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Standard XII

Physics

Dimensional Analysis

If r = 5× 1...

Question

If $r = 5 \times 10^{- 4} m, ρ = 10^{3} k g m^{- 3}, g = 10 m / s^{2}, T = 0.11 N m^{- 1}$ , the radius of the drop when it detaches from the dropper is approximately.

1.4 \times 10^{- 3} m

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3.3 \times 10^{- 3} m

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2.0 \times 10^{- 3} m

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4.1 \times 10^{- 3} m

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Solution

The correct option is C $2.0 \times 10^{- 3} m$
Just before dettaching
force due to surface tension = weight of the liquid in the drop
$T (2 π r) = (\frac{4}{3} π R^{3}) ρ g$
substituting the values we get
$R = 2.0 \times 10^{- 3} m$

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Similar questions

Q. If

r = 5 \times 10^{- 4} m ρ = 10^{3} K g m^{- 3}, g = 10 m / s^{2}, T = 0.11 N m^{- 1}

the radius of the drop when it detached from the dropper is approximately:

Q. Find the accelerations

a_{1}, a_{2}

and

a_{3}

of the three blocks as shown in figure, if a horizontal force of

10 N

is applied on the

3 kg

block. The coefficient of static friction (

μ

) for all contact surfaces are shown in the figure. Take

g = 10 {m/s}^{2}

Q. A curve has a radius of

50 m

and a banking angle (

θ

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15^{\circ}

. What is the maximum safe speed for a car on this curve if the road does not offer any friction? Take

g = 10 {m/s}^{2}

Q. The scientist "Q" was experimenting to revive a giant called the Mountain. He accidentally dropped a vial of ammonia on the ground. He saw ammonia react with oxygen and produce NO and water.

Let's say the rate of formation of NO is 3.6 x 10^-3

M s^{- 1}

Calculate

Maester Qyburn was experimenting to revive a giant called The Mountain. He accidentally dropped a vial of ammonia on the ground. He saw ammonia react with oxygen and produce NO and water.

Let's say the rate of formation of NO is 3.6 x 10^-3 Ms^-1

Calculate

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If r=5×10−4m,ρ=103kg m−3,g=10 m/s2,T=0.11 Nm−1, the radius of the drop when it detaches from the dropper is approximately.

The correct option is C 2.0×10−3mJust before dettachingforce due to surface tension = weight of the liquid in the dropT(2πr)=(43πR3)ρgsubstituting the values we get R=2.0×10−3m

If $r = 5 \times 10^{- 4} m, ρ = 10^{3} k g m^{- 3}, g = 10 m / s^{2}, T = 0.11 N m^{- 1}$ , the radius of the drop when it detaches from the dropper is approximately.

The correct option is C $2.0 \times 10^{- 3} m$
Just before dettaching
force due to surface tension = weight of the liquid in the drop
$T (2 π r) = (\frac{4}{3} π R^{3}) ρ g$
substituting the values we get
$R = 2.0 \times 10^{- 3} m$