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Solution :

Let v be the speed of the solid sphere at the bottom of the incine. Applying principle of conservation of energy, we get <br> `(1)/(2)mv^(2)+(1)/(2)Iw^(2)=mgh` <br> As `I=(2)/(5)mr^(2),(1)/(2)mv^(2)+(1)/(2)((2)/(5)mr^(2))omega^(2)=mgh` <br> as `romega=v,(1)/(2)mv^(2)+(1)/(5)mv^(2)=mgh` <br> `v=sqrt((10)/(7)gh)`. <br> As h is same in two cases, v must be same i.e., It will reach the bottom with the same speed. Time taken to roll down the two planes will also be the same, as their height is the same.