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physics ncert solutions class 11th Physics Class 11th Mechanical Properties of Solids

9.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

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Vishal Baghel | Contributor-Level 10

5 months ago

Cross-sectional area of wire A, $a_{1}$ = 1 ${m m}^{2}$ = 1 $* 10^{- 6} m^{2}$

Cross-sectional area of wire B, $a_{2}$ = 2 ${m m}^{2}$ = 2 $* 10^{- 6} m^{2}$

Young's modulus for steel, $Y_{1}$ = 2 $* 10^{11}$ N/ $m^{2}$

Young's modulus for aluminium, $Y_{2}$ = 7 $* 10^{10}$ N/ $m^{2}$

Stress in the wire = $\frac{F o r c e}{a r e a}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_{1}}{a_{1}}$ = $\frac{F_{2}}{a_{2}}$ or $\frac{F_{1}}{F_{2}}$ = $\frac{a_{1}}{a_{2}}$ = $\frac{1}{2}$ ………(i)

Where $F_{1}$ is the force exerted on steel wire and $F_{2}$ is the force exerted on aluminium wire

Taking a moment around the point of suspension, we get

$F_{1} y$ = $F_{2} (1.05 - y)$

$\frac{F_{1}}{F_{2}}$ = $\frac{(1.05 - y)}{y}$ ……(ii)

Using equation (i) and (ii), we can

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Cross-sectional area of wire A, $a_{1}$ = 1 ${m m}^{2}$ = 1 $* 10^{- 6} m^{2}$

Cross-sectional area of wire B, $a_{2}$ = 2 ${m m}^{2}$ = 2 $* 10^{- 6} m^{2}$

Young's modulus for steel, $Y_{1}$ = 2 $* 10^{11}$ N/ $m^{2}$

Young's modulus for aluminium, $Y_{2}$ = 7 $* 10^{10}$ N/ $m^{2}$

Stress in the wire = $\frac{F o r c e}{a r e a}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_{1}}{a_{1}}$ = $\frac{F_{2}}{a_{2}}$ or $\frac{F_{1}}{F_{2}}$ = $\frac{a_{1}}{a_{2}}$ = $\frac{1}{2}$ ………(i)

Where $F_{1}$ is the force exerted on steel wire and $F_{2}$ is the force exerted on aluminium wire

Taking a moment around the point of suspension, we get

$F_{1} y$ = $F_{2} (1.05 - y)$

$\frac{F_{1}}{F_{2}}$ = $\frac{(1.05 - y)}{y}$ ……(ii)

Using equation (i) and (ii), we can write

$\frac{1}{2} = \frac{(1.05 - y)}{y}$ or y = 2.1 – 2y or y = 2.1/3 = 0.7 m

We know Young's modulus Y = Stress / Strain

Y = (F/A)/ Strain or Strain = (F/A)/Y

In case of Steel wire, ${S t r a i n}_{s t e e l}$ = $\frac{F_{1}}{a_{1} * Y_{1}}$

In case of Aluminium wire, ${S t r a i n}_{a l u}$ = $\frac{F_{2}}{a_{2} * Y_{2}}$

Since both the strains are equal, we get

$\frac{F_{1}}{F_{2}}$ = $\frac{a_{1} * Y_{1}}{a_{2} * Y_{2}}$ = $\frac{1 * 10^{- 6} * 2 * 10^{11}}{2 * 10^{- 6} * 7 * 10^{10}}$ = 1.429

From (ii) we get

$\frac{(1.05 - y)}{y}$ = 1.429

y = 0.432m

In order to produce an equal strain in the two wires. The mass should be suspended at a distance of 0.432m from the end A

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