A light rod of length 200,cm is suspended fro | vedclass

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

Question

(A) Let $T_1$ be the tension in the steel wire and $T_2$ be the tension in the brass wire. Let $A_1 = 0.1\,cm^2$ and $A_2 = 0.2\,cm^2$ be their respec

Vedclass · Accepted Answer

$\frac{4}{3}\,m$ from steel wire

Vedclass · Answer

(A) Let $T_1$ be the tension in the steel wire and $T_2$ be the tension in the brass wire. Let $A_1 = 0.1\,cm^2$ and $A_2 = 0.2\,cm^2$ be their respective cross-sectional areas.Since the stresses are equal,$\frac{T_1}{A_1} = \frac{T_2}{A_2}$.Substituting the values,$\frac{T_1}{0.1} = \frac{T_2}{0.2}$,which gives $T_2 = 2T_1$.For the translatory equilibrium of the rod,$T_1 + T_2 = W$,where $W$ is the weight hung.Substituting $T_2 = 2T_1$,we get $T_1 + 2T_1 = W$,so $T_1 = \frac{W}{3}$ and $T_2 = \frac{2W}{3}$.Let $x$ be the distance of the weight $W$ from the steel wire. For rotational equilibrium about the point where the weight is hung,the torques must balance: $T_1 x = T_2 (L - x)$,where $L = 200\,cm = 2\,m$.Substituting the values,$\frac{W}{3} x = \frac{2W}{3} (2 - x)$.Dividing by $\frac{W}{3}$,we get $x = 2(2 - x) = 4 - 2x$.Thus,$3x = 4$,which gives $x = \frac{4}{3}\,m$ from the steel wire.

Similar Questions

$A$ steel wire can sustain $100\,kg$ weight without breaking. If the wire is cut into two equal parts, each part can sustain a weight of ......... $kg$.

In a series combination of copper and steel wires of same length and same diameter,a force is applied at one of their ends while the other end is kept fixed. The combined length is increased by $2 \ cm$. The wires will have ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module