Question 49: Units , Dimensions and Error Analysis

Question

The position of a particle moving on $x$-axis is given by $x(t)=A \sin t+B \cos ^2 t+C t^2+D$, where $t$ is time. The dimension of $\frac{A B C}{D}$ is

Accepted Answer

Let's analyze the position function step by step.
The particle's position is given by:
$$x(t) = A \sin t + B \cos^2 t + C t^2 + D,$$
where $ t $ represents time, which has the dimension $ T $.
Since the overall dimensions of position $ x(t) $ must be length $ (L) $, each term in the expression must also have dimensions of length.
Term $ A \sin t $:
The sine function, $ \sin t $, is dimensionless (its argument must be dimensionless, and by convention, if $ t $ is in an appropriate unit where the argument is dimensionless, the function itself is dimensionless).
Therefore, $ A $ must carry the dimension of length:
$$ [A] = L. $$
Term $ B \cos^2 t $:
Similarly, the cosine function is dimensionless, and so is its square.
Thus, $ B $ must have the dimension of length:
$$ [B] = L. $$
Term $ C t^2 $:
Here, $ t^2 $ has the dimension $ T^2 $.
To have the overall term with the dimension $ L $, we require:
$$ [C] 	imes [T^2] = L. $$
So, the dimension of $ C $ must be:
$$ [C] = L \, T^{-2}. $$
Term $ D $:
This is a constant term added to position, so it must also have the dimension of length:
$$ [D] = L. $$
Now, we are asked to find the dimension of:
$$ \frac{ABC}{D}. $$
The dimensions of $ A $, $ B $, and $ C $ are:
$$ [A] = L, \quad [B] = L, \quad [C] = L \, T^{-2}. $$
Multiplying them together:
$$ [ABC] = L 	imes L 	imes (L \, T^{-2}) = L^3 \, T^{-2}. $$
Since $ [D] = L $, dividing by $ D $ gives:
$$ \frac{[ABC]}{[D]} = \frac{L^3 \, T^{-2}}{L} = L^2 \, T^{-2}. $$
Thus, the dimension of $ \frac{A B C}{D} $ is:
$$ \boxed{L^2 T^{-2}}. $$
The correct answer is Option C.

Answer

$\mathrm{L}$

Answer

$\mathrm{L}^3 \mathrm{~T}^{-2}$

Answer

$\mathrm{L}^2 \mathrm{~T}^{-2}$

Answer

$\mathrm{L}^2$

Units , Dimensions and Error Analysis

The position of a particle moving on x-axis is given by x(t)=A \sin t+B \cos ^2 t+C t^2+D, where t is time. The dimension of \frac{A B C}{D} is

Choose the correct answer: