Kinematics Question 304
A force $ \vec{F}=3\hat{i}+c\hat{j}+2\hat{k} $ acting on a particle causes a displacement $ \vec{S}=-4\hat{i}+2\hat{j}-3\hat{k} $ in its own direction. If the work done is 6J, then the value of c will be [DPMT 1997]
Options:
A) $ {{( A^{2}+B^{2}+\frac{AB}{\sqrt{3}} )}^{1/2}} $
B) $ A+B $
C) $ {{(A^{2}+B^{2}+\sqrt{3}AB)}^{1/2}} $
D) $ {{(A^{2}+B^{2}+AB)}^{1/2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ |\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{A^2B^2 - (\overrightarrow{A}.\overrightarrow{B})^2} $
$ AB\sin \theta =\sqrt{3}AB\cos \theta $
$ \Rightarrow $ $ \tan \theta =\sqrt{3} $ $ \theta =60{}^\circ $
Now $ |\overrightarrow{R}|=|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{A^{2}+B^{2}+2AB\cos \theta } $
$ =\sqrt{A^{2}+B^{2}+AB} $
$ ={{(A^{2}+B^{2}+AB)}^{1/2}} $
BETA
