Let the two rods each have length $1\,\text{m}$. The lower left end is fixed at the floor, the lower right end is a roller moving along the floor, and the top joint carries the $2\,\text{kg}$ weight.

We need the vertical speed of the top joint (and hence the weight) as the roller moves right with constant speed.

1. Geometry of the mechanism

Let the fixed lower pivot be at $(0,0)$.

Let the roller be at $(x,0)$, where $x$ increases as the roller moves to the right.

Let the top joint be at $(x/2, y)$ by symmetry, since it is connected to both lower ends by equal rods of length $1$.

From the rod length condition,

$\left(\frac{x}{2}\right)^2 + y^2 = 1$

So,

$y = \sqrt{1-\frac{x^2}{4}}$

This is the height of the weight above the ground.

2. Differentiate to get vertical speed

Differentiate

$y = \left(1-\frac{x^2}{4}\right)^{1/2}$

with respect to time:

$\frac{dy}{dt} = \frac{1}{2}\left(1-\frac{x^2}{4}\right)^{-1/2}\left(-\frac{x}{2}\right)\frac{dx}{dt}$

Hence,

$\frac{dy}{dt} = -\frac{x}{4\sqrt{1-x^2/4}}\frac{dx}{dt}$

Since the roller moves right with constant speed, $\dfrac{dx}{dt}=v$ = constant.

Thus,

$\left|\frac{dy}{dt}\right| = \frac{xv}{4\sqrt{1-x^2/4}}$

This is the speed of the weight.

3. Does this speed increase or decrease?

Let

$f(x)=\frac{x}{\sqrt{1-x^2/4}}$

Then the upward/downward speed magnitude is proportional to $f(x)$.

As the roller moves right, $x$ increases. We check whether $f(x)$ increases with $x$.

Differentiate:

$f(x)=x\left(1-\frac{x^2}{4}\right)^{-1/2}$

$f'(x)=\left(1-\frac{x^2}{4}\right)^{-1/2}+x\left(-\frac12\right)\left(1-\frac{x^2}{4}\right)^{-3/2}\left(-\frac{x}{2}\right)$

$f'(x)=\left(1-\frac{x^2}{4}\right)^{-1/2}+\frac{x^2}{4}\left(1-\frac{x^2}{4}\right)^{-3/2}$

This is positive for allowed values of $x$. Therefore the speed magnitude increases as $x$ increases.

So when the roller moves right, the weight's speed is not constant and not decreasing; it is increasing.

4. Check option D

When the weight is $0.4\,\text{m}$ above the ground,

$y=0.4$

Using

$\left(\frac{x}{2}\right)^2+y^2=1$

$\frac{x^2}{4}+0.16=1$

$\frac{x^2}{4}=0.84$

$x^2=3.36$

$x\approx 1.833$

Now,

$\left|\frac{dy}{dt}\right| = \frac{xv}{4y}$

since $\sqrt{1-x^2/4}=y$.

So,

$\left|\frac{dy}{dt}\right|=\frac{1.833\,v}{4\times 0.4}=\frac{1.833}{1.6}v\approx 1.146v$

This is not $\frac34 v$.

So option D is false.

5. Evaluate options

• A: Constant speed — False
• B: Decreasing speed — False
• C: Increasing speed — True
• D: speed is $\frac34$ of roller speed when height is $0.4$ m — False

Final Answer

The weight moves up with increasing speed.

So the correct option is

$\boxed{\text{C}}$

The stored answer is B, which does not match this result.