⇦ | ⇨ |
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: - A short linear object of length L lies on the axis of a spherical mirror of focal
- Light from the constellation Virgo is observed to increase in wavelength by 0.4%
- A block of mass m is placed on a smooth wedge of inclination θ
- The resistance of a wire is 5 ohm at 50° C and 6 ohm at 100° C. The resistance
- The frequency of light ray having the wavelength 3000Å is
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A short linear object of length L lies on the axis of a spherical mirror of focal
- Light from the constellation Virgo is observed to increase in wavelength by 0.4%
- A block of mass m is placed on a smooth wedge of inclination θ
- The resistance of a wire is 5 ohm at 50° C and 6 ohm at 100° C. The resistance
- The frequency of light ray having the wavelength 3000Å is
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)