If speed (V),acceleration (A),and force (F) a | vedclass

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

Question

(D) Young's modulus $(Y)$ is defined as the ratio of stress to strain. Since strain is dimensionless,the dimension of $Y$ is the same as that of stres

Vedclass · Accepted Answer

$V^{-4} A^2 F$

Vedclass · Answer

(D) Young's modulus $(Y)$ is defined as the ratio of stress to strain. Since strain is dimensionless,the dimension of $Y$ is the same as that of stress (Force/Area).$[Y] = [F] / [L^2]$.We are given fundamental units $V, A, F$.We know $F = M \cdot A$,so $M = F \cdot A^{-1}$.Also,$V = L \cdot T^{-1}$ and $A = L \cdot T^{-2}$.Dividing $A$ by $V$: $A/V = (L \cdot T^{-2}) / (L \cdot T^{-1}) = T^{-1}$,so $T = V \cdot A^{-1}$.Now,$L = V \cdot T = V \cdot (V \cdot A^{-1}) = V^2 \cdot A^{-1}$.Therefore,$L^2 = (V^2 \cdot A^{-1})^2 = V^4 \cdot A^{-2}$.Substituting these into the expression for $Y$:$[Y] = [F] / [L^2] = F / (V^4 \cdot A^{-2}) = F \cdot V^{-4} \cdot A^2$.

If speed $(V)$,acceleration $(A)$,and force $(F)$ are considered as fundamental units,the dimension of Young's modulus will be

Similar Questions

If momentum $(P)$,area $(A)$ and time $(T)$ are taken to be the fundamental quantities,then the dimensional formula for energy is:

$(P+\frac{a}{V^2})(V-b)=RT$ represents the equation of state of some gases. Where $P$ is the pressure,$V$ is the volume,$T$ is the temperature and $a, b, R$ are the constants. The physical quantity,which has the same dimensional formula as that of $\frac{b^2}{a}$,will be

Assertion $(A)$: Energy per unit volume and angular momentum can be added dimensionally.
Reason $(R)$: Physical quantities having same dimensions can be added or subtracted.

Consider two physical quantities $A$ and $B$ related to each other as $E = \frac{B - x^2}{At}$,where $E, x,$ and $t$ have dimensions of energy,length,and time respectively. The dimension of $AB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module