Question

# A particle $$A$$ has charge $$+q$$ and particle $$B$$ has charge $$+4q$$ with each of them having the same mass $$m$$. When allowed to move from rest through same electrical potential difference, the ratio of their speeds $$\dfrac { V_A }{V_B }$$ will become

A
2:1
B
1:2
C
1:4
D
4:1

Solution

## The correct option is B $$1:2$$Since kinetic energy imparted to a system because of accelerated voltage V$$KE = qV$$____(1) and $$Kinetic energy(KE) = \dfrac{1}{2} mv^2$$ ____(2)From (1) and (2)$$\dfrac{K_A}{K_B} = \dfrac{q V}{4 qV} = \dfrac{1/2 m V^2_{A}}{1/2 mV^2_{B}}$$$$\dfrac{V_{A}^2}{V_B^2} = \dfrac{1}{4} = \dfrac{1}{2}$$PhysicsNCERTStandard XII

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