99 Percentile Qs Bank for JEE MainPhysicsLaws of Motion
Two massless springs of spring constant K₁ and K₂ are connected one after the other forming a single chain, suspended vertically and certain mass is attached to the free end. If e₁ and e₂ are their respective extensions and f is their stretching force, the total extension produced is
Options
- Af (K₁-K₂ )
- Bf ( rac 1 K₁ - 1 K₂ )
- Cf (K₁+K₂ )
- Df ( 1 K₁ + 1 K₂ )
Correct answer
D. f ( 1 K₁ + 1 K₂ )
Step-by-step solution
The total restoring force is related to extension as follows: f=-K x ---(1) The extension in the springs are related to the restoring force as follows: f=-K₁ e₁---(2) and f=-K₂ e₂---(3) The total extension of the springs is: x=e₁+e₂---(4) Using equation (1),(2) and (3) aligned & (- f K )= (- f K₁ )+ (- f K₂ ) & 1 K = ( 1 K₁ + 1 K₂ )---(5) aligned The total extension using equation (4) and (5)is given by, x= f K =f ( 1 K₁ + 1 K₂ )