A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $$60$$° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $$2$$−$$3/10s$$ , then the height of the top of the inclined plane, in meters, is $$____________$$.Take $$g=10m/s2$$.

Question

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an  angle $$60$$° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $$2$$−$$3/10s$$ , then the height of the top of the inclined plane, in meters, is  $$____________$$.Take $$g=10m/s2$$.

Infinity Learn · Accepted Answer

$$ac=gsin$$⁡θ$$1+ICMR2$$ aring $$=gsin$$⁡θ$$2$$,adisc $$=2gsin$$⁡θ$$3$$ For ring hsin⁡θ$$=12gsin$$⁡θ$$2t12$$⇒$$t1=4hgsin2$$⁡θ$$=16h3g$$ For disc, hsin⁡θ$$=122gsin$$⁡θ$$3t22$$⇒$$t2=3hgsin2$$⁡θ$$=4hg$$ Given :$$t2$$−$$t1=16h3g$$−$$4hg=2$$−$$310$$⇒$$h43$$−$$2=2$$−$$3$$⇒$$h=(2$$−$$3)3(4$$−$$23)=32$$⇒$$h=0.75m$$

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