| ⇦ | 
| ⇨ |
An open knife edge of mass 200g is dropped from height 5m on a cardboard. If the knife edge penetrates distance 2m into the cardboard, the average resistance offered by the cardboard to the knife edge is
Options
(a) 7N
(b) 25N
(c) 35N
(d) None of these
Correct Answer:
7N
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
- A train is moving along a straight path with uniform acceleration
- In the given (V – T) diagram, what is the relation between pressure P₁ and P₂?
- If force(F), Velocity(V) and time(T) are taken as fundamental units, then the dimensions
- In adiabatic expansion, product of PV
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
- A train is moving along a straight path with uniform acceleration
- In the given (V – T) diagram, what is the relation between pressure P₁ and P₂?
- If force(F), Velocity(V) and time(T) are taken as fundamental units, then the dimensions
- In adiabatic expansion, product of PV
Topics: Laws of Motion (103)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
We can use work energy theorem ,
Total kinetic energy lost is K=mg(h+d)
Work done by the resistance force F is W=Fd
By equating , we get Force =
mg(h+d)/d = [{(200/1000)×(5+2)}/2] N = [0.2×7÷2] N = 7N (ans)