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At two points on a horizontal tube of va...

At two points on a horizontal tube of varying cross-section, the radii are 1 cm and 0.4 cm, velocities of the fluid are `v_1` and `v_2` and the pressure difference `(P_1-P_2)` between these point is 4.9 cm of water Then the value of `sqrt((v_2)^2-(v_1)^2` is

A

3.13 cm/sec

B

98 cm/sec

C

9.8 cm/sec

D

60 cm/sec

Text Solution

Verified by Experts

The correct Answer is:
B
by Bernouli's theorem
`P_1+1/2rhov_1^2=P_2+1/2rhov_2^2`
Where `rho` is the density of the liquid.
`:." "(P_1-P_2)=1/2rho(V_2^2-v_1^2)`
`:." "v_2^2-v_1^2=(2(P_1-P_2))/rho`
But `P_1-P_2=4.9xxrhoxxg`
`:." "v_2^2-v_1^2=(4.9xxrhoxxgxx2)/rho`
`=4.9xx2xx980=98xx98`
`:." "sqrt(v_2-v_1)=98cm//sec`
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