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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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Force acting on a particle is $(2i+3j)N$ . Work done by the force is zero, when the particle is moved on the line $3y+kx =5$ value of k is

\[(a)\;2 \quad (b)\;4 \quad (c)\;6 \quad (d)\;8 \]

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Answer: 2
Work done is zero if force is perpendicular to the displacement .
Force is represented by the vector $=(2 i+3j)N$, where the slope of the line joining (2,3) to origin ($y=\large\frac{3}{2}$$x+c$) is 3/2
Equation of the line, the particle is moved on is $3y+kx =5$
$\Rightarrow y=\large\frac{-k}{3}$$x+\large\frac{5}{3}$
Work done by force is zero, hence force is perpendicular to displacement $\rightarrow m_1m_2=-1$
$\Rightarrow \bigg(\large\frac{3}{2}\bigg)\bigg(\large\frac{-k}{3}\bigg)$$=-1 \rightarrow$ $k=2$
answered Jul 16, 2013 by meena.p
edited Aug 25, 2014 by balaji.thirumalai

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