Two sources A and B are sending notes of freq | vedclass

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(B) Let the frequency of the sources be $n = 680 \, Hz$ and the speed of sound be $v = 340 \, ms^{-1}$.The listener moves from $A$ towards $B$ with ve

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$2.5$

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(B) Let the frequency of the sources be $n = 680 \, Hz$ and the speed of sound be $v = 340 \, ms^{-1}$.The listener moves from $A$ towards $B$ with velocity $u$.As the listener moves away from source $A$, the apparent frequency $n^{\prime}$ heard from $A$ is:$n^{\prime} = n \left( \frac{v - u}{v} \right) = 680 \left( \frac{340 - u}{340} \right) = 2(340 - u)$As the listener moves towards source $B$, the apparent frequency $n^{\prime \prime}$ heard from $B$ is:$n^{\prime \prime} = n \left( \frac{v + u}{v} \right) = 680 \left( \frac{340 + u}{340} \right) = 2(340 + u)$The beat frequency is the difference between the two apparent frequencies:$n^{\prime \prime} - n^{\prime} = 10$$2(340 + u) - 2(340 - u) = 10$$680 + 2u - 680 + 2u = 10$$4u = 10$$u = 2.5 \, ms^{-1}$

List-$I$	List-$II$
$(A)$ Magnetic field intensity	$(i)$ $Wb$
$(B)$ Magnetic flux	(ii) $Wb \cdot m^{-2}$
$(C)$ Magnetic pole strength	(iii) $A \cdot m$
$(D)$ Magnetic induction	(iv) $A \cdot m^{-1}$

Process	Elasticity Type
$(i)$ Suspension fibre of galvanometer	$(a)$ Linear
$(ii)$ Bending of beam	$(b)$ Shear
$(iii)$ Cutting piece of paper	$(c)$ Bulk
$(iv)$ Mechanical waves in fluid	$(d)$ Shear

Two sources $A$ and $B$ are sending notes of frequency $680 \, Hz$. A listener moves from $A$ towards $B$ with a constant velocity $u$. If the speed of sound in air is $340 \, ms^{-1}$, what must be the value of $u$ so that the listener hears $10$ beats per second (in $ \, ms^{-1}$)?

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