The fraction of a floating object of volume V₀ and density d₀ above the surface of liquid of density d will be

\(\frac{d_{0}}{d-d_{0}}\)
\(\frac{d-d_{0}}{d}\)
\(\frac{d_{0}}{d}\)
\(\frac{d_{0}d}{d+d_{0}}\)

Solution

If a water drop is kept between two glass plates, then its shape is

None of these

Solution

Angle of contact is acute.

A liquid is allowed to flow into a tube of truncated cone shape. Identify the correct statement from the following

The speed is high at the wider end and high at the narrow end.
The speed is low at the wider end and high at the narrow end.
The speed is same at both ends in a stream line flow.
The liquid flows with uniform velocity in the tube.

Solution

The theorem of continuity is valid.

∴ A₁v₁ρ= A₂v₂ρ as the density of the liquid can be taken as uniform.

∴ A₁v₁ = A₂v₂

⇒Smaller the area, greater the velocity.

In the figure, the velocity V₃ will be

Zero
4 ms^-1
1 ms^-1
3 ms^-1

Solution

According to equation of continuity

A₁V₁= A₂V₂ + A₃V₃

⇒4 × 0.2 = 2 × 0.2 + 0.4 × V₃ ⇒ V₃= 1 m/s.

There is a hole in the bottom of tank having water. If total pressure at bottom is 3 atm (1 atm = 10⁵N/m²) then the velocity of water flowing from hole is

\(\sqrt{400}\) m/s
\(\sqrt{600}\) m/s
\(\sqrt{60}\) m/s
None of these

Solution

Horizontal tube of non-uniform cross-section has radii of0.1 m and 0.05 m respectively at M and N for a stream line flow of liquid the rate of liquid flow is

continuously changes with time
greater at Mthan at N
greater at N than at M
same at M and N

Solution

According to principle of continuity, for a stream line flow of fluid through a tube of non-uniform cross-section the rate of flow of fluid (Q) is same at every point in the tube.

i.e., Av = constant ⇒ A₁v₁= A₂v₂

Therefore, the rate of flow of fluid is same at M and N.

1 m³ water is brought inside the lake upto 200 metres depth from the surface of the lake. What will be change in the volume when the bulk modulus of elastically of water is 22000 atmosphere?

(density of water is 1 × 10³ kg/m³ atmosphere pressure= 10⁵ N/m² and g = 10 m/s²)

8.9 × 10^-3 m³
7.8 × 10^-3 m³
9.1 × 10^-4 m³
8.7 × 10^-4 m³

Solution

A big drop of radius R is formed by 1000 small droplets of water. The radius of small drop is

^R⁄₁₀
^R⁄₁₀₀
^R⁄₅₀₀
^R⁄₁₀₀₀

A small ball (menu) falling under graivty in a viscous medium experiences a drag force proportional to the instantaneous speed u such that F_drang=Ku. Then the terminal speed of the ball within the viscous medium is

^k⁄_mg
^mg⁄_k
\(\sqrt{\frac{mg}{k}}\)
\(\left ( \frac{mg}{k} \right )^{2}\)

Water flows in a stream line manner through a capillary tube of radius a, the pressure difference being P and the rate flow Q. If the radius is reduced to ^a⁄₂ and the pressure is in creased to 2P,the rate of flow becomes

4Q
Q
^Q⁄₂
^Q⁄₈

Solution

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