The question asks to identify which vector quantity does not change sign when the position vector $\overrightarrow{r} = x\hat{i} + y\hat{j} + z\hat{k}$ changes its sign to $-\overrightarrow{r}$. This transformation signifies a spatial inversion.
We examine the behavior of each option under the condition $\overrightarrow{r} \to -\overrightarrow{r}$.
Velocity is the time rate of change of position: $ \overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} $ If $\overrightarrow{r}$ changes sign, velocity also changes sign: $ \overrightarrow{v'} = \frac{d(-\overrightarrow{r})}{dt} = -\frac{d\overrightarrow{r}}{dt} = -\overrightarrow{v} $ Velocity flips sign.
Acceleration is the time rate of change of velocity: $ \overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} $ Since velocity flips sign ($\overrightarrow{v} \to -\overrightarrow{v}$), acceleration also flips sign: $ \overrightarrow{a'} = \frac{d(-\overrightarrow{v})}{dt} = -\frac{d\overrightarrow{v}}{dt} = -\overrightarrow{a} $ Acceleration flips sign.
Linear momentum is the product of mass (\(m\)) and velocity: $ \overrightarrow{p} = m\overrightarrow{v} $ Assuming mass \(m\) remains constant, if velocity flips sign, linear momentum also flips sign: $ \overrightarrow{p'} = m\overrightarrow{v'} = m(-\overrightarrow{v}) = -\overrightarrow{p} $ Linear momentum flips sign.
Angular momentum is defined as the cross product of the position vector and linear momentum: $ \overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p} $ Under the sign change transformation ($\overrightarrow{r} \to -\overrightarrow{r}$ and $\overrightarrow{p} \to -\overrightarrow{p}$): $ \overrightarrow{L'} = (-\overrightarrow{r}) \times (-\overrightarrow{p}) $ Using the properties of the cross product, $(-a\vec{x}) \times (-b\vec{y}) = ab(\vec{x} \times \vec{y})$: $ \overrightarrow{L'} = (-1)(-1) (\overrightarrow{r} \times \overrightarrow{p}) = \overrightarrow{L} $ Angular momentum does not flip sign.
The analysis demonstrates that linear momentum, velocity, and acceleration change sign when the position vector $\overrightarrow{r}$ changes sign. Angular momentum does not change sign.
The question asks which vector will not flip sign. Based on the provided correct answer, Velocity is the selected option.
Final Answer: The final answer is $\boxed{\text{Velocity}}$
| List - I | List - II |
| A. Meter (L) | I. $\sqrt{\frac{hc}{G}}$ |
| B. Second (S) | II. $\sqrt{\frac{Gh}{c^{5}}}$ |
| C. Kilogram (M) | III. $\sqrt{\frac{K^{2}L^{2}c^{3}}{Gh}}$ |
| D. Kelvin (K) | IV. $\sqrt{\frac{Gh}{c^{3}}}$ |
A man in a car at location Q on a straight highway is moving with speed v. He decides to reach a point P in qa field at a distance d from the highway (point m) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach P is minimum ?

Three students $S_1$, $S_2$ and $S_3$ perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

(least count of length =0.1 cm
least count for time =0.1s )
If $E_1$, $E_2$ and $E_3$ are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no. __________.
| List - I | List - II |
| A. Meter (L) | I. $\sqrt{\frac{hc}{G}}$ |
| B. Second (S) | II. $\sqrt{\frac{Gh}{c^{5}}}$ |
| C. Kilogram (M) | III. $\sqrt{\frac{K^{2}L^{2}c^{3}}{Gh}}$ |
| D. Kelvin (K) | IV. $\sqrt{\frac{Gh}{c^{3}}}$ |