Question

A two meter long rod is suspended with the help of two wires of equal length. One wire is of steel and its cross-section is 0.1 $c{m}^2$ and another wire is of brass and its cross-section area is 0.2 $c{m}^2$. If a load W is suspended from the rod and introduced in both the wires is same then the ratio of tensions in them will be

• A. (a) will depend on the position of W
• B. (b) $T_1/T_2=2$
• C. (c) $T_1/T_2=1$
• D. (d) $T_1/T_2=0.5$

Answer 1

The correct option is A (a) will depend on the position of W

Given,$A_S=0.1c{m}^2$

$A_B=0.2c{m}^2$

$T_{1}x=T_2.(l-x)$

Here 'x' is not given

So it will depend on the position of W.