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, $$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{t^3}/6} & {{t^2}/2} & 0 \cr t & 1 & 0 \cr } } \right|$$

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