Question

Same spring is attached with 2kg, 3kg and 1 kg blocks in three different cases as shown in the figure. If $x_1,x_2\text{and}{x}_3$ be the extensions in the spring in these three cases then

• A. $x_1=0,x_3>x_2$
• B. $x_2>x_1>x_3$
• C. $x_3>x_1>x_2$
• D. $x_1>x_2>x_3$

Answer 1

The correct option is B $x_2>x_1>x_3$

Let us find the acceleration of the system in all three cases. So, we have acceleration
$a=\frac{\text{(Supporting Force-Opposing force)}}{\text{Total mass}}$
Thus, In (case 1)
$a=\frac{2g-2g}{4}=0$
$\therefore T=k{x}_1=2g\Rightarrow x_1=\frac{2g}{k}$

In (case 2)
$a=\frac{3g-2g}{5}=\frac{g}{5}$
$\therefore T=k{x}_2=3g+3\times \frac{g}{5}=\frac{18g}{5}\Rightarrow x_2=\frac{18g}{5k}$

In (case 3)
$a=\frac{2g-g}{3}=\frac{g}{3}$
$\therefore T=k{x}_3=g-\frac{g}{3}=\frac{2g}{3}\Rightarrow x_3=\frac{2g}{3k}$
Now, finding the ratio of $x_1$, $x_2$ and $x_3$ we have
$x_1:x_2:x_3=2g:3.6g:0.67g$
Thus, $x_2>x_1>x_3$