}^{2}q{B}^{2}}{2V} = \dfrac{{\left(3 \times {10}^{- 2}\right)}^{2} \times \left(2 \times {10}^{- 6}\right) \times {\left(4 \times {10}^{- 3}\right)}^{2}}{2 \times 100} = 144 \times {10}^{- 18}kg$