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BITSAT 2019 Paper

Section: Physics

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Question : 16 of 150

Marks: +1, -0

A force of 2

\hat{i}

\hat{j}

\hat{k}

N acts on a body for 4 second, produces a displacem ent of (3

\hat{i}

\hat{j}

\hat{k}

) m . The power used is

9.5 W
7.5 W
6.5 W
4.5 W

Solution:

We know Power

=\frac{\text{Workdone}}{\text{time}}

- - - (A)

Also work done

=

Force

\times

displacement- - - - - (B)

Work done is the dot product of the vector quantities

W \cdot D =[2\hat{i}+3\hat{j}+4\hat{k}] \cdot [3\hat{i}+4\hat{j}+5\hat{k}]

=6+12+20

=38

Substituting the above found value in

A

Power

=\frac{38}{4}=9.5\text{ W}

t =4

seconds (given)

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