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Physics NEET Questions — Free MCQ Practice

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The Young's modulus of a wire is Y. If the energy per unit volume is E, then the strain will be

A.

$\sqrt{\frac{2 E}{Y}}$

B.

$\sqrt{2 E Y}$

C.

$E Y$

D.

$\frac{E}{Y}$

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Explanation

(a)

$E n e r g y p e r u n i t v o l u m e = \frac{1}{2} \times Y \times {(s t r a i n)}^{2} ∴ s t r a i n = \sqrt{\frac{2 E}{Y}}$

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If a spring extends by x on loading, then the energy stored by the spring is (if T is tension in the spring and k is spring constant)

A.

$\frac{T^{2}}{2 x}$

B.

$\frac{T^{2}}{2 k}$

C.

$\frac{2 x}{T^{2}}$

D.

$\frac{2 T^{2}}{k}$

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Explanation

(b) $U = \frac{F^{2}}{2 K} = \frac{T^{2}}{2 K}$

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A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. It is stretched by an amount x. The work done is -

A.

$\frac{Y x A}{2 L}$

B.

$\frac{Y x^{2} A}{L}$

C.

$\frac{Y x^{2} A}{2 L}$

D.

$\frac{2 Y x^{2} A}{L}$

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Explanation

(c)

$W = \frac{1}{2} \times \frac{A Y}{L} \times {(x)}^{2} W h e r e x i s d e f o r m a t i o n$

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A stretched rubber has

A.

Increased kinetic energy

B.

Increased potential energy

C.

Decreased kinetic energy

D.

Decreased potential energy

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Explanation

(b)

Due to stretching, potential energy is stored in the rubber band which is released on releasing the rubber. Thus, potential energy is increased.

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When load of 5kg is hung on a wire then extension of 3m takes place, then work done will be

A.

75 joule

B.

60 joule

C.

50 joule

D.

100 joule

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Explanation

(a) $W = \frac{1}{2} F l = \frac{1}{2} \times M g \times l = \frac{1}{2} \times 5 \times 10 \times 3 = 75 J$

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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

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

Physics MCQs for NEET — Practice Questions with Answers

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