Question

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

• A. 3
• B. 2
• C. 1
• D. 4

Answer 1

The correct option is B 2

Step 1, Given data

Vector = $2\overrightarrow{i}+3\overrightarrow{j}$

The particle is moving on the line 3 y + k x=5

Step 2, Finding the value of k

From the vector, we can say that $tan\theta =\frac{3}{2}$ it is also the slope of the line which can be drawn along with the vector.

The given line can be written as $y=-\frac{k}{3}x+\frac{5}{3}$

Slope = $\frac{-k}{3}$

According to the question the work done is zero $\therefore$ force is perpendicular to the displacement
We know that if two lines are perpendicular to each other then the product of their slope is -1

So,

$\frac{3}{2}\times \frac{-k}{3}=-1$

$Or,k=2$

Hence the value of k is 2.