QJEEAdv18Ph627a: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph627a Solution: Consider a body of mass at rest at the origin at time A force…

Given, force applied on body

$\overrightarrow F = (\alpha t\widehat i + \beta \widehat j)$

Here, $\alpha$ = 1.0 N s ^{$-$ 1} and $\beta$ = 1.0 N. Therefore,

$\overrightarrow F = t\widehat i + \widehat j$

By Newton's second law of motion, F = ma. Given m = 1.0 kg. Therefore,

$\overrightarrow F = \overrightarrow a \Rightarrow \overrightarrow a = (t\widehat i + \widehat j)$

We know $\overrightarrow a = {{d\overrightarrow v } \over {dt}} \Rightarrow \overrightarrow v = \int {\overrightarrow a dt}$

$\Rightarrow \overrightarrow v = \int {(t\widehat i + \widehat j)dt = {{{t^2}} \over 2}\widehat i + t\widehat j}$ ..... (1)

Also, $\overrightarrow v = {{\overrightarrow {dr} } \over {dt}} \Rightarrow \overrightarrow r = \int {\overrightarrow v dt}$

$\Rightarrow \overrightarrow r = \int {\left( {{{{t^2}} \over 2}\widehat i + t\widehat j} \right)dt = {{{t^3}} \over 6}\widehat i + {{{t^2}} \over 2}\widehat j}$ ..... (2)

Now, $$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ t^3/6& t^2/2& 0\\ t& 1& 0\end{array}\right|$$

$\overrightarrow \tau = \widehat i(0) - \widehat j(0) + \widehat k\left( {{{{t^3}} \over 6} - {{{t^3}} \over 2}} \right) = {{ - 1} \over 3}{t^3}\widehat k$ N m

At t = 1.0 s, $\overrightarrow \tau = {{ - 1} \over 3}\widehat k$ N m

Hence, option (A) is correct.

Now, from Eq. (1), $\overrightarrow v = {{{t^2}} \over 2}\widehat i + t\widehat j$

At t = 1.0 s, $\overrightarrow v = \left( {{1 \over 2}\widehat i + \widehat j} \right)$ m s ^{$-$ 1} = $= {1 \over 2}(\widehat i + 2\widehat j)$ m s ^{$-$ 1}

Hence, option (C) is correct.