Physics - NLM Question with Solution | TestHub

PhysicsNLMMedium2 minQB

PhysicsMediummultiple choice

A force of 160 N stretches a spring 0.050 m beyond its unstretched length of 0.2 m.

Answer:

B, D

Solution:

Spring constant, $k = \frac{Force}{deformation} = \frac{160}{0.05} = 3200 N / m$

Also any force, $F = kx$ (x-deformation)

for length $ℓ = 0.15 m$ is $∣ x ∣ = 0.05 m$

for $ℓ = 0.3 m$ , $∣ x ∣ = 0.1 m \Rightarrow F = 320 N \Rightarrow$ (B) correct

for $ℓ = 0.275 m$ , $∣ x ∣ = 0.075 m$ , $F = 240 N \Rightarrow$ (D) correct

Stream:JEESubject:PhysicsTopic:NLM

⏱ 2mℹ️ Source: QB

Loading discussions...