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

Byju's Answer

Standard XII

Physics

Atmospheric Pressure

A rectangular...

Question

A rectangular plate $10 c m \times 5 c m$ is placed horizontally below the water surface in a trough at a depth of $2 m$ . If the atmospheric pressure is $1.013 \times 10^{5} N / m^{2}$ , then the total thrust on the plate is: $(g = 10 m / s^{2})$ .

100 N

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

606.5 N

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

303.25 N

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

909.75 N

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

Open in App

Solution

The correct option is B $606.5 N$
$F_{T h} = (P_{0} + ρ g h) A r e a F_{T h} = (1.01 \times 10^{5} + 1000 \times 10 \times 2) 10 \times 5 {c m}^{2} F_{T h} = 12.1 \times 10^{4} \times 50 {c m}^{2} F_{T h} = 606.5 N$

Suggest Corrections

Similar questions

Q. A thin square plate of side

4 m

is placed horizontally

2 m

below the surface of water. Calculate the thrust on one of the surface of the plate.
(Given: Atmospheric pressure,

P_{a t m} = 1.013 \times 10^{5} N / m^{2}

Density of water,

ρ_{w}

1000 k g / m^{3}

,
Acceleration due to gravity,

(g) = 9.8 m s^{- 1}

)

Q. A square plate of side

30 m

is placed at

1 m

below the surface of water. The atmospheric pressure is

1.013 \times 10^{5} {N m}^{- 2}

. Calculate the total thrust on the plate. (Given: Density of water =

10^{3} {kg m}^{- 3}, g = 10 {m s}^{- 2})

Q. A thin square plate of side

4 m

is placed horizontally

2 m

below the surface of water. Calculate the thrust on one of the surface of the plate.
(Given: Atmospheric pressure,

P_{a t m} = 1.013 \times 10^{5} N / m^{2}

Density of water,

ρ_{w}

1000 k g / m^{3}

,
Acceleration due to gravity,

(g) = 9.8 m s^{- 1}

)

If the atmospheric pressure is $10^{5} \frac{N}{m^{2}}$ and the density of water $10^{3} \frac{k g}{m^{3}}$ . What will be the depth below the surface of water if pressure is equal to thrice the atmospheric pressure?

Q. What is the pressure 200 m below the surface of the ocean if the sp. gravity of sea water is 1.03 : [Atmospheric pressure

= 1.013 \times 10^{5} N / m^{2}

Join BYJU'S Learning Program

A rectangular plate 10 cm×5 cm is placed horizontally below the water surface in a trough at a depth of 2 m. If the atmospheric pressure is 1.013×105N/m2, then the total thrust on the plate is: (g=10 m/s2).

The correct option is B 606.5 N FTh=(P0+ρgh)AreaFTh=(1.01×105+1000×10×2)10×5cm2FTh=12.1×104×50cm2FTh=606.5N

A rectangular plate $10 c m \times 5 c m$ is placed horizontally below the water surface in a trough at a depth of $2 m$ . If the atmospheric pressure is $1.013 \times 10^{5} N / m^{2}$ , then the total thrust on the plate is: $(g = 10 m / s^{2})$ .

The correct option is B $606.5 N$
$F_{T h} = (P_{0} + ρ g h) A r e a F_{T h} = (1.01 \times 10^{5} + 1000 \times 10 \times 2) 10 \times 5 {c m}^{2} F_{T h} = 12.1 \times 10^{4} \times 50 {c m}^{2} F_{T h} = 606.5 N$