Question

Q:- If momentum is increased by 13% than k.e increase by %
1. 37%
2.27%
3.47%
4.Non of these.​

Answer 1

if momentum is increased by 13% , than we have to find percentage increment in kinetic energy ..

solution : we know, the relation between kinetic energy and linear momentum is given by, K.E = $\frac{P^2}{2m}$

where , P is momentum and m is mass of the particle.

let initial momentum is P

∴ initial kinetic energy , K.E = $\frac{P^2}{2m}$ ...(1)

momentum is increased by 13%.

new momentum of the particle is 1.13P

∴ new kinetic energy of the particle, K.E' = $\frac{(1.13P)^2}{2m}=\frac{1.2769P^2}{2m}$

⇒ K.E' = 1.2769K.E [ from eq (1) ]

now percentage change in kinetic energy = $\frac{K.E'-K.E}{K.E}\times100$

= $\frac{1.2769K.E-K.E}{K.E}\times100$

= 27.69%

therefore the kinetice energy increased by 27 % (nearest value of 27.69%) so the correct option (2) is correct.

Answer 2

Answer:

27%

Explanation:

you can use this formula for any question