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If the splash is hear 4.23 seconds after a stone is dropped into a well, 78.4 metres deep, find the velocity of sound in air. |
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Answer» Here, depth of the well, `s=78.4metres` Total time after which splash is heard `=4.23s`, If `t_(1)=` time taken by stone to hit the water surface in the well. `t_(2)=` time taken by splash of sound to reach the top of the well. then, `t_(1)+t_(2)=4.23s` ...(i) Now, for downwards journey of stone, `u=0, a=9.8ms^(-2), s=78.4m, t=t_(1)=?` As `s=ut+(1)/(2)at^(2)` `:.78.4=0+(1)/(2)xx9.8t_(1)^(2)=4.9t_(1)^(2)` `t_(1)^(2)=(78.4)/(49.9)=16 or t_(1)=sqrt(16)=4s` From (i), `t_(2)=4.23-t_(1)=4.23-4=0.23` second If v is the velocity of sound in air,then `v=(distance (s))/(time(t_(2)))=(78.4)/(0.23)=34.87ms^(-1)` |
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