UPSEE Solved Model Paper 5

cos−1(

1

√3

)
at = g sin θ
ac =

u2

l

=

2glcosθ

l

= 2g cos θ

a is the resultant acceleration
Since a is horizontal , it makes an angle θ with ac
So,
tan θ =

acsin90°

a1+accos90°

=

2gcosθ

gsinθ

=

2

tanθ

or tan2θ = 2 ⇒ sec2θ = 3
⇒ cos θ =

1

√3

The value of θ for which the resultant acceleration of the bob is directed horizontally is
θ = cos−1(

1

√3

)