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The sound from a very high burst of fireworks takes 5 s to arrive at the observer. The burst occurs 1662 m above the observer and travels vertically through two stratifier layers of air, the top one of thickness `d_1` at `0^@C` and the bottom one of thickness `d_2` at `20^@C`. Then (assume velocity of sound at `0^@C` is `330(m)/(s)`)A. `d_1=342m`B. `d_2=1320m`C. `d_1=1485m`D. `d_2=342m` |
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Answer» Correct Answer - A Time taken is given by `T=t_1+t_2=(d_1)/(v_1)+(d_2)/(v_2)` `v_1=v_(0^@C)=330(m)/(s)` `v_2=(330+0.6t)=342(m)/(s)` `d=1662m` `T=(d_1)/(330)+((d-d_1))/(342)=5s` `(d_1(342-330))/(330xx342)+(d)/(342)=5s` `12d_1=5(342xx330)-330xx1662` `d_1=1320m` `d_2=342m` |
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