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A ball is dropped into a well. The water level in the well is at a depth 'h' below the top. If the speed of sound is V, then after what time the splash of sound is heard?

$\begin {array} {1 1} (1)\;h\bigg[ \sqrt{\large\frac{2}{gh}}- \large\frac{1}{v} \bigg] & \quad (2)\;h \bigg[ \large\frac{2}{g}+ \large\frac{1}{v}\bigg] \\ (3)\;h \bigg[ \large\frac{2}{g}- \large\frac{1}{v}\bigg] & \quad (4)\;h\bigg[ \sqrt{\large\frac{2}{gh}}+ \large\frac{1}{v} \bigg] \end {array}$

2 Answers

(4) $h\bigg[ \sqrt{\large\frac{2}{gh}} + \large\frac{1}{v} \bigg]$
answered Nov 7, 2013 by pady_1
 
1.Time taken for ball to reach water below Using h=1/2gt^2 t=root(2h/g).......1 2.Time taken for sound to return back Speed of sound =c. Therefore using distance/speed is time taken.... t2=h/c.......2 3.adding 1 and 2 We get root (2h/g )+h/c=T Taking h common...... T=h(root(2/gh)+1/c) (Try last step...u wil get it right away.... Thnx
answered Oct 7 by aryanchhetri123
 

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