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is the ans

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Q ;A uniform rod is pivoted at its upper end hangs vertically .it is displaced through an angle of 60 and then released.Find the magnitude of the force acting on a particle of mass (dm) at the tip of the rod when the rod makes the angle of 37 with the vertical?????

wt about other qestions mainly 68,6763

with ans

A1B1C1 R THE INTERIOR POINTS OF THE SIDES AC ,AB,BC RESPECTIVELY PROVE THAT:

AREA A1B1C1/AREAOFABC 1/COSEC^2A+COSEC^2B+COSEC^2C

COS3B+COS3C+COS3A=1

COS3B+COS3C+COS3A-1=0

COS3(B+C)/2.COS3(B-C)/2-SIN^3A/2=0

SOLVING WE GET

COS3C/2.COS3B/2.COS3A/2=0

EITHER3C/2=

C=

/3 THEREFORE TRIANGLE IS ACUTE ANGLED

A1B1C1 R THE INTERIOR POINTS OF THE SIDES AC ,AB,BC RESPECTIVELY PROVE THAT:

AREA A1B1C1/AREAOFABC 1/COSEC^2A+COSEC^2B+COSEC^2C

now, no. of collision in 1 daY =24*3600/2

=43200

wts the concept behind this step an how is it taken

no Ra is not R1

WT IS NO OF COLLISION IN UR SOLOF Q34 WTIS 2.03ND HOW 2MINCOMES SO THAT DIFFERENCE IS .04SEC

unable to understand this step

m=mg(1.2 sin 37)

=mg(1.2)*3/5

=0.9(g/l)

instead it should be

I*=MgSIN37*L WHERE L=1m ND I=ML^2/3

=9/5g NOT WT U HAD WRITTEN

COSA +COSB=2(1-COSC)

COSA+B/2.COSA-B/2=2SIN^2C/2

SINC/2COSA-B/2=2SIN^2C/2

COSA-B/2=2SINC/2=2COSA+B/2

COSA/2.COSB/2+SINA/2SINB/2=2(COSA/2COSB/2-SINA/2SINB/2)

3SINA/2SINB/2=COSA/2COSB/2

TANA/2TANB/2=1/3

S-b.S-c/S.S-a *

S-c.S-a/S.S-b=1/3

S-c/S=1/3

3S-3c=S

2S=3c

a+b+c=3c

a+b=2c hence sides r in AP

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