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Standard XII

Physics

Shear Strain

We have two w...

Question

We have two wires $W_{1}$ and $W_{2}$ . Both are made of same length. The radius of cross-section of $W_{2}$ is twice that of $W_{2}$ . Same load suspended from both of them. If the strain in $W_{1}$ be $4$ , then that in $W_{2}$ will be

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Solution

The correct option is A $1$

$\begin{matrix} using, \frac{F}{A} & = Y \frac{Δ l_{1}}{l_{1}} \frac{F}{π r^{2}} = Y (4) ⟶ for W_{1} \Rightarrow F & = 4 π r^{2} Y \end{matrix}$

$\begin{matrix} for W_{2} ⟶ \frac{4 π r^{2} Y}{4 π r^{2}} = Y \frac{Δ l_{2}}{l_{2}} \Rightarrow \frac{Δ l_{2}}{l_{2}} = 1 Hence Answer is (A) \end{matrix}$

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We have two wires W1 and W2. Both are made of same length. The radius of cross-section of W2 is twice that of W2. Same load suspended from both of them. If the strain in W1 be 4, then that in W2 will be

The correct option is A 1 using, FA=YΔl1l1Fπr2=Y(4)⟶ for W1⇒F=4πr2Y for W2⟶4πr2Y4πr2=YΔl2l2⇒Δl2l2=1 Hence Answer is (A)

We have two wires $W_{1}$ and $W_{2}$ . Both are made of same length. The radius of cross-section of $W_{2}$ is twice that of $W_{2}$ . Same load suspended from both of them. If the strain in $W_{1}$ be $4$ , then that in $W_{2}$ will be