You visited us 1 times! Enjoying our articles? Unlock Full Access!

Young's Modulus of Elasticity

The force con...

Question

The force constant of a wire is $k$ and that of another wire of the same material is $2 k$ . When both the wires are stretched, then work done is

W_{2} = 1.5 W_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

W_{2} = 2 W_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

W_{2} = W_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

W_{2} = 0.5 W_{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $W_{2} = 0.5 W_{1}$
$\begin{matrix} In wire 1, force constant = k in wree 2, force constant = 2 K Now, using Hooke's law \begin{matrix} F & = K x_{1} W_{1} & = \frac{1}{2} K (Δ x)^{2} = \frac{1}{2} (K) {(\frac{F}{K})}^{2} W_{1} & = \frac{F^{2}}{2 K} - (1) \end{matrix} \end{matrix}$
$\begin{matrix} For wire 2, F = (2 k) (x_{2}) \begin{matrix} w_{2} = & \frac{1}{2} (2 k) {(x_{2})}_{2}^{2} = \frac{1}{2} (2 k) {(\frac{F}{2 k})}^{2} = \frac{F^{2}}{4 k} - (2) Woing equation (1) and (2) we get w_{2} & = 0.5 w_{1} \end{matrix} \end{matrix}$

Suggest Corrections

Similar questions

The force constant of a wire is k and that of another wire is 2k. When both the wires are stretched through same distance, then the work done

W_{1}

W_{2}

W_{2}

W_{2}

W_{1}

4

W_{2}

(0, 0)

(a, a)

\to F = (3^i + 4^j)

1

O P

2

O Q P

W_{1}

W_{2}

W_{1}

W_{2}

W_{2}

K_{1}

K_{2}

(K_{1} > K_{2})

W_{1}

W_{2}

Join BYJU'S Learning Program