For given values to Tand m, resonance length in winter is x_(1) cm, and in summer, it is x_(2) cm. Then

For given values to Tand m, resonance length in winter is x_(1) cm, an

1.	For given values to Tand m, resonance length in winter is x_(1) cm, and in summer, it is x_(2) cm. Then
Answer» `x_(1) = x_(2)` `x_(2) gt x_(1)` `x_(2) lt x_(1)` cannot say Solution :`N = (upsilon)/(LAMBDA) = (1)/(2 L) sqrt((T)/(m))` Is summer, `upsilon` is more. Therefore, resonance LENGTH `(l)` is more , and vice- VERSA. Therefore, `x_(2) gt x_(1)`

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For given values to Tand m, resonance length in winter is x_(1) cm, and in summer, it is x_(2) cm. Then

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