In a system of units,if force (F),acceleratio | vedclass

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

Question

(B) Let the dimensional formula of energy $(E)$ be expressed as $E = k F^a A^b T^c$,where $k$ is a dimensionless constant.The dimensions of the quanti

Vedclass · Accepted Answer

$F A T^2$

Vedclass · Answer

(B) Let the dimensional formula of energy $(E)$ be expressed as $E = k F^a A^b T^c$,where $k$ is a dimensionless constant.The dimensions of the quantities are:$[E] = [M L^2 T^{-2}]$$[F] = [M L T^{-2}]$$[A] = [L T^{-2}]$$[T] = [T]$Substituting these into the equation:$[M L^2 T^{-2}] = [M L T^{-2}]^a [L T^{-2}]^b [T]^c$$[M L^2 T^{-2}] = [M^a L^{a+b} T^{-2a-2b+c}]$Comparing the powers of $M$,$L$,and $T$ on both sides:For $M$: $a = 1$For $L$: $a + b = 2 \implies 1 + b = 2 \implies b = 1$For $T$: $-2a - 2b + c = -2 \implies -2(1) - 2(1) + c = -2 \implies -4 + c = -2 \implies c = 2$Thus,the dimensional formula of energy is $F^1 A^1 T^2$,which is $F A T^2$.

In a system of units,if force $(F)$,acceleration $(A)$,and time $(T)$ are taken as fundamental units,then the dimensional formula of energy is:

Similar Questions

If velocity $v$,acceleration $A$,and force $F$ are chosen as fundamental quantities,then the dimensional formula of angular momentum in terms of $v, A$,and $F$ would be

The position of a body with acceleration '$a$' is given by $x = K{a^m}{t^n}$,where $t$ is time. Find the dimensions of $m$ and $n$.

The position of a particle moving on the $x-$axis is given by $x(t) = A \sin t + B \cos^2 t + Ct^2 + D$,where $t$ is time. The dimension of $\frac{ABC}{D}$ is $-$

Which of the following is dimensionally correct?

$1 \, J$ of energy is to be converted into a new system of units in which length is measured in $10 \, m$,mass in $10 \, kg$,and time in $1 \, minute$. The numerical value of $1 \, J$ in the new system is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module