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A positive charge $$+Q$$ is fixed at a point $$A$$. another positively charged particle of mass $$m$$ and charge $$+q$$ is projected from a point $$B$$ with velocity $$u$$ as shown in figure. The point $$B$$ is at the lage distance from $$A$$ and at distance $$d$$ from the line $$AC$$. The initial velocity is parallel to the line $$AC$$. The point $$C$$ is at very large distance from $$A$$. The minimum distance (in meter) of $$+q$$ from $$+Q$$ during the motion is $$d(1+\sqrt {A})$$. Find the value of $$A$$.
(Take $$Qq=4\pi { \varepsilon  }_{ 0 }m{ u }^{ 2 }d$$ and $$d=(\sqrt{2}-1)$$ meter)
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A
3
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B
2
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C
4
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D
5
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Solution

The correct option is B $$2$$
The path of the particle will be as shown in figure. At the point of minimum distance $$(D)$$ the velocity of the particle will be perpendicular to its position vector w.r.t $$+Q$$
$$\quad \cfrac { 1 }{ 2 } m{ u }^{ 2 }+0=\cfrac { 1 }{ 2 } m{ v }^{ 2 }+\cfrac { kQq }{ { r }_{ min } } .......(1)$$
Torque on $$q$$ about $$Q$$ is zero hence angular momentum about $$Q$$ will be conserved
$$\Rightarrow mv{ r }_{ min }=mud.......(2)$$
By eqn $$(2)$$ in $$(1)$$
$$\cfrac { 1 }{ 2 } m{ u }^{ 2 }=\cfrac { 1 }{ 2 } m{ \left( \cfrac { ud }{ { r }_{ min } }  \right)  }^{ 2 }+\cfrac { kQq }{ { r }_{ min } } $$
$$\Rightarrow \cfrac { 1 }{ 2 } m{ u }^{ 2 }\left( 1-\cfrac { { d }^{ 2 } }{ { { r }_{ min } }^{ 2 } }  \right) =\cfrac { m{ u }^{ 2 }d }{ { r }_{ min } } \quad (\because \quad kQq=m{ u }^{ 2 }d)$$
$$\Rightarrow { { r }_{ min } }^{ 2 }-2{ r }_{ min }d-{ d }^{ 2 }=0$$
$$\Rightarrow { r }_{ min }=\cfrac { 2d\pm \sqrt { 4{ d }^{ 2 }+4{ d }^{ 2 } }  }{ 2 } =d(1\pm \sqrt { 2 } )$$
Distance cannot be negative
$$\therefore \quad { r }_{ min }=d(1 + \sqrt { 2 } )$$

Physics

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