Question

A light rigid bar is suspended horizontally from two vertical wires, one of steel and one of brass, as shown in figure. Each wire is 2.00 m long. The diameter of the steel wire is 0.60 mm and the length of the bar AB is 0.20 m. When a mass of 10 kg is suspended from the centre of AB bar remains horizontal.
i) What is the tension in each wire ?
ii) Calculate the extension of the steel wire and the energy stored in it.
iii) Calculate the diameter of the brass wire. ​

Answer 1

$\Large\bf{\color{cyan}CaLcUlAtIoN,}$

➡ See the attachment free body diagram.

$\bf\blue{As\:shown \:in\:diagram,}$

=》T₁ + T₂ = mg

$\bf\red{Where,}$

• m is the mass of AB bar = 10 kg

• g is gravitational acceleration = 9.8 m/s²

=》 T₁ + T₂ = 10 × 9.8

=》 T₁ + T₂ = 98 N ------(1)

$\bf\pink{Now,}$

✅ Consider torque at point O (Centre/mid pt. of AB bar).

=》 (T₁ × r) - (T₂ × r) = 0

$\bf\orange{Here,}$

• r is the length of OA for T₁ & OB for T₂. But both r are same, because O is the midpoint of AB.

=》 r (T₁ - T₂) = 0

=》 T₁ - T₂ = 0

=》 T₁ = T₂ = $\bf{\dfrac{98}{2}\:=\:49\:N}$

$\Large\bold\therefore$ The tension in each wire is 49 N.

[NOTE ➡ If we take g = 10 m/s², then the value of tension will be different.]

$\bf\purple{We\:know \:that,}$

$\red\bigstar\:\:\bf\orange{Young's\:Modulus\:(E)\:=\:\dfrac{Stress}{Strain}\:}$

$:\implies\:\:\bf{E\:=\:\dfrac{\frac{Force}{Area}}{\frac{\triangle{l}}{l}}\:}$

$:\implies\:\:\bf{E\:=\:\dfrac{\frac{Force}{\pi{r^2}}}{\frac{\triangle{l}}{l}}\:}$

$\bf\green{Where,}$

• E be the Young's modulus of steel = 2.0 × 10¹¹ Pa

• Force = T₁ = 49 N

• r = radius of steel wire = $\bf{\dfrac{Diameter}{2}\:=\:\dfrac{0.60}{2}\:}$= 0.30 mm = 0.0003 m

• l = 2.00 m.

$:\implies\:\:\bf{E\:=\:\dfrac{Force}{\pi{r^2}}\times{\dfrac{l}{\triangle{l}}}\:}$

$:\implies\:\:\bf{\triangle{l}\:=\:\dfrac{Force}{\pi{r^2}}\times{\dfrac{l}{E}}\:}$

$:\implies\:\:\bf{\triangle{l}\:=\:\dfrac{49}{\pi\times{(0.0003)^2}}\times{\dfrac{2}{2\times{10^{11}}}}\:}$

$:\implies\:\:\bf{\triangle{l}\:=\:\dfrac{49}{\pi\times{9\times{10^{-8}}}}\times{1\times{10^{-11}}}\:}$

$:\implies\:\:\bf{\triangle{l}\:=\:\dfrac{49}{9\pi}\times{10^8}\times{10^{-11}}\:}$

$:\implies\:\:\bf{\triangle{l}\:=\:1.737\times{10^{-3}}\:m}$

$\Large\bold\therefore$ The extension of the steel wire is 1.737 × 10⁻³ m.

$\bf{\color{olive}We\:know \:that,}$

$\purple\bigstar\:\:\bf\green{Energy\:=\:\dfrac{1}{2}\times{Stess}\times{Strain}\times{Volume}\:}$

$:\implies\:\:\bf{Energy\:=\:\dfrac{1}{2}\times{\dfrac{F}{A}}\times{\dfrac{\triangle{l}}{l}}\times{A\:.\:l}\:}$

$:\implies\:\:\bf{Energy\:=\:\dfrac{1}{2}\times{F}\times{\triangle{l}}\:}$

$:\implies\:\:\bf{Energy\:=\:\dfrac{1}{2}\times{49}\times{1.737\times{10^{-3}}}\:}$

$:\implies\:\:\bf\blue{Energy\:=\:42.5\times{10^{-3}}\:}$

$\Large\bold\therefore$ The energy stored is 42.5 × 10⁻³ J

$\bf\orange{As\:we\:know,}$

☆ Extension is same in both steel and brass wire. Because the resultant tension is same in both.

