Laws Of Motion Question 400

Question: String is massless and pulley is smooth in the adjoining figure total mass or left hand side of the pulley side of the pulley is m1 and on right hand side is m2 . Friction coefficient between block B and the wedge is μ=12 and θ=30 Select thewrong answer

Options:

A) Block B will slide down if m1=m2

B) Block B may remain stationary with respect to wedge, for suitable values of m1 and m2 with m1>m2

C) Block B cannot remain stationary with respect to wedge in any case

D) Block B will slide down if m1>m2

Show Answer

Answer:

Correct Answer: B

Solution:

If m1=m2 , block will slide down when

μgcosθ<gsinθ or12(32)<12

If m1=m2 , block will slide down when

μ(g+a)cosθ<(ga)sinθ

or 12(32)<12 where a=(m2m1m1+m2)g

therefore in all cases block will slide down



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक