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NEET
Exam
NEET
Subject
Physics
Chapter
Units and Dimensions

NEET · Physics · Units and Dimensions

The velocity v of a particle at time t is given by the equation v =...

The velocity vv of a particle at time tt is given by the equation v=At+Bt+C+Dv = At + \dfrac{B}{t + C} + D. Match the constants in List-I with their dimensional formulas in List-II.

List-IList-II
(A) Constant A(I) [L][L]
(B) Constant B(II) [T][T]
(C) Constant C(III) [LT−1][L T^{-1}]
(D) Constant D(IV) [LT−2][L T^{-2}]


Choose the correct answer from the options given below:

Options

  1. A(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  2. B(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
  3. C(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  4. D(A)-(IV), (B)-(I), (C)-(III), (D)-(II)

Correct answer

B. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Step-by-step solution

According to the principle of homogeneity of dimensions, only quantities with the same dimensions can be added or subtracted.

In the denominator (t+C)(t + C), the constant CC is added to time tt. Therefore, [C]=[t]=[T][C] = [t] = [T].

Each term in the equation must have the same dimensions as velocity vv, which is [LT−1][L T^{-1}].

For the term DD: [D]=[v]=[LT−1][D] = [v] = [L T^{-1}].

For the term AtAt: [A][t]=[v]⇒[A][T]=[LT−1]⇒[A]=[LT−2][A][t] = [v] \Rightarrow [A][T] = [L T^{-1}] \Rightarrow [A] = [L T^{-2}].

For the term Bt+C\dfrac{B}{t + C}: [B][T]=[LT−1]⇒[B]=[L]\dfrac{[B]}{[T]} = [L T^{-1}] \Rightarrow [B] = [L].

Thus, the correct matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).

Answer: (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
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