$\longmapsto\:\:\bf{\triangle{l}_{brass}\:=\:1.737\times{10^{-3}}\:}$

$\bf\red{So,}$

$:\implies\:\:\bf{E_{brass}\:=\:\dfrac{Force}{\pi{r^2}}\times{\dfrac{l}{\triangle{l}}}\:}$

$:\implies\:\:\bf{\pi{r^2}\:=\:\dfrac{Force}{E_{brass}}\times{\dfrac{l}{\triangle{l}}}\:}$

$\bf\purple{Where,}$

• $\bf{E_{brass}}$ = 1.0 × 10¹¹ Pa

• Force = T₂ = 49 N

• l = 2.00 m

• $\bf{\triangle{l}}$ = 1.737 × 10⁻³ m

$:\implies\:\:\bf{\pi\:{\Big(\dfrac{Diameter}{2}\Big)^2}\:=\:\dfrac{49}{1.0 \times 10^{11}}\times{\dfrac{2}{1.737\times{10^{-3}}}}\:}$

$:\implies\:\:\bf{\pi\times{\dfrac{(Diameter)^2}{4}}\:=\:49 \times 10^{-11}\times{1.15\times{10^{3}}}\:}$

$:\implies\:\:\bf{(Diameter)^2\:=\:56.35 \times 10^{-8}\times{\dfrac{4}{\pi}}\:}$

$:\implies\:\:\bf{(Diameter)^2\:=\:56.35 \times 10^{-8}\times{\dfrac{4}{3.14}}\:}$

$:\implies\:\:\bf{(Diameter)^2\:=\:56.35 \times 10^{-8}\times{1.27}\:}$

$:\implies\:\:\bf{Diameter\:=\:\sqrt{71.5 \times 10^{-8}}\:}$

$:\implies\:\:\bf\green{Diameter\:=\:8.45 \times 10^{-4}\:m}$

$\Large\bold\therefore$ The diameter of the brass wire is 8.45 × 10⁻⁴ m.

Answer 2

Answer:

=》T₁ + T₂ = mg

\begin{gathered}\bf\red{Where,} \\ \end{gathered}

Where,

m is the mass of AB bar = 10 kg

g is gravitational acceleration = 9.8 m/s²

=》 T₁ + T₂ = 10 × 9.8

=》 T₁ + T₂ = 98 N ------(1)

\begin{gathered}\bf\pink{Now,} \\ \end{gathered}

Now,

✅ Consider torque at point O (Centre/mid pt. of AB bar).

=》 (T₁ × r) - (T₂ × r) = 0

\begin{gathered}\bf\orange{Here,} \\ \end{gathered}

Here,

r is the length of OA for T₁ & OB for T₂. But both r are same, because O is the midpoint of AB.

=》 r (T₁ - T₂) = 0

=》 T₁ - T₂ = 0

=》 T₁ = T₂ = \bf{\dfrac{98}{2}\:=\:49\:N}

2

98

=49N

\Large\bold\therefore∴ The tension in each wire is 49 N.

[NOTE ➡ If we take g = 10 m/s², then the value of tension will be different.]

\begin{gathered}\bf\purple{We\:know \:that,} \\ \end{gathered}

Weknowthat,

\begin{gathered}\red\bigstar\:\:\bf\orange{Young's\:Modulus\:(E)\:=\:\dfrac{Stress}{Strain}\:} \\ \end{gathered}

★Young

′

sModulus(E)=

Strain

Stress

\begin{gathered}:\implies\:\:\bf{E\:=\:\dfrac{\frac{Force}{Area}}{\frac{\triangle{l}}{l}}\:} \\ \end{gathered}

:⟹E=

l

△l

Area

Force

\begin{gathered}:\implies\:\:\bf{E\:=\:\dfrac{\frac{Force}{\pi{r^2}}}{\frac{\triangle{l}}{l}}\:} \\ \end{gathered}

:⟹E=

l

△l

πr

2

Force

\begin{gathered}\bf\green{Where,} \\ \end{gathered}

Where,

E be the Young's modulus of steel = 2.0 × 10¹¹ Pa

Force = T₁ = 49 N

r = radius of steel wire = \begin{gathered}\bf{\dfrac{Di

hope it's helpful to you