⇦ | ⇨ |
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: - The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second
- The input signal given to a CE amplifier having a voltage gain of 150 is
- The kinetic energy of particle moving along a circle of radius R depends
- A transformer is used to light a 100 W and 110 V lamp from a 220 V mains.
- A body weighs 50 grams in air and 40 grams in water. How much would it weigh in a liquid
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second
- The input signal given to a CE amplifier having a voltage gain of 150 is
- The kinetic energy of particle moving along a circle of radius R depends
- A transformer is used to light a 100 W and 110 V lamp from a 220 V mains.
- A body weighs 50 grams in air and 40 grams in water. How much would it weigh in a liquid
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