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NEET
Exam
NEET
Subject
Physics
Chapter
Units and Dimensions

NEET · Physics · Units and Dimensions

Assertion (A): If x and y are the distance along x and y axes...

Assertion (A): If xx and yy are the distance along xx and yy axes respectively then the dimensions of (d3ydx3\frac{d^3 y}{d x^3}) is (\mathrm{M}^0 \mathrm{~L}^{-2} \mathrm{~T}^0). Reason (R): Dimensions of (\int_a^b y d x) is (\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^0).

Options

  1. AAssertion (A) is true, Reason (R) is truc; Reason (R) is a correct explanation for Assertion (A).
  2. BAssertion (A) is truc, Reason (R) is truc; Reason (R) is not a correct explanation for Assertion (A).
  3. CAssertion (A) is true, Reason (R) is false.
  4. DAssertion (A) is falsc, Reason (R) is true.

Correct answer

B. Assertion (A) is truc, Reason (R) is truc; Reason (R) is not a correct explanation for Assertion (A).

Step-by-step solution

Assertion (A): "If xx and yy are distances along xx and yy axes respectively, then the dimensions of (d3ydx3\frac{d^3 y}{d x^3}) is M0L−2T0."M⁰ L^{-2} T^{0 . "} - Here, yy has dimensions of length (L). - xx also has dimensions of length (L). - Differentiating once: (dydx→\frac{d y}{d x} \rightarrowL / L=) dimensionless. - Differentiating twice: (d2ydx2→\frac{d^2 y}{d x^2} \rightarrow) dimension =1/L=1 / L. - Differentiating thrice: (d3ydx3→\frac{d^3 y}{d x^3} \rightarrow) dimension =1/L2=L−2=1 / L²=L^{-2}. So, dimensions = M0L−2T0M⁰ L^{-2} T⁰. (1) Assertion (A) is true.
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