NIMCET 2018 Question Paper with solutions

Here, velocity vector of bird is
$\overrightarrow{\mathit{\upsilon }}$ = $10\hat{\mathit{i}}+6\hat{\mathit{j}}+\hat{\mathit{k}}$ or $\left|\overrightarrow{{\mathit{\upsilon }}_{\mathit{x}}}\right|$ = 10 km/h
$\left|\overrightarrow{{\mathit{\upsilon }}_{\mathit{y}}}\right|$ = 6 km/h
$\left|\overrightarrow{{\mathit{\upsilon }}_{\mathit{z}}}\right|$ = 1 km/h
Initial position vector of the bird
$\overrightarrow{\mathit{O}\mathit{A}}$ = $\hat{\mathit{i}}+2\hat{\mathit{j}}+3\hat{\mathit{k}}$
When bird reach a point in space that is 13 m high from the ground then final position vector of the bird will be
$\overrightarrow{\mathit{O}\mathit{B}}$ = $\hat{\mathit{i}}+2\hat{\mathit{j}}+13\hat{\mathit{k}}$
∴ $\overrightarrow{\mathit{A}\mathit{B}}$ = $\overrightarrow{\mathit{O}\mathit{B}}-\overrightarrow{\mathit{O}\mathit{A}}$ = 10 $\hat{\mathit{k}}$
or $\left|\overrightarrow{\mathit{A}\mathit{B}}\right|$ = 10 m
Time taken by bird to cover a distance of 10 m along the vertical is
t = $\left|\overrightarrow{\mathit{A}\mathit{B}}\right|/|\overrightarrow{{\mathit{V}}_{\mathit{x}}}|$ = $\frac{10}{1\times \frac{5}{18}}$ = $\frac{10\times 18}{5}$ = 36 seconds