A stone of mass m is tied to one end of a wire of length L. The diameter of the wire is D and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle θ with the vertical. If Young's modulus of the wire is Y, then the increase in the length of the wire is

Question

A stone of mass m is tied to one end of a wire of length L. The diameter of the wire is D and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle θ with the vertical. If Young's modulus of the wire is Y, then the increase in the length of the wire is (A) (4 mgL/π D2Y) (B) (4 mgL/π D2Y sin θ) (C) (4 mgL/π D2Y cos θ) (D) (4 mgL/π D2Y tan θ)

Tardigrade Admin · Accepted Answer

The situation is as shown in the figure. For vertical equilibrium of stone

T cos θ = m g

or

T = \frac{m g}{cos θ}

\dots (i)

As

Y

= \frac{T}{A}

\frac{L}{Δ L}

∴

Δ L

= \frac{T L}{A Y}

(U s in g (i))

= \frac{m gL}{cos θ ( π D ^{2} /4 ) Y}

= \frac{4 m gL}{π D ^{2} Y cos θ}

$\frac{4 m gL}{π D ^{2} Y}$

$\frac{4 m gL}{π D ^{2} Y s in θ}$

$\frac{4 m gL}{π D ^{2} Y cos θ}$

$\frac{4 m gL}{π D ^{2} Y t an θ}$

Q. A stone of mass m is tied to one end of a wire of length L. The diameter of the wire is D and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle θ with the vertical. If Young's modulus of the wire is Y, then the increase in the length of the wire is

πD2Y4mgL​

πD2Ysinθ4mgL​

πD2Ycosθ4mgL​

πD2Ytanθ4mgL​

The situation is as shown in the figure. For vertical equilibrium of stone Tcosθ=mg or T=cosθmg​ …(i) As Y =AT​ ΔLL​ ∴ ΔL =AYTL​ (Using(i)) =cosθ(πD2/4)YmgL​ =πD2Ycosθ4mgL​

$\frac{4 m gL}{π D ^{2} Y}$

$\frac{4 m gL}{π D ^{2} Y s in θ}$

$\frac{4 m gL}{π D ^{2} Y cos θ}$

$\frac{4 m gL}{π D ^{2} Y t an θ}$