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

9.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10-2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

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Vishal Baghel | Contributor-Level 10

5 months ago

Length of the mild steel wire, l = 1.0 m

Area of cross-section, A = 0.5 $\times 10^{- 2}$ ${c m}^{2}$ = 0.5 $\times 10^{- 6} m^{2}$

A mass of 100 gm is suspended at the midpoint.

m = 100 gm= 0.1 kg

Due to the weight, the wire dips, as shown in the figure.

Original length = XZ, depression = l

The final length of the wire after it dips = XO + OZ

Increase in length of the wire, Δl = (XO + OZ) – XZ ……(i)

From Pythagoras theorem

XO = OZ = $\sqrt{{0.5}^{2} + l^{2}}$

From equation (i)

Δl = 2 $\times \sqrt{{0.5}^{2} + l^{2}}$ - 1.0 = 2 $\times 0.5 \times \sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0 = $\sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0

Neglecting the smaller terms, we can write, Δl = $\frac{l^{2}}{0.5}$

We know, Strain = $\frac{I n c r e a s e i n l e n g t h}{O r i g i n a l l e n g t h}$

Let T be the tension in the wire, then

mg = 2T

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Length of the mild steel wire, l = 1.0 m

Area of cross-section, A = 0.5 $\times 10^{- 2}$ ${c m}^{2}$ = 0.5 $\times 10^{- 6} m^{2}$

A mass of 100 gm is suspended at the midpoint.

m = 100 gm= 0.1 kg

Due to the weight, the wire dips, as shown in the figure.

Original length = XZ, depression = l

The final length of the wire after it dips = XO + OZ

Increase in length of the wire, Δl = (XO + OZ) – XZ ……(i)

From Pythagoras theorem

XO = OZ = $\sqrt{{0.5}^{2} + l^{2}}$

From equation (i)

Δl = 2 $\times \sqrt{{0.5}^{2} + l^{2}}$ - 1.0 = 2 $\times 0.5 \times \sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0 = $\sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0

Neglecting the smaller terms, we can write, Δl = $\frac{l^{2}}{0.5}$

We know, Strain = $\frac{I n c r e a s e i n l e n g t h}{O r i g i n a l l e n g t h}$

Let T be the tension in the wire, then

mg = 2T $\cos ? θ$

From the figure

$\cos ? θ$ = $\frac{O Y}{O X}$ = $\frac{0.5}{\sqrt{{0.5}^{2} + l^{2}}}$ = $\frac{0.5}{0.5 \times \sqrt{1 + {(\frac{l}{0.5})}^{2}}}$

Expanding the expression and eliminating the higher terms:

$\cos ? θ$ = $\frac{l}{0.5 \times \sqrt{1 + {(\frac{1}{2}) (\frac{l}{0.5})}^{2}}}$

Since ( 1+ $\frac{l^{2}}{0.5}$ ) = 1 for small l, $\cos ? θ$ = $\frac{1}{0.5}$

Then T = $\frac{m g}{2 (\frac{l}{0.5})}$ = $\frac{m g \times 0.5}{2 l}$ = $\frac{m g}{4 l}$

Stress = $\frac{T e n s i o n}{A r e a}$ = $\frac{m g}{4 l \times A}$

Young’s modulus = $\frac{S t r e s s}{S t r a i n}$

Y = $\frac{m g \times 0.5}{? l \times A \times l^{2}}$

l= $\sqrt{\frac{m g \times 0.5}{4 Y A}}$

For steel, Y = 2 $\times 10^{11}$ Pa, then

l = $\sqrt{\frac{0.1 \times 9.8 \times 0.5}{4 \times 2 \times 10^{11} \times 0.5 \times 10^{- 6}}}$ = 0.0106 m

Hence, the depression at the midpoint is 0.0106 m

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9.19 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10-2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

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