| ⇦ | 
| ⇨ |
A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 sec. Then the coefficient of friction is
Options
(a) 0.01
(b) 0.02
(c) 0.03
(d) 0.06
Correct Answer:
0.06
Explanation:
From equation of motion, v = u – at
⇒ o = u – μgt
⇒ μ = u/gt
= 6/(10×10)
= 0.06
Related Questions: - An electric dipole of moment ‘p’ is placed in an elecatric field of intensity ‘E’
- A body is thrown vertically upward in air when air resistance is taken into account
- The given graph represents V-I characteristic for a semiconductor device
- Two point objects of masses 1.5 g and 2.5 g respectively are at a distance of 16 cm
- During an adiabatic process, the cube of the pressure is found to be inversely
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An electric dipole of moment ‘p’ is placed in an elecatric field of intensity ‘E’
- A body is thrown vertically upward in air when air resistance is taken into account
- The given graph represents V-I characteristic for a semiconductor device
- Two point objects of masses 1.5 g and 2.5 g respectively are at a distance of 16 cm
- During an adiabatic process, the cube of the pressure is found to be inversely
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
Leave a Reply