You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Hydrology & Irrigation

Indirect Methods of River Gauging

A current met...

Question

A current meter is used to measure the velocity at a certain depth in a flow cross-section . The date obtained is as follows-
then what are the number of revolutions corresponding to the velocity 0.35 m/s?

$\begin{matrix} Number of & Velocity revolutions (s^{- 1}) & (m / s) 24 & 0.3 50 & 0.6 \end{matrix}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 30
For current meter -

$V = a N_{1} + b$
$0.30 = 24 a + b$
and
$0.6 = 50 a + b$

Solving (i) and (ii) , we get

a = 0.011538, b = 0.023

$∴ For V = 0.35 m / s$

$0.35 = 0.011538 \times N_{s} + 0.023$

$N_{s} = 28.3 ≃ 30$

Suggest Corrections

Similar questions

Q. The number of revolutions of a current meter in 50 seconds were found to be 12 and 30 corresponding to the velocities of 0.25 and 0.46 m/s respectively. What velocity (in m/s) would be indicated by 50 revolutions of that current meter in one minute?

Q. An ideal liquid flowing through a pipe

A

of cross-section

0.2 m^{2}

with velocity

10 m/s

enters a

T

- junction. One side of the

T

-junction

B

has cross-section area

0.1 m^{2}

and the other side

C

has cross-section area

0.05 m^{2}

. If the velocity of water in

C

15 m/s,

then in

B

the velocity is

Q. An ideal liquid flowing through a pipe A of cross-section

0.2 m^{2}

with velocity

10 m / s

enters a T-junction. One side of the T-junction B has cross-section area

0.1 m^{2}

and the other side C hascross-section area

0.05 m^{2}

. If the velocity of water in C is

15 m / s

then in B the velocity is

A, B

and

C

are three objects each moving with constant velocity.

A^{'}

s speed is

10 m/s

in a direction

P Q

. The velocity of

B

relative to

A

6 m/s

at an angle of

{cos}^{- 1} (15 / 24)

P Q

. The velocity of

C

relative to

B

12 m/s

in the direction

Q P

, then find the magnitude of velocity of

C

Q. The angular velocity of a particle moving in a circle of radius

50 c m

is increased in

5

min from

100

revolutions per minute to

400

revolutions per minute. Find the tangential acceleration of the particle.

Join BYJU'S Learning Program

A current meter is used to measure the velocity at a certain depth in a flow cross-section . The date obtained is as follows- then what are the number of revolutions corresponding to the velocity 0.35 m/s? Number ofVelocityrevolutions (s−1)(m/s)240.3500.6

The correct option is C 30For current meter - V=aN1+b 0.30=24a+b and 0.6=50a+b Solving (i) and (ii) , we get a = 0.011538, b = 0.023 ∴For V=0.35 m/s 0.35=0.011538×Ns+0.023 Ns=28.3≃30

The correct option is C 30
For current meter -

$V = a N_{1} + b$
$0.30 = 24 a + b$
and
$0.6 = 50 a + b$

Solving (i) and (ii) , we get

a = 0.011538, b = 0.023

$∴ For V = 0.35 m / s$

$0.35 = 0.011538 \times N_{s} + 0.023$

$N_{s} = 28.3 ≃ 30$