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[ ] [ ] (1/logx)dx can be evaluated but involves higher level of mathematics, higher than jee level.
don bother abt it.
n deedee u will not get the ans your way. why dont u chec urself before posting
somthing wrong????
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this is the nth time im seeing qn numbers from hc verma.!!!
not all hav hcverma
take some time 2post ur question completely
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The ridge is gonna exert some force on the bloc
but we can conserve angular momentum abt the ridge, as that force doesnt give any torque abt the ridge
Just before impact angl momentum abt the ridge = M*v*a/2
Final angular momentum = I*W
were I is the moment of inertia abt the ridge and W the angular velocity as seen from the frame of the ridge.
=[Ma^2/6+Ma^2/2 ]*W
equate both
n u get
W=3v/4a
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(y-1)^2=4(x-1)
vertex at (1,1)
x>1---> 3 normals can b drawn.
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Abt bottommost point net torque is zero
so u can conserve angular momentum abt lowermost point.
suppose inital angular veloc is 
and final angular veloc is  0
I=I 0 +mrv
and once pure rollin starts v=r 0
plug in for I and solve
2/3mr^2=5/3mr^20
0=2/5
so v=2/50r
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Conserve angular momentum abt lowermost point
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its just (a+b).[(b+c)x(c+a)]
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gokul, mayb im wrong bt
acc 2 u angular veloc of the rod is 2v/ 3l u used B 's veloc / dist of B from IAORnow do d samething for A
l/2= 3v
gives =2 3v/l
wich is not poss.
so ders som bug in ur working
correct me if im wrong
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mg(l/2) = 1/2 *I *^2
Karthik , the rod is rotating about the hinge
so u gotta take I =ml^{2}/3 ??
By conserv of energy,
mgl/2=1/2*ml^{2}/3*^2
gives =3g/l
Now if F is d force by hinge on d rod
at vertical postition,
F-mg=m/*l/2*^2
F=5/2mg
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Disclaimer:
I donno abc of mods.. All I've seen is just 1 or 2 sums bein solvd by mods so heres my atempt. Der must surely be an easy way out
Basically u gotta find the remainder wen 43 ^211 is divided by 1000
43^3 ends wid 507 . 43^6=49 mod 1000
now write 43^211 =43^210*43
(43^6)^35*43
so the abv thing mod 1000 will be
49^35 mod 1000 *43 mod 1000
(50-1)^35 mod 1000 *43 mod 1000
= 249 *43 mod 1000
wich is 707
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Hint: ob write 97 as 100-3??
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See this
a similiar question
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