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# 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 $0.1 \mathrm{cm}^{2}$ and another wire is of brass and its cross-section area is $0.2 \mathrm{cm}^{2}$. If a load $\mathrm{W}$ is suspended from the rod and introduced in both the wires is same then the ratio of tensions in them will be(a) will depend on the position of $\mathrm{W}$(b) $\mathrm{T}_{1} / \mathrm{T}_{2}=2$(c) $\mathrm{T}_{1} / \mathrm{T}_{2}=1$(d) $\mathrm{T}_{1} / \mathrm{T}_{2}=0.5$

Last updated date: 19th Sep 2024
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Hint: We know that tension is nothing but the drawing force acting on the body when it is hung from objects like chain, cable, string etc. It is represented by T. The direction of tension is the pull which is given the name tension. Thus, the tension will point away from the mass in the direction of the string or rope. In case of the hanging mass, the string pulls it upwards, so the string or rope exerts an upper force on the mass and the tension will be on the upper side. The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire. Based on this concept we have to solve this question.

$\mathrm{A}_{\mathrm{S}}=0.1 \mathrm{cm}^{2}$
$\mathrm{A}_{\mathrm{B}}=0.2 \mathrm{cm}^{2}$
$\mathrm{T}_{1} \mathrm{x}=\mathrm{T}_{2} \cdot(\mathrm{l}-\mathrm{x})$