Kinematics Question 260

Question: The torque of the force $ \overrightarrow{F}=(2\hat{i}-3\hat{j}+4\hat{k})N $ acting at the point $ \overrightarrow{r}=(3\hat{i}+2\hat{j}+3\hat{k}) $ m about the origin be [CBSE PMT 1995]

Options:

A) $ 6\hat{i}-6\hat{j}+12\hat{k} $

B) $ 17\hat{i}-6\hat{j}-13\hat{k} $

C) $ -6\hat{i}+6\hat{j}-12\hat{k} $

D) $ -17\hat{i}+6\hat{j}+13\hat{k} $

Show Answer

Answer:

Correct Answer: B

Solution:

$ \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F} $

$ = \begin{vmatrix} \hat{i} & \hat{j} & {\hat{k}} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \\ \end{vmatrix} $

$ =[ (2\times 4)-(3\times -3) ]\hat{i}+[ (2\times 3)-(3\times 4) ]\hat{j} $

$ +[ (3\times -3)-(2\times 2) ]\hat{k} $

$ =17\hat{i}-6\hat{j}-13\hat{k} $



NCERT Chapter Video Solution

Dual Pane