To solve this problem, we can apply Hooke's law for elastic materials, which states that the extension of an elastic material is directly proportional to the applied force (tension). Mathematically, we can write this as:
ΔL=k⋅T\Delta L = k \cdot T
Where:
• ΔL\Delta L is the extension of the string (change in length),
• TT is the applied tension,
• kk is the proportionality constant (elastic constant) of the string.
Given:
• When the tension T=4 NT = 4 \, \text{N}, the length of the string is aa.
• When the tension T=5 NT = 5 \, \text{N}, the length of the string is bb.
• We are asked to find the length of the string when the tension is 9 N9 \, \text{N}.
Step 1: Determine the relationship between tension and length
Since the extension is directly proportional to the tension, the change in length (ΔL\Delta L) for different tensions can be written as:
ΔL1ΔL2=T1T2\frac{\Delta L_1}{\Delta L_2} = \frac{T_1}{T_2}
Let’s define the original length of the string as L0L_0. Then:
• For T=4 NT = 4 \, \text{N}, the length of the string is L0+ΔL1=aL_0 + \Delta L_1 = a,
• For T=5 NT = 5 \, \text{N}, the length of the string is L0+ΔL2=bL_0 + \Delta L_2 = b.
Thus:
a−L0b−L0=45\frac{a - L_0}{b - L_0} = \frac{4}{5}
This simplifies to:
5(a−L0)=4(b−L0)5(a - L_0) = 4(b - L_0)
Expanding both sides:
5a−5L0=4b−4L05a - 5L_0 = 4b - 4L_0
Rearranging terms:
5a−4b=L05a - 4b = L_0
Step 2: Find the length of the string when the tension is 9 N
Now, let’s calculate the length of the string when the tension is T=9 NT = 9 \, \text{N}. Using the same proportional relationship:
ΔL3ΔL2=95\frac{\Delta L_3}{\Delta L_2} = \frac{9}{5}
Thus, the change in length when the tension is 9 N is:
ΔL3=95⋅(b−L0)\Delta L_3 = \frac{9}{5} \cdot (b - L_0)
So the total length of the string when T=9 NT = 9 \, \text{N} is:
L3=L0+ΔL3=L0+95(b−L0)L_3 = L_0 + \Delta L_3 = L_0 + \frac{9}{5} (b - L_0)
Substitute L0=5a−4bL_0 = 5a - 4b:
L3=(5a−4b)+95(b−(5a−4b))L_3 = (5a - 4b) + \frac{9}{5} (b - (5a - 4b))
Simplify the second term:
L3=5a−4b+95(b−5a+4b)L_3 = 5a - 4b + \frac{9}{5} (b - 5a + 4b) L3=5a−4b+95(5b−5a)L_3 = 5a - 4b + \frac{9}{5} (5b - 5a) L3=5a−4b+95⋅5(b−a)L_3 = 5a - 4b + \frac{9}{5} \cdot 5(b - a) L3=5a−4b+9(b−a)L_3 = 5a - 4b + 9(b - a) L3=5a−4b+9b−9aL_3 = 5a - 4b + 9b - 9a L3=−4a+5bL_3 = -4a + 5b
Final Answer:
Thus, the length of the string when the tension is 9 N is given by 5b−4a5b - 4a.
The correct option is B.
