Ncert Exercises & Solutions

A rod of length \(l\) and negligible mass is suspended at its two ends by two wires of steel (wire \(A\)) and aluminium (wire \(B\)) of equal lengths (figure). The cross-sectional areas of wires \(A\) and \(B\) are \(1.0~\text{mm}^2\) and \(2.0~\text{mm}^2\) respectively.
\((Y_{\text{Al}}=70\times10^9~\text{N/m}^2\) and \(Y_{\text{steel}}=200\times10^9~\text{N/m}^2)\)

(a)	Mass \(m\) should be suspended close to wire \(A\) to have equal stresses in both wires.
(b)	Mass \(m\) should be suspended close to \(B\) to have equal stresses in both wires.
(c)	Mass \(m\) should be suspended in the middle of the wires to have equal stresses in both wires.
(d)	Mass \(m\) should be suspended close to wire \(A\) to have equal strain in both wires.

The correct statements are:
1. (b), (c)
2. (a), (d)
3. (b), (d)
4. (c), (d)

(3) Hint: The stress and strain in the wires depend on Young's modulus of the wires.

Step 1: Find the stresses in the wires.

Let the mass is placed at x from end B.

Let

T_{A}

and

T_{B}

be the tensions in wire A and wire B respectively.

For the rotational equilibrium of the system,

Τ_{ζ} = 0

(Total torque = 0)

\Rightarrow

T_{B} x - T_{A} (l - x) = 0

\Rightarrow

\frac{T_{B}}{T_{A}} = \frac{l - x}{x}

...(i)

Stress in wire A,

= S_{A} = \frac{T_{A}}{a_{A}}

Stress in wire B,

S_{B} = \frac{T_{B}}{a_{B_{}}}

where

a_{A} and a_{B}

are cross-sectional areas of wires A and B respectively.

Step 2: Find the location of the mass for equal stresses in the wires.

According to the question,

a_{B} = 2 a_{A}

Now, for equal stresses,

S_{A} = S_{B}

\Rightarrow \frac{T_{A}}{a_{A}} = \frac{T_{B}}{a_{B}} \Rightarrow \frac{T_{B}}{T_{A}} = \frac{a_{B}}{a_{A}} = 2

\Rightarrow \frac{l - x}{x} = 2 \Rightarrow \frac{l}{x} - 1 = 2

\Rightarrow x = \frac{l}{3} \Rightarrow l - x = l - l / 3 = \frac{2 l}{3}

Hence, the mass m should be placed closer to B.

Step 3: Find the location of the mass for equal strain in the wires.

For equal strain,

{(strain)}_{A} = {(strain)}_{B}

\Rightarrow \frac{(Y_{A})}{S_{A}} = \frac{Y_{B}}{S_{B}} (Where Y_{A} and Y_{B} are Young' s moduli)

\Rightarrow \frac{Y_{steel}}{T_{A} / a_{A}} = \frac{Y_{Al}}{T_{B} / a_{B}}

\Rightarrow \frac{Y_{steel}}{Y_{Al}} = \frac{T_{A}}{T_{B}} \times \frac{a_{B}}{a_{A}} = (\frac{x}{l - x}) (\frac{2 a_{A}}{a_{A}})

\Rightarrow \frac{200 \times 10^{9}}{70 \times 10^{9}} = \frac{2 x}{l - x} \Rightarrow \frac{20}{7} = \frac{2 x}{l - x}

\Rightarrow \frac{10}{7} = \frac{x}{l - x} \Rightarrow 10 l - 10 x = 7 x

\Rightarrow 17 x = 10 l \Rightarrow x = \frac{10 l}{17}

l - x = l - \frac{10 l}{17} = \frac{7 l}{17}

Hence, the mass m should be placed closer to wire A.

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