The dimensions of K in the equation W = frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation $W = \frac{1}{2}K{x^2}$,where $W$ is work and $x$ is displacement.Dimensional formula of work $W$ is $[M^1 L^2 T^{-2}]$.Dimensi

Vedclass · Accepted Answer

${M^1}{L^0}{T^{ - 2}}$

Vedclass · Answer

(A) Given the equation $W = \frac{1}{2}K{x^2}$,where $W$ is work and $x$ is displacement.Dimensional formula of work $W$ is $[M^1 L^2 T^{-2}]$.Dimensional formula of displacement $x$ is $[L^1]$.Rearranging the equation for $K$,we get $K = \frac{2W}{x^2}$.Substituting the dimensions: $[K] = \frac{[M^1 L^2 T^{-2}]}{[L^1]^2} = \frac{[M^1 L^2 T^{-2}]}{[L^2]} = [M^1 L^0 T^{-2}]$.Thus,the dimensions of $K$ are $[M^1 L^0 T^{-2}]$.

The dimensions of $K$ in the equation $W = \frac{1}{2}K{x^2}$ are:

Similar Questions

The dimensional formula of capacitance is . . . . . . . Take $Q$ as the dimensional formula of charge.

Write the magnitude and dimensional formula of $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

Write the dimensional formula of $\frac{k}{m}$,where $k$ is the spring constant and $m$ is the mass.

The dimension of Planck's constant is equal to that of:

Dimensions of the coefficient of viscosity are

Vedclass Test Series

Exam Paper Generator

Online Exam Module