Given that the displacement of an oscillating particle is given by [m0L0T-1] . The dimensional formula for (ABCD) is

Question

Given that the displacement of an oscillating particle is given by [m0L0T-1] . The dimensional formula for (ABCD) is (A)  [m0L-1T-2]  (B)  [m0L0T0]  (C)  1.5μ  (D)  μ

Tardigrade Admin · Accepted Answer

y=A sin (Bx+Ct+D) As each term inside the bracket is dimension less, so NaCl OH- H2O C and D is dimensionless textC textu text2 ( textSCN ) text2. K6=(RT02/1000Δ Hv)

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

$[m^{0} L^{- 1} T^{- 2}]$

$[m^{0} L^{0} T^{0}]$

$1.5 μ$

$μ$

$y = A sin (B x + Ct + D)$ As each term inside the bracket is dimension less, so $N a Cl$ $O H^{-}$ $H_{2} O$ C and D is dimensionless $C u_{2}$ $(SCN)_{2} .$ $K_{6} = \frac{R T _{0}^{2}}{1000Δ H _{v}}$

Q. Given that the displacement of an oscillating particle is given by [m0L0T−1] . The dimensional formula for (ABCD) is

[m0L−1T−2]

[m0L0T0]

1.5μ

μ

y=Asin(Bx+Ct+D) As each term inside the bracket is dimension less, so NaCl OH− H2​O C and D is dimensionless Cu2​ (SCN)2​. K6​=1000ΔHv​RT02​​

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

$[m^{0} L^{- 1} T^{- 2}]$

$[m^{0} L^{0} T^{0}]$

$1.5 μ$

$μ$

$y = A sin (B x + Ct + D)$ As each term inside the bracket is dimension less, so $N a Cl$ $O H^{-}$ $H_{2} O$ C and D is dimensionless $C u_{2}$ $(SCN)_{2} .$ $K_{6} = \frac{R T _{0}^{2}}{1000Δ H _{v}}$