1ans is 3/4
THE TIME TAKEN BY A BODY TO SLIDE DOWN A ROUGH 45 DEGREE INCLINED PLANE IS TWICE THAT REQUIRED TO SLIDE DOWN A SMOOTH 45 DEGREE INCLINED PLANE.THE COEFFICIENT OF KINETIC FRICTION BETWEEN THE OBJECT AND ROUGH PLANE IS GIVEN BY?
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UP 0 DOWN 0 0 15
15 Answers
1yes answer is 3/4 i forgot square root :P :P
time inversely proportional to square root of acceleration
39Let L be the length of the incline.
Case I : Smooth Incline
Net force bringing the block down = mgsin@ where @ = 45°.
Acceleration = gsin@
Assuming the block starts from rest,
L = 0 + 12gsin@ t12
=> 2Lgsin@ = t12
Case II : Rough Incline
Net force bringing the block down = mgsin@ - Fk
= mgsin@ - μmgcos@
Acceleration = gsin@ - μgcos@
So t22 = 2Lgsin@ - μgcos@
But we know that 2t1 = t2
=> 4t12 = t22
=> 4 x 2Lgsin@ = 2Lgsin@ - μgcos@
=> 4(sin@ - μcos@) = sin@
=> 3sin@ = 4μcos@
=> 34tan@ = μ
=> 34 x tan(45) = μ
Tried and tested method....shortcut istamaal karoge toh chullu bhar paani mein doob maroge(like anand says :P)
1T(rough)/T(smooth)=√a(smooth)/a(rough)
=g/√2 * √{1/1-u}=2
now solving we get u=3/4
1we will protest!!!!!!!
pink for pritish no pink for me...........we are in a democracy LOL!!!! bahut nainsaafi hai re..........
1we want justice we want justice :P :P :P :P..........inspired from SIMON GO BACK!!!!! pink karne wale waapis jao aur uske aage peeche waale post bhi pink karo LOL!!!!!!!
1easier method no shortcut
s=ut-1/2at^2 since u=0
1/2(gsin@)t^2=1/2(gsin@-ugcos@)4t^2
gsin@=4gsin@-4ugcos@
or u =3/4 there was no shortcut mehthod involved
1are yaar school ka exam nahi ho raha hai ki no shortcuts chilla rahe ho
are pritish CHACHA doosron ko gumrah mat kar nahin to tumhare upar case kar dunga.
waise bhi iit mein option mark karna hai chahe tu shortcut se laa ya puri duniya ka longcut le kar.
khair jo bhi ho answer is 3/4