Question

Given that the displacement of an oscillating particle is given by $ [{{m}^{0}}{{L}^{0}}{{T}^{-1}}] $ . The dimensional formula for (ABCD) is

• A. $ [{{m}^{0}}{{L}^{-1}}{{T}^{-2}}] $
• B. $ [{{m}^{0}}{{L}^{0}}{{T}^{0}}] $
• C. $ 1.5\mu $
• D. $ \mu $

Answer 1

Answer: (B)

$ y=A\sin (Bx+Ct+D) $ As each term inside the bracket is dimension less, so $ NaCl $ $ O{{H}^{-}} $ $ {{H}_{2}}O $ C and D is dimensionless $ \text{C}{{\text{u}}_{\text{2}}} $ $ {{\left( \text{SCN} \right)}_{\text{2}}}. $ $ {{K}_{6}}=\frac{RT_{0}^{2}}{1000\Delta {{H}_{v}}} $