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Mechanical Properties of Fluids NEET Questions — Free MCQ Practice

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Assertion & Reason type questions Read the assertion and reason carefully to mark the correct option out of the options given below Assertion : To empty an oil tank two holes so it will made. Reason : Oil will come out of two holes so it will be emptied faster.

A. If both assertion and reason are true and the reason is the correct explanation of the assertion. B. If both assertion and reason are true but reason is not the correct explanation of the assertion. C. If assertion is true but reason is false. D. IF the assertion and reason are false. E. If assertion is flase but reason is true.

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Assertion & Reason type questions Read the assertion and reason carefully to mark the correct option out of the options given below Assertion : A bubble comes from the bottom of a lake to the top. Reason : Its radius increases.

A. If both assertion and reason are true and the reason is the correct explanation of the assertion. B. If both assertion and reason are true but reason is not the correct explanation of the assertion. C. If assertion is true but reason is false. D. IF the assertion and reason are false. E. If assertion is flase but reason is true.

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A Small spherical body of radius r is falling under gravity in a viscous medium. Due to friction the medium gets heated. How does the late of heating depend on radius of body when it attains terminal velocity!

A. $ r^2 $ B. $ r^3 $ C. $ r^4 $ D. $ r ^5 $

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Explanation

$ Rate of heat produced = (Viscous force F) \times (Velocity V) $ $ { dQ \over dt} = 6 \pi nr v^2 \left[ { 2 \over g } { ( \rho - \rho_o)r^2 g \over n } \right] ^2 $

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The upper edge of a gate in a dam runs along water surface. The gate is 2 m high and 3 m wide and is hinged along a horizontal line through its center. Calculate the torque about hinge. [This question is only for Dropper and XII batch]

A.

$10^{4} N m$

B.

$2 \times 10^{4} N m$

C.

$3 \times 10^{4} N m$

D.

$4 \times 10^{4} N m$

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Explanation

The torque acting on the gate is given by the product of the force acting on it and the perpendicular distance between the line of action of the force and the hinge. The force acting on the gate is the weight of the gate, which is equal to its mass times the acceleration due to gravity.

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A gas undergoes a process in which its pressure P and volume V are related as ${VP}^{n} = constant$ . The bulk modulus for the gas in the process is
[Only for Dropper and XII batch]

A.

$\frac{P}{n}$

B.

$P^{1 / n}$

C.

nP

D.

$P^{n}$

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Explanation

$(3) D i f f e r e n t i a t e V P^{n} = C w . r . t o V u \sin g m u l t i p l i c a t i o n r u l e : V \frac{d}{d V} (P^{n}) + P^{n} \frac{d}{d V} (V) = 0 V n P^{n - 1} \frac{d P}{d V} + P^{n} . 1 = 0 V n P^{n - 1} d P + P^{n} . d V = 0 n V P^{n - 1} d P = - P^{n} . d V \Rightarrow \frac{d P}{d V} = - \frac{P^{n}}{n V P^{n - 1}} = - \frac{P}{n V} B u l k M o d u l u s = - \frac{d P}{\frac{d V}{V}} = - (- \frac{P}{\frac{n V}{V}}) = \frac{P}{n}$

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A tube of length L is filled completely with an incompressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $ω$ . The force exerted by liquid at the other end is

A.

$\frac{{Mω}^{2} L}{2}$

B.

${MLω}^{2}$

C.

$\frac{{Mω}^{2} L}{4}$

D.

$\frac{M L^{2} ω^{2}}{2}$

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Explanation

$Force on element of width dx = (\frac{M}{L} dx) (x) ω^{2} Total force F = \int_{0}^{L} \frac{M}{L} {xω}^{2} dx = \frac{{MLω}^{2}}{2}$

Choice A is correct.

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A cubical vessel of height 1 m is full of water. Find the work done in pumping out whole water.

A.

49 J

B.

490 J

C.

4900 J

D.

49000 J

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Explanation

$W o r k d o n e = P E o f t h e w a t e r = m g \frac{h}{2} = 4900 J$

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The flow rate from a tap of diameter 1.25 cm is 3 lit/min. The coefficient of viscosity of water is $10^{- 3}$ Pas. The nature of flow is :

A.

Turbulent

B.

Laminar

C.

Neither laminar nor turbulent

D.

data inadequate

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Explanation

$L e t t h e s p e e d o f t h e f l o w b e v . T h e d i a m e t e r o f t h e t a p = d = 1.25 c m = 1.25 \times 10^{- 2} m D e n s i t y o f w a t e r = ρ = 10^{3} k g m^{- 3} V i s \cos i t y = η = 10^{- 3} P a s T h e v o l u m e o f t h e w a t e r f l o w i n g o u t p e r s e c o n d i s Q = v \times π d^{2} / 4 \Rightarrow v = 4 Q / d^{2} π R e y n o l d s n u m b e r i s g i v e n b y : R = \frac{ρ v d}{η} = \frac{4 ρ Q}{π d η} = \frac{4 \times 10^{3} \times Q}{(3.14 \times 1.25 \times 10^{- 2} \times 10^{- 3})} = 1.019 \times 10^{8} Q Q = 3 L / m i n = \frac{3 \times 10^{- 3}}{60} = 5 \times 10^{- 5} m^{3} / s e c R = 5095 T h e f l o w w i l l b e t u r b u l e n t .$

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Water flowing from a hose pipe fills a 15-liter container in one minute. The speed of water from the free opening of radius 1 cm is (in ms $^{- 1}$ ) :

A.

2.5

B.

$\frac{π}{2.5}$

C.

$\frac{2.5}{π}$

D.

5 $π$

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Explanation

$V o l u m e o f t h e w a t e r = 15 l i t r e = 15 \times 10^{- 3} m^{3} V o l u m e = A r e a \times l e n g t h = A r e a \times v e l o c i t y \times t i m e V e l o c i t y = \frac{15 \times 10^{- 3}}{π \times {(1 \times 10^{- 2})}^{2} \times 60} = \frac{2.5}{π} m / s$

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Work of 6.0 x 10 $^{- 4}$ Joule is required to be done in increasing the size of a soap film from 10cm x 6cm to 10cm x 11cm. The surface tension of the film is :

A.

5 x 10 $^{- 2}$ N/m

B.

6 x 10 $^{- 2}$ N/m

C.

1.5 x 10 $^{- 2}$ N/m

D.

1.2 x 10 $^{- 1}$ N/m

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Explanation

$S u r f a c e T e n s i o n = \frac{S u r f a c e E n e r g y}{A r e a} S u r f a c e E n e r g y = S u r f a c e T e n s i o n \times A r e a W o r k d o n e = c h a n g e i n s u r f a c e e n e r g y = S u r f a c e T e n s i o n \times C h a n g e i n a r e a S u r f a c e T e n s i o n = \frac{6 \times 10^{- 4}}{2 \times 10 \times 5 \times 10^{- 4}} = \frac{6}{100} = 6 \times 10^{- 2} N / m$

Mechanical Properties of Fluids MCQs for NEET — Physics Questions with Answers

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