Question

Please help, show your working.​

Answer 1

Let Mass m is hung to the initial spring of unstreached length L and Force Constant k.

Also,

Let the the extension in the spring be = x.

At Balanced Condition :-

$\sf mg + k \mathtt x = 0 \\ \\ \large:\longmapsto \boxed{\pmb{ \boxed{k = \frac{mg}{x} }}} \: \: - - (1)$

Note : Negative sign is ignored because it only represents the direction of the Force

We Know That ,

Extension in the spring is directly proportional to length of unstretched spring,

Hence,

$\sf \mathtt x \propto L \\ \\ \large :\longmapsto \bf x = pL\\ \pink{( \bf for \: some \: const. \: p)}$

Putting Value of x in (1)

$\large:\longmapsto \boxed{ \pmb{ \boxed{k = \frac{mg}{pL} }}} \: \: - - - (2)$

When The spring is cut into two parts of unstreached length $\sf l_1 \: And \: l_2$

and Force Constant $\sf k_1 \: And \: k_2$

We Have,

$l_1 = \eta \: l_2$

Obviously,

$l_1 + l_2 = L \: \: - - - (3) \\ \\ :\longmapsto \eta \: l_2 + l_2 = L \\ \\ :\longmapsto l_2( \eta + 1) = L \\ \\ \large :\longmapsto \boxed{ \pmb{ \boxed{l_2 = \frac{L}{ \eta + 1} }}} \: \: - - - (4)$

Putting (4) in (3) We Get :

$l_1 =L - \frac{L}{ \eta + 1} \\ \\ = \frac{\eta \: L + \cancel{ L } - \cancel{ L}}{ \eta + 1} \\ \\ \large :\longmapsto \boxed{ \pmb{ \boxed{l_1= \frac{ \eta \: L}{ \eta + 1} }}} \: \: - - - (5)$

From (2) :-

$\sf k_1 = \dfrac{mg}{ p \mathtt l_1}$

Using (4) :-

$\sf k_1 = \dfrac{mg}{p \big( \dfrac{ \eta \: L}{ \eta + 1} \big)} \\ \\ = \sf\frac{mg}{pL} \bigg( \frac{ \eta + 1}{ \eta} \bigg) \\ \\ \purple{ \large :\longmapsto  \underline {\boxed{{\pmb{ k_1 = k \bigg( \frac{ \eta + 1}{ \eta} \bigg)} }}}}$

Also,

$\sf k_2= \dfrac{mg}{ p \mathtt l_2}$

Using (5) :-

$\sf k_2 = \dfrac{mg}{p \big( \dfrac{ L}{ \eta + 1} \big)} \\ \\ = \sf\frac{mg}{pL} ( \eta + 1) \\ \\ \purple{ \large :\longmapsto  \underline {\boxed{{\pmb{ k_2 = k ( \eta + 1) } }}}}$

$\large\underline \pink{ \underline{\pmb{ \frak{ \text{H}ence \: Option \: 4 \: is \: Correct} }}}$

Answer 2

See the Answer in picture attached

hope it helps you