An elastic spring under a tension of 3 ,N has | vedclass

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(D) Let the natural length of the spring be $\ell$ and the spring constant be $K$. The tension $T$ in a spring is given by $T = K(L - \ell)$, where $L

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Let the natural length of the spring be $\ell$ and the spring constant be $K$. The tension $T$ in a spring is given by $T = K(L - \ell)$, where $L$ is the stretched length.For the first case: $3 = K(a - \ell)$ --- $(1)$For the second case: $2 = K(b - \ell)$ --- $(2)$We need to find the tension $T'$ for length $L' = (3a - 2b)$.$T' = K(L' - \ell) = K(3a - 2b - \ell)$We can rewrite this as: $T' = K[3(a - \ell) - 2(b - \ell)]$Substitute the values from equations $(1)$ and $(2)$:$T' = 3[K(a - \ell)] - 2[K(b - \ell)]$$T' = 3(3) - 2(2)$$T' = 9 - 4 = 5 \,N$.

An elastic spring under a tension of $3 \,N$ has a length $a$. Its length is $b$ under a tension of $2 \,N$. For its length $(3a - 2b)$, the value of tension will be . . . . . . $N$.

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