IIT JEE Advanced 2008 Question Paper 2

The minimum speed V at position A, which is required by the bob to reach the top end at position B, is √5gL. From conservation of energy, we get

1

2

mv2 =

1

2

m(v′)2 + mgl (1 - cos θ)
T - mg cos θ =

mv2

l

Therefore,
(v′)2 = v2 + 2gl cos θ - 2gl
(

v

2

)2 = v2 + 2gl (cosθ - 1)

3v2

4

= 2gl (1- cos θ)

15gl

4

= 2gl (1- cos θ)
⇒ cos θ = 1 -

15

8

= −

7

8

= - 0.87
Since we have cos (

3π

4

) =

−1

√2

= - 0.71, we get

3π

4

< θ < π.