Correct Answer - Option 3 : 50

__CONCEPT:__

Kinetic Energy

- The energy possessed by a particle by the virtue of its motion is called kinetic energy. It is given by,

\(⇒ KE=\frac{1}{2}mv^{2}\)

Linear Momentum:

- It is defined as the product of the mass of an object, m, and its velocity, v. It, is a vector quantity.

⇒ P = mv

Where KE = kinetic energy, P = linear momentum, m = mass and v = velocity

__CALCULATION:__

Given when %ΔP = 1%, %ΔKE = x% and when %ΔKE = 300%, %ΔP = y%

- We know that the
**relation between the kinetic energy and the momentum** is given as,

\(⇒ KE=\frac{P^{2}}{2m}\) -----(1)

Let P_{1} = P and KE_{1} = KE

For first case when %ΔP = 1% and %ΔKE = x%,

\(⇒ P_{2}=P+\frac{1}{100}P\)

\(⇒ P_{2}=\frac{101}{100}P\) -----(2)

\(⇒ KE_{2}=KE+\frac{x}{100}KE\)

\(⇒ KE_{2}=\frac{(100+x)}{100}KE\) -----(3)

By equation 1, equation 2, and equation 3,

\(⇒ KE_{2}=\frac{P_{2}^{2}}{2m}\)

\(⇒ \frac{(100+x)}{100}KE=\frac{101^{2}}{100^2}\frac{P^2}{2m}\)

\(⇒(100+x)KE=\frac{101^{2}}{100}KE\)

⇒ 100 + x = 102.01

⇒ x = 2.01 -----(4)

For second case when %ΔP = y% and %ΔKE = 300%,

\(⇒ P'_{2}=P+\frac{y}{100}P\)

\(⇒ P'_{2}=\frac{100+y}{100}P\) -----(5)

\(⇒ KE'_{2}=KE+\frac{300}{100}KE\)

\(⇒ KE'_{2}=4KE\) -----(6)

By equation 1, equation 5, and equation 6,

\(⇒ KE'_{2}=\frac{(P_{2}')^{2}}{2m}\)

\(⇒ 4KE=\frac{\left ( 100+y \right )^{2}}{100^{2}}\frac{P^{2}}{2m}\)

\(⇒ 4KE=\frac{\left ( 100+y \right )^{2}}{100^{2}}KE\)

⇒ 100 + y = 200

⇒ y = 100 -----(7)

By dividing equation 7 and equation 4,

\(\Rightarrow \frac{y}{x}=\frac{100}{2.01}\)

\(\Rightarrow \frac{y}{x}=50\)

- Hence, option 3 is correct.