1.

A tunning fork produces 4 beats/s both with 50 and 40 cm of a stretched wire of sonometer. The frequency of fork is :

Answer»

36 Hz
50 Hz
90 Hz
110 Hz

Solution :For stretched string `V_(1) -(1)/(2 l_(1)) sqrt((T)/(m))`
Then `v.= ((V)/(V - U_(s)) )` V
`rArr 1000 = ((350)/(350 - 50)) v rArr v = (1000 xx 300 )/(350 )`
When the source is moving away from the OBSERVER then
` v.. = ((V)/(V + U_(s)) ) ` V
` rArr"" v.. = ((350)/(350 + 50) ) xx (1000 xx 300)/(350)`
v.. = 750 Hz
correct choice is (d).
` v_(2) = (1)/(2l_(2) sqrt((T)/(m))`
`therefore (v_(1))/(v_(2)) = (l_(2))/(l_(1)) = (40)/(50) = (4)/(5) " " rArr ""V_(1) = (4)/(5) v_(2)`
LET the unknonw frequency of TUNNING fork = v
`therefore "" v- v_(1) = 4 `
`v_(2) - v = 4 `
Adding `"" v_(2) - v_(1)= 8 `
`v_(2) - (4)/(5) v_(2)= 8 `
`(1)/(5) v_(2) = 8"" rArr v_(2) = 40` Hz.
`therefore "" 40 -v =4"" rArr "" v = 36 ` Hz.
Hence the correct choice is (a) .


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