A) \[~x+y=6\]
B) \[x+4y=6\]
C) \[y+4x=6\]
D) \[x-y=6\]
E) \[x+y=8\]
Correct Answer: C
Solution :
Work done by the gravitational field is zero, when displacement is perpendicular to gravitational field. Here, gravitational field, \[I=4\,\hat{i}+\hat{j}\] If \[{{\theta }_{1}}\] is the angle which I makes with positive x-axis, then \[\tan {{\theta }_{1}}=\frac{1}{4}\] or \[{{\theta }_{1}}={{\tan }^{-1}}\left( \frac{1}{4} \right)=14{}^\circ 6\] If \[{{\theta }_{2}}\] is the angle which the line\[y+4x=6\]makes with positive x-axis, then \[{{\theta }_{2}}={{\tan }^{-1}}(-4)=75{}^\circ 56\] So, \[{{\theta }_{1}}+{{\theta }_{2}}=90{}^\circ \]ie, the line \[y+4x=6\]is perpendicular to I.
You need to login to perform this action.
You will be redirected in
3 sec
