Dimensions of position vector derivatives – Rankers Physics

Dimensions: Practice Problem & Solution

Assertion (A): If \(vec{r}\) is the position vector then dimensions of \(\frac{d^2vec{r}}{dt^2}\) is \([M^0L^1T^{-2}]\). Reason (R): Dimensions of \(int \left(\frac{d^2vec{r}}{dt^2}\right) dt\) is \([M^0L^1T^{-1}]\) where \(\vec{r} \rightarrow\) position vector, \(t \rightarrow\) time.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Dimensions. Understanding the underlying formula is key to arriving at the correct answer below:

The second derivative of position with respect to time represents acceleration with dimensions \([L T^{-2}]\). Integrating this acceleration with respect to time yields velocity with dimensions \([L T^{-1}]\). Both statements are true, but R is not the explanation of A.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. The diameter of a spherical bob, when measured with vernier callipers yielded the following values: 3.33 cm, 3.32 cm, 3.34 cm, 3.33 cm and 3.32 cm. The mean diameter to appropriate significant figures is:
2. The mechanical quantity, which has dimensions of reciprocal of mass \( \text{M}^{-1} \) is
3. The dimension of Planck constant equals to that of: [2001]
4. The ratio of the dimensions of Planck's constant and that of the moment of inertia is the dimension of: [2005]
5. The time dependence of a physical quantity (p) is given by \(p = p_0 \text{exp } (-\alpha t^2)\), where (alpha) is constant and (t) is the time. The constant \(\alpha\): [1993]