An object can be hanged to a 1.2 m long wire of diameter 2 mm, and the wire does not break. However, if the object is slightly displaced and begins to slowly swing, the wire breaks. If another piece of wire having the same length and material but wider diameter is used, the speed of the same'swinging object at lowest point can even be 6 m / s. Calculate the least diameter (in mm ) of this wider wire.

Question

An object can be hanged to a 1.2 m long wire of diameter 2 mm, and the wire does not break. However, if the object is slightly displaced and begins to slowly swing, the wire breaks. If another piece of wire having the same length and material but wider diameter is used, the speed of the same'swinging object at lowest point can even be 6 m / s. Calculate the least diameter (in mm ) of this wider wire. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

m g=σ(π((d1/2))2) m g+(m v2/l)=σ(π((d2/2))2) So, 1+(v2/l g)=((d2/d1))2 ⇒ 1+(36/12)=((d2/2 mm ))2 ⇒ d2=4 mm

mg=σ(π(2d1​​)2) mg+lmv2​=σ(π(2d2​​)2) So, 1+lgv2​=(d1​d2​​)2 ⇒1+1236​=(2mmd2​​)2 ⇒d2​=4mm

$m g = σ (π (\frac{d _{1}}{2})^{2})$
$m g + \frac{m v ^{2}}{l} = σ (π (\frac{d _{2}}{2})^{2})$
So, $1 + \frac{v ^{2}}{l g} = (\frac{d _{2}}{d _{1}})^{2}$
$\Rightarrow 1 + \frac{36}{12} = (\frac{d _{2}}{2 mm})^{2}$
$\Rightarrow d_{2} = 4 mm$