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NEET NEET Questions — Free MCQ Practice

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W is the work done in forming a bubble of radius r, the work is done in forming a bubble of radius 2r will be:-

A.

4W

B.

3W

C.

2W

D.

W

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Explanation

$(1) W = T \times A = T \times 4 {πr}^{2} W \infty r^{2}$

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Equation of continuity based on :

A.

Conservation of mass

B.

Conservation of energy

C.

Conservation of angular momentum

D.

None of these

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Explanation

It is based on the conservation of mass

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Bernoulli's theorem is based on :

A.

conservation of energy

B.

conservation of mass

C.

conservation of momentum

D.

conservation of angular momentum

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Explanation

Conceptual

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Two spherical soap bubbles of radii r₁ and r₂ in vaccum collapse under isothermal condition. The resulting bubble has radius equal to :

A.

$\frac{r_{1} + r_{2}}{2}$

B.

$\frac{r_{1} r_{2}}{r_{1} - r_{2}}$

C.

$\sqrt{r_{1} r_{1}}$

D.

$\sqrt{r_{1}^{2} + r_{2}^{2}}$

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Explanation

$(4) for isothermal condition PV = P_{1} V_{1} + P_{2} V_{2} \frac{4 T}{r} \frac{4}{3} {πr}^{3} = \frac{4 T}{r_{1}} \frac{4}{3} {πr}_{1}^{3} + \frac{4 T}{r_{2}} . \frac{4}{3} {πr}_{2}^{3} r^{2} = r_{1}^{2} + r_{2}^{2}$

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The energy needed in breaking of a drop of liquid of radius R into n drops of radius r is given by (T is surface tension and P is atmospheric pressure) :

A.

$(4 {πr}^{2} n - 4 {πR}^{2}) T$

B.

$(4 / 3 {πr}^{2} n - 4 / 3 {πR}^{3}) T$

C.

$(4 {πR}^{2} n - 4 {πr}^{2}) T$

D.

$(8 {πr}^{2} n - 8 {πR}^{2}) T$

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Explanation

$S u r f a c e a r e a o f l a r g e s p h e r e = 4 {πR}^{2} S u r f a c e a r e a o f n s m a l l s p h e r e s = n 4 π r^{2} Work done = T \times ∆ A = T \times (n 4 {πr}^{2} - 4 {πR}^{2})$

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A rectangular film of liquid is extended from (4 cm $\times 2 c m$ ) to $(5 c m \times 4 \times c m)$ . If the work done is $3 \times 10^{- 4}$ J,the value of the surface tension of the liquid is

A.

$0.250 N m^{- 1}$

B.

$0.125 N m^{- 1}$

C.

$0.2 N m^{- 1}$

D.

$8.0 N m^{- 1}$

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Explanation

(b) Increases in surface energy= work done in area $\times$ Surface tension

$∵$ Increase in surface area,

$∆ A = (5 \times 4 - 4 \times 2) \times 2 (∵ f l i m h a s t w o s u r f a c e s)$

= $(20 - 8) \times 2 c m^{2} = 24 c m^{2} = 24 \times 10^{- 4} m^{2}$

So, work done, W=T. $∆ A$

$3 \times 10^{- 4} = T \times 24 \times 10^{- 4} ∴ T = \frac{1}{8} = 0.125 N / m$

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Three liquids of densities $ρ_{1}, ρ_{2} a n d ρ_{3}$ (with $p_{1} > p_{2} > p_{3}$ ), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact $θ_{1}, θ_{2} a n d θ_{3}$ obey

A.

$\frac{π}{2} > θ_{1} > θ_{2} > θ_{3} \geq 0$

B.

$0 \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}$

C.

$\frac{π}{2} < θ_{1} < θ_{2} < θ_{3} < π$

D.

$π > θ_{1} > θ_{2} > θ_{3} > \frac{π}{2}$

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Explanation

(b) According to ascent formula for capillary tube,

$h = \frac{2 T \cos θ}{p g r} ∴ \frac{\cos θ_{1}}{p_{1}} = \frac{\cos θ_{2}}{p_{2}} = \frac{\cos θ_{3}}{p_{3}} T h u s, \cos θ \propto ρ ∴ p_{1} > p_{2} > p_{3} ∴ \cos θ_{1} > \cos θ_{2} > \cos θ_{3} 0 \leq θ_{1} < θ_{2} < θ_{3} < \frac{π}{2}$

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Two non-mixing liquids of densities $ρ$ and n $ρ$ (n>1) are put in a container. The height of each liquid is h. A solid cylinder floats with its axis vertical and length pL $(p < 1)$ in the denser liquid. The density of the cylinder is d. The cylinder floats with its axis vertical and length pL (p<1) in the denser liquid. The density d is equal to

A.

{2+(n+1)p} $ρ$

B.

{2+(n-1)p} $ρ$

C.

{1+(n-1)p} $ρ$

D.

{1+(n+1)p} $ρ$

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Explanation

According to Archimedes' principle, the buoyant force on the cylinder is equal to the weight of the liquid displaced by the cylinder. The volume of the liquid displaced is pLÏ€r^2, where r is the radius of the cylinder. The weight of this liquid is pLÏ€r^2Ïn. This weight must equal the weight of the cylinder, which is pLÏ€r^2d. Therefore, d = Ïn.

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The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 x 10^-11 Pa^-1 and density of water is 10³kg/m³. What fractional compression of water will be obtained at the bottom of the ocean?

A.

0.8x10^-2

B.

1.0x10^-2

C.

1.2x10^-2

D.

1.4x10^-2

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Explanation

Given d=2700 m

ρ=10³kg/m³

Compressibility=45.4 x 10^-11 Pa^-1 per pascal

The pressure at the bottom of ocean is given by

P=ρgd=10³x10x2700=27x10⁶Pa

So, fractional compression=compressibility x pressure

=45.4 x 10^-11x 27 x10⁶
=1.2x10^-2

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A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m². Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (P_air=1.2kg/m³)

A.

4.8x10⁵N,downwards

B.

4.8x10⁵N,upwards

C.

2.4x10⁵N,upwards

D.

2/4x10⁵N,downwards

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Explanation

From Bernoulli's theorem

p₁+ $\frac{1}{2}$ ρv₁²=p₂+ $\frac{1}{2}$ ρv₂²

where, p₁ and p₂ are pressure inside and outside the roof and v₁ v₂ are velocities of wind inside and outside the roof, Neglect the width of the roof.

Pressure difference is

p₁-p₂= $\frac{1}{2}$ ρ(v₂²-v₁²)

= $\frac{1}{2}$ x1.2(40²-0)=960N/m²

Force acting on the roof is given by

F=(p₁-p₂)A=960 x 250=24x10⁴N=24X10⁵N

As the pressure inside the roof is more than outside to it. So the force will act m the upward direction, i.e. F=24x10⁵N,upward

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