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NEET NEET Questions — Free MCQ Practice

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The work per unit volume to stretch the length by 1% of a wire with constant cross sectional area will be. $[Y = 9 \times 10^{11} N / m^{2}]$

A.

$9 \times 10^{11} J$

B.

$4.5 \times 10^{7} J$

C.

$9 \times 10^{7} J$

D.

$4.5 \times 10^{11} J$

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Explanation

(b) $U = \frac{1}{2} \times Y \times {(S t r a i n)}^{2} = \frac{1}{2} \times 9 \times 10^{11} \times {(\frac{1}{100})}^{2} = 4.5 \times 10^{7} J$

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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

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

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When a cylindrical tube is dipped vertically into a liquid the angle of contact is 140^o. When the tube is dipped with an inclination of 40^o, the angle of contact is-

A.

$100 °$

B.

$140 °$

C.

$180 °$

D.

$60 °$

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Explanation

The angle of contact is independent of the tilting angle.

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Two liquid drops have their diameters as 1 mm and 2 mm. The ratio of excess pressure in them is :

A.

1:2

B.

2:1

C.

4:1

D.

1:4

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Explanation

$(2) ∆ P = \frac{2 T}{r} \frac{∆ P_{1}}{∆ P_{2}} = \frac{r_{2}}{r_{1}} = \frac{2}{1}$

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If a soap bubble of radius 3 cm coalesce with another soap bubble of radius 4 cm under isothermal conditions, the radius of the resultant bubble formed is in cm-

A.

7

B.

1

C.

5

D.

12

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Explanation

$P_{1} = \frac{4 T}{r_{1}} & V_{1} = \frac{4}{3} {πr}_{1}^{3} P_{2} = \frac{4 T}{r_{2}} & V_{2} = \frac{4}{3} {πr}_{2}^{3} Two bubbles coelesce isothermally : PV = P_{1} V_{1} + P_{2} V_{2} \frac{4 T}{r} \times \frac{4}{3} {πr}^{3} = \frac{4 T}{r_{1}} \times \frac{4}{3} {πr}_{1}^{3} + \frac{4 T}{r_{2}} \times \frac{4}{3} {πr}_{2}^{3} r^{2} = {r_{1}}^{2} + {r_{2}}^{2} r = \sqrt{{r_{1}}^{2} + {r_{2}}^{2}} = 5 c m$

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$Water flows through a non - uniform tube of area of cross section s A, B, and C whose value s are 25, 15, and 35 {cm}^{2} respectively . The ratio of the velocities of water at the sections A, B, and C is -$

A.

5:3:7

B.

7:3:5

C.

21:35:15

D.

1:1:1

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Explanation

$By equation of continuity Av = const . A_{1} v_{1} = A_{2} v_{2} = A_{3} v_{3} Since A_{1} : A_{2} : A_{3} = 25 : 15 : 35 v_{1} : v_{2} : v_{3} = 21 : 35 : 15$

NEET Practice Questions (MCQs) with Answers & Solutions

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