1ohk [1]
The max value of m {in kg} so that the arrangement shown in the figure is in equlibrium is given by:
1. 2
2. 2.5
3. 3
4. 4.5
given : - M = 10 kg and μ=.1 for all surfaces
-
UP 0 DOWN 0 0 22
22 Answers
11I think the options are wrong.
I tried all the options but none of them provide equilibrium
1may be possible bhaiya... bcoz i got this ques in my coaching's practice sheet!!
62what is the source of this one?
it seems ki there is a mistake in it!
62yes surbhi.. me too am getting the same..
i am trying to see this one!
1bhaiya mera bhi yahi fbd bana hai. and answer aa raha hai .5 kg
62http://targetiit.com/iit_jee_forum/posts/friction_problem_1303.html
yes subash it is from the same link..
I am also getting m=0.5 but there may be something wrong and common to all of us ;)
let me see it again :)
622T-N-f2=Ma
N=ma
mg-T-f1=2ma
a=0
so
N=0
2T=f2
mg-T=f1
mg=f1+f2/2
in the limitingcase,
f1=0 because it is just accelerating and N=0
f2=0.1 x 10 x 10 = 10
mg=5
g= 0.5
1no sir, i think their must not ne normal reaction b/w two blocks, because in that case , their will be a net horizontal force on m towards right.!!!& it will not be in eq,
correct me if wrong...
11http://targetiit.com/iit_jee_forum/posts/2/friction_problem_1303.html
same figure bhaiyya :p
1yeah mera bhi yahi aa raha hai .. but not in the option... [2]... r we doing ny mistake?
11No u cannot use image link . You have to upload the image instead
