Ncert Exercises & Solutions

A rod of length

l

$l$ and negligible mass is suspended at its two ends by two wires of steel (wire

A

$A$ ) and aluminium (wire

B

$B$ ) of equal lengths (figure). The cross-sectional areas of wires

A

$A$ and

B

$B$ are

1.0 {mm}^{2}

$1.0~\text{mm}^2$ and

2.0 {mm}^{2}

$2.0~\text{mm}^2$ respectively.

(Y_{Al} = 70 \times 10^{9} {N/m}^{2}

$(Y_{\text{Al}}=70\times10^9~\text{N/m}^2$ and

Y_{steel} = 200 \times 10^{9} {N/m}^{2})

$Y_{\text{steel}}=200\times10^9~\text{N/m}^2)$

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

The correct statements are:

1.	(b), (c)	3.	(b), (d)
2.	(a), (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}

$T_{A}$ and

T_{B}

$T_{B}$ be the tensions in wire A and wire B respectively.

For the rotational equilibrium of the system,

T_{ζ} = 0

$Τ_{ζ} = 0$ (Total torque = 0)

\Rightarrow

$\Rightarrow$

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

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

\Rightarrow

$\Rightarrow$

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

$\frac{T_{B}}{T_{A}} = \frac{l - x}{x}$ ...(i)

Stress in wire A,

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

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

Stress in wire B,

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

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

where

a_{A} and a_{B}

$a_{A} and a_{B}$ 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}

$a_{B} = 2 a_{A}$

Now, for equal stresses,

S_{A} = S_{B}

$S_{A} = S_{B}$

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

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

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

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

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

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

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}

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

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

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

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

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

\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}})

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

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

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

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

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

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

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

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

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