Question

Plz Solve This Question With Proper Explanation!​

Answer 1

The potential energy stored in a stretched spring due to its elasticity is,

$\displaystyle\longrightarrow\sf{U=\dfrac{1}{2}kx^2}$

where,

• $\sf{k=}$ Spring Constant
• $\sf{x=}$ Extension or Compression

In case of the same spring the spring constant $\sf{k}$ must be constant.

So we get,

$\displaystyle\longrightarrow\sf{U\propto x^2}$

Therefore,

$\displaystyle\longrightarrow\sf{\dfrac{U_1}{U_2}=\dfrac{\left(x_1\right)^2}{(x_2)^2}}$

Here,

• $\sf{x_1=5\ cm}$

• $\sf{U_1=E}$

• $\sf{x_2=10\ cm}$

Then,

$\displaystyle\longrightarrow\sf{\dfrac{E}{U_2}=\dfrac{5^2}{10^2}}$

$\displaystyle\longrightarrow\sf{\dfrac{E}{U_2}=\dfrac{25}{100}}$

$\displaystyle\longrightarrow\sf{\dfrac{E}{U_2}=\dfrac{1}{4}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{U_2=4E}}}$

Hence (b) is the answer.