QJEEM24Pha602: JEE Main 2024 Physics - Class 11 JEEM24Pha602 Solution: An elastic spring under tension of has a length . Its length is …

To determine the tension for the elastic spring's length of $(3a - 2b)$ , we can use Hooke's law which states that the force exerted by a spring is proportional to the extension or compression of the spring from its natural length. Mathematically, Hooke's law is given by:

$F = k \cdot \Delta x$

where:

• $F$ is the force exerted by the spring.
• $k$ is the spring constant (a measure of the stiffness of the spring).
• $\Delta x$ is the displacement from the natural length.

Let's denote the natural (unstretched) length of the spring as $l_0$ . Given that the spring's length is $a$ under a tension of $3 \mathrm{~N}$ , and its length is $b$ under a tension of $2 \mathrm{~N}$ , we can write the equations as:

$3 \mathrm{~N} = k \cdot (a - l_0)$

$2 \mathrm{~N} = k \cdot (b - l_0)$

Next, we need to find the length displacement $(3a - 2b)$ in terms of the natural length $l_0$ . We express the displacements as:

$x = (3a - 2b) - l_0$

To solve for this, let's express $a - l_0$ and $b - l_0$ from the given conditions:

$a - l_0 = \frac{3 \mathrm{~N}}{k}$

$b - l_0 = \frac{2 \mathrm{~N}}{k}$

Substitute these into the displacement equation:

$x = (3a - 2b) - l_0 = 3 \left( \frac{3 \mathrm{~N}}{k} + l_0 \right) - 2 \left( \frac{2 \mathrm{~N}}{k} + l_0 \right) - l_0$

On simplifying, we get:

$x = 3 \frac{3 \mathrm{~N}}{k} + 3 l_0 - 2 \frac{2 \mathrm{~N}}{k} - 2 l_0 - l_0$

$x = \frac{9 \mathrm{~N}}{k} + 3 l_0 - \frac{4 \mathrm{~N}}{k} - 3 l_0$

$x = \frac{5 \mathrm{~N}}{k}$

Finally, by Hooke's Law, the force corresponding to this displacement is:

$F = k \cdot x = k \cdot \frac{5 \mathrm{~N}}{k} = 5 \mathrm{~N}$

Therefore, the tension required for the elastic spring's length $(3a - 2b)$ is 5 N.