Two wires A and B of the same material have r | vedclass

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

Question

(A) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F L}{A \Delta l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

(A) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F L}{A \Delta l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta l$ is the change in length.Rearranging for force,we get $F = \frac{Y A \Delta l}{L}$.Since the material is the same,$Y$ is constant. For the same change in length,$\Delta l$ is constant.Thus,$F \propto \frac{A}{L}$.Since $A = \pi r^2$,we have $F \propto \frac{r^2}{L}$.Given the ratios: $\frac{r_A}{r_B} = \frac{2}{1}$ and $\frac{L_A}{L_B} = \frac{4}{1}$.Therefore,the ratio of forces is $\frac{F_A}{F_B} = \left( \frac{r_A}{r_B} ight)^2 	imes \left( \frac{L_B}{L_A} ight)$.Substituting the values: $\frac{F_A}{F_B} = \left( \frac{2}{1} ight)^2 	imes \left( \frac{1}{4} ight) = 4 	imes \frac{1}{4} = 1$.So,the ratio is $1:1$.

Two wires $A$ and $B$ of the same material have radii in the ratio $2:1$ and lengths in the ratio $4:1$. The ratio of the normal forces required to produce the same change in the lengths of these two wires is

Similar Questions

Which of the following is an organometallic compound?

The major product of the following reaction is

Which kingdom includes unicellular protozoans and primary aquatic eukaryotic organisms?

If $f(0)=0, f(1)=1, f(2)=2$ and $f(x)=f(x-2)+f(x-3)$ for $x=3, 4, 5, \ldots$, then $f(9)$ is equal to

Which one of the following reactions is an example of oxidative decarboxylation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module