A time dependent force $$F = 6t$$ acts on a particle of mass 1 kg. If the particle starts from the rest, the work done by the force during the first 1 sec will be:

We have a force that changes with time according to the relation $$F = 6t \;{\rm N}$$. The mass of the particle is given as $$m = 1\ {\rm kg}$$ and the particle starts from rest, so its initial velocity is $$u = 0\ {\rm m\,s^{-1}}$$ at $$t = 0\ {\rm s}$$.

Because the force is the only agent acting, Newton’s second law $$F = ma$$ connects force and acceleration. Stating the formula first, $$F = ma \;\Rightarrow\; a = \dfrac{F}{m}.$$ Substituting the given expressions, we get

$$a = \dfrac{6t}{1} = 6t\ {\rm m\,s^{-2}}.$$

Acceleration is the time derivative of velocity, that is $$a = \dfrac{dv}{dt}.$$ So we write

$$\dfrac{dv}{dt} = 6t.$$

To obtain velocity, we integrate both sides with respect to time from the initial instant $$t = 0$$ (where $$v = 0$$) to a general time $$t$$:

$$\int_{0}^{v} dv = \int_{0}^{t} 6t'\,dt'.$$

The left‐hand side simply gives $$v - 0 = v$$. The right‐hand side integrates as follows:

$$\int_{0}^{t} 6t'\,dt' = 6\left[\dfrac{t'^2}{2}\right]_{0}^{t} = 3t^2.$$

Hence the velocity as a function of time is

$$v = 3t^2\ {\rm m\,s^{-1}}.$$

Next, displacement $$s$$ is the integral of velocity with respect to time, using $$v = \dfrac{ds}{dt}$$. Performing the integration from $$t = 0$$ to $$t = 1\ {\rm s}$$ (the interval of interest), we get

$$s = \int_{0}^{1} v\,dt = \int_{0}^{1} 3t^2\,dt.$$

Carrying out the integration:

$$\int_{0}^{1} 3t^2\,dt = 3\left[\dfrac{t^3}{3}\right]_{0}^{1} = 1\ {\rm m}.$$

Now, the work done by a force is the integral of the force along the displacement. Since everything is along one line, we can write work $$W$$ in the time‐integral form $$W = \int F\,v\,dt$$, because $$F\,ds = F\,v\,dt$$ and $$ds = v\,dt.$$ Substituting the expressions for $$F$$ and $$v$$, we obtain

$$W = \int_{0}^{1} (6t)(3t^2)\,dt = \int_{0}^{1} 18t^3\,dt.$$

Integrating term by term:

$$\int_{0}^{1} 18t^3\,dt = 18\left[\dfrac{t^4}{4}\right]_{0}^{1} = 18 \times \dfrac{1}{4} = 4.5\ {\rm J}.$$

For verification, we may also use the work-energy theorem, which states that the net work done equals the change in kinetic energy, $$W = \Delta K = \dfrac{1}{2} m v^2 - \dfrac{1}{2} m u^2.$$ At $$t = 1\ {\rm s}$$ we have already found $$v = 3\ {\rm m\,s^{-1}}$$, so

$$\Delta K = \dfrac{1}{2}(1)(3)^2 - 0 = \dfrac{9}{2} = 4.5\ {\rm J},$$

exactly matching the result obtained above.

Hence, the correct answer is Option B.