A uniform elastic string has length a1 when tension is T1 and length a2 when tension is T2. The amount of work done in stretching it from its natural length to a length (a1 + a2) is :

1 2 a 1 T 1 − a 2 T 2 2 T 1 − T 2 a 1 − a 2

1 2 a 1 T 1 − a 2 T 2 2 T 1 − T 2 a 1 − a 2

1 2 a 1 T 1 + a 2 T 2 2 T 1 − T 2 a 1 − a 2

1 2 T 1 + T 2 2 T 1 − T 2 a 1 − a 2

1 2 T 1 + T 2 T 1 − T 2 a 1 − a 2

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A uniform elastic string has length a₁ when tension is T₁ and length a₂ when tension is T₂. The amount of work done in stretching it from its natural length to a length (a₁ + a₂) is :

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$\frac{1}{2} \frac{{(a_{1} T_{1} - a_{2} T_{2})}^{2}}{(T_{1} - T_{2}) (a_{1} - a_{2})}$

$\frac{1}{2} \frac{{(a_{1} T_{1} + a_{2} T_{2})}^{2}}{(T_{1} - T_{2}) (a_{1} - a_{2})}$

$\frac{1}{2} \frac{{(T_{1} + T_{2})}^{2}}{(T_{1} - T_{2}) (a_{1} - a_{2})}$

$\frac{1}{2} \frac{(T_{1} + T_{2})}{(T_{1} - T_{2}) (a_{1} - a_{2})}$

answer is A.

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Amount of work done = change in PE

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