Consider a closed surface 'S' surrounding volume V. if →r is the position vector of a point inside s with →n the unit normal on 'S', the valuse of the integral ∯5→r.^nds is
A
3V
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B
5V
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C
10V
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D
15V
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Solution
The correct option is D 15V From Gauss Divergence Theorem ∯→F.^nds=∭vdiv→Fdv ⇒∯5→r.nds=∭vdiv→rdv
= 5∭v(3)dv
=15V