A thin ring, a disk and an annular cylinder, of same mass M, are released from a point 3.6 m from the ground up an inclined plane of 30^{o} inclination. The ring and the disk have the same radius R. For the annular cylinder, the outer radius is R and the inner radius is R/2. Times taken by the ring, disk and annular cylinder, respectively, to reach the ground are in the ratio,

1.√2 : √1.5 : √1.6

2. √1.4 : √1.5 : √2

3.√2 : √1.5 : √1.5

4.√1.4 : √1.5 : √1.6

Case of ring I = MR^{2}

Mgy = ½Mv^{2} + ½Iω^{2}

=> Mgy = ½Mv^{2} + ½(MR^{2} )ω^{2}

=> Mgy = ½Mv^{2} + ½(MR^{2} )(v/R)^{2}

=> Mgy = ½Mv^{2} + ½Mv^{2}

=> v^{2} = gy

=> v^{2 } = gy

=> v = √(gy)

Time taken,

Case of disk I = ½MR^{2}

Mgy = ½Mv^{2} + ½Iω^{2}

=> Mgy = ½Mv^{2} + ½(½MR^{2})ω^{2}

=> Mgy = ½Mv^{2} + 1/4MR^{2}(v/R)^{2}

=> Mgy = 3/4Mv^{2}

=> v^{2} = 4/3gy

=> v^{ } = √(4/3)gy

Case of annular cylinder I = ½(r_{1}^{2} + r_{2}^{2})

r_{1} = R

r_{2} = ½R

I = 5/8R^{2}

Mgy = ½Mv^{2} + ½Iω^{2}

=> Mgy = ½Mv^{2} + ½(5/8R^{2})ω^{2}

=> Mgy = ½Mv^{2} + ½(5/8MR^{2} )(v/R)^{2}

=> Mgy = ½Mv^{2} + 5/16Mv^{2}

=> 13/16 v^{2} = gy

=> v^{2 } = (16/13)gy

=> v =√{(16/13)gy}

Taking the ratios of time t in three cases

We get,

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