Question

AU-tube is of non uniform cross-section. The area of cross-sections of two sides of tube are A and 2A (see fig.). It contains non-viscous liquid of mass m. The liquid is displaced slightly and free to oscillate. Its time period of oscillations is ______.

Options

• T = 2π`sqrt("m"/(3rho"gA"))` 

• T = 2π`sqrt("m"/(2rho"gA"))` 

• T = 2π`sqrt("m"/(rho"gA"))` 

• None of these

Answer 1

AU-tube is of non uniform cross-section. The area of cross-sections of two sides of tube are A and 2A (see fig.). It contains non-viscous liquid of mass m. The liquid is displaced slightly and free to oscillate. Its time period of oscillations is `underlinebb("T" = 2πsqrt("m"/(3rho"gA")))`.

Explanation:

If the liquid in the left side limb is moved slightly by y, the right limb's liquid will expand by y/2.

The restoring force

F = -PA

= `-rho"g"((3"y")/2)xx2"A"`

= 3ρgA(-y)

a = `"F"/"m"` = 3ρgA(-y)/m

on comparing with, a = -ω²y, we get

ω = `sqrt((3rho"gA")/"m")` and T = 2π`sqrt("m"/(3rho"gA"))`