Question

Answer this I will mark as brainliest it's urgent please ​

Answer 1

Answer:

• The Mass (M₂) required to Break the Wire is 120 Kg.

Given:

1. Mass Suspended (M₁) = 30 Kg.

Explanation:

As Given Diameter is Twice than the Initial Diameter.

as We Know,

D ∝r

• D Denotes Diameter.
• r Denotes Radius.

Therefore,

Radius is Twice than the Initial Radius.

∴2r₁ = r₂

$\rule{300}{1.5}$

From Breaking Stress Formula,

$\large\bigstar\: {\boxed{\tt S = \dfrac{F}{A}}}$

$\bold{Here}\begin{cases}\text{S Denotes Breaking Stress} \\ \text{F Denotes Breaking Force} \\ \text{A Denotes Area}\end{cases}$

Now,

$\large{\boxed{\tt S = \dfrac{F}{A}}}$

Substituting the values,

$\large{\tt \longmapsto S = \dfrac{M \times g}{A}}$

• As the Force is Caused due to the Weight of the wire.

$\large{\tt \longmapsto S = \dfrac{M \times g}{ \pi (r)^2}}$

• As the Area will be πr² ( Circular base )

$\large{\tt \longmapsto S = \dfrac{M_1 \times g}{ \pi (r_1)^2}}$

Substituting,

$\large{\tt \longmapsto S = \dfrac{30 \: Kg\times g}{ \pi (r_1)^2}}$

$\large{\tt \longmapsto S = \dfrac{30 \times g}{ \pi r_1^2}}$

$\large\longmapsto{\underline{\boxed{\tt S = \dfrac{30g}{r_1^2 \pi}}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, This Breaking Stress is There,

From Breaking Stress Formula,

$\large\bigstar\: {\boxed{\tt S = \dfrac{F}{A}}}$

$\bold{Here}\begin{cases}\text{S Denotes Breaking Stress} \\ \text{F Denotes Breaking Force} \\ \text{A Denotes Area}\end{cases}$

Now,

$\large{\boxed{\tt S = \dfrac{F}{A}}}$

Substituting the values,

$\large{\tt \longmapsto S = \dfrac{M_2 \times g}{A}}$

• As the Force is Caused due to the Weight of the wire.

$\large{\tt \longmapsto S = \dfrac{M \times g}{ \pi (r)^2}}$

• As the Area will be πr² ( Circular base )

$\large{\tt \longmapsto S = \dfrac{M_2 \times g}{ \pi (r_2)^2}}$

Substituting,

$\large{\tt \longmapsto \dfrac{30g}{\pi r_1^2} = \dfrac{M_2 \times g}{ \pi (r_2)^2}}$

$\large{\tt \longmapsto \Bigg[\dfrac{30}{r_1^2}\Bigg] \times \dfrac{g}{\pi} = \Bigg[\dfrac{M_2 }{(r_2)^2}\Bigg] \times \dfrac{g}{\pi}}$

$\large{\tt \longmapsto \Bigg[\dfrac{30}{r_1^2}\Bigg] \times \cancel{\dfrac{g}{\pi}} = \Bigg[\dfrac{M_2 }{(r_2)^2}\Bigg] \times \cancel{\dfrac{g}{\pi}}}$

$\large{\tt \longmapsto \Bigg[\dfrac{30}{r_1^2}\Bigg] = \Bigg[\dfrac{M_2 }{(r_2)^2}\Bigg]}$

As we Know,

• 2r₁ = r₂.

$\large{\tt \longmapsto \Bigg[\dfrac{30}{r_1^2}\Bigg] = \Bigg[\dfrac{M_2 }{(2 r_1)^2}\Bigg]}$

$\large{\tt \longmapsto \dfrac{30}{r_1^2} = \dfrac{M_2 }{4 r_1^2}}$

$\large{\tt \longmapsto \dfrac{30}{\cancel{r_1^2}} = \dfrac{M_2 }{4 \cancel{r_1^2}}}$

$\large{\tt \longmapsto 30 = \dfrac{M_2}{4}}$

$\large{\tt \longmapsto M_2 = 30 \times 4}$

$\huge{\boxed{\boxed{\tt M_2 = 120 \: Kg}}}$

∴ The Mass (M₂) required to Break the Wire is 120 Kg.

$\rule{300}{1.5}$