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Q.

Force acting on a particle is $(2 \hat{i} + 3 \hat{j}) N$ . Work done by this force is zero, when a particle is moved along the line 3y + kx = 5. Here the value of k is

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a

2

b

8

c

6

d

4

answer is A.

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Detailed Solution

$\frac{x}{(5 / k)} + \frac{y}{(5 / 3)} = 1$
x intercept : $(\frac{5}{k})$ ; y intercept : $(\frac{5}{3})$
$\vec{S} = \frac{5}{K} \hat{i} + \frac{5}{3} \hat{j} W = 0 \Rightarrow (\frac{5}{k}) 2 - (\frac{5}{3}) 3 = 0 \Rightarrow k = 2$
Alternate method
$m_{1} . m_{2} = - 1, (\frac{3}{2}) (- \frac{K}{3}) = - 1 \Rightarrow k = 2$

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