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Work, Energy, and Power physics ncert solutions class 11th Physics Class 11th Work, Energy and Power

6.21 The blades of a windmill sweep out a circle of area A.

(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ?

(b) What is the kinetic energy of the air ?

(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ?

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Payal Gupta | Contributor-Level 10

5 months ago

6.21 Given, the area of the windmill sweep = A, Wind velocity = v

The volume of air passing through the blade = Av

Let the density of air be $?$ , the mass of air passing through the blade = $? A v$

(a) The mass of air passing through the blade in time t = $? A v t$

(b) The kinetic energy of air = $(\frac{1}{2}) m v^{2}$ = $(\frac{1}{2}) ? A v t v^{2}$ = $(? A t v^{3})$ /2 …. (1)

(c) Area, A = 30 $m^{2}$ , v = 36 km/h = 10 m/s, density of air be $?$ = 1.2 kg/ $m^{3}$

Total wind energy, from eqn. (1) = 18 kW

Electrical energy = 25 % of wind energy = 0.25 $* 18 = 4.5 k W$

6.21 Given, the area of the windmill sweep = A, Wind velocity = vThe volume of air passing through the blade = AvLet the density of air be <math><mi>? </mi></math> , the mass of air passing through the blade = <math><mi>? </mi><mi>A</mi><mi>v</mi></math> (a) The mass of air passing through the blade in time t = <math><mi>? </mi><mi>A</mi><mi>v</mi><mi>t</mi></math>  (b) The kinetic energy of air = <math><mfenced separators="|"><mrow><mrow><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfenced><mi>m</mi><msup><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = <math><mfenced separators="|"><mrow><mrow><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfenced><mi>? </mi><mi>A</mi><mi>v</mi><mi>t</mi><msup><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = <math><mo> (</mo><mi>? </mi><mi>A</mi><mi>t</mi><msup><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup><mo>)</mo></math> /2 …. (1)  (c) Area, A = 30 <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> , v = 36 km/h = 10 m/s, density of air be <math><mi>? </mi></math> = 1.2 kg/ <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math>Total wind energy, from eqn. (1) = 18 kWElectrical energy = 25 % of wind energy = 0.25 <math><mo>*</mo><mn>18</mn><mo>=</mo><mn>4.5</mn><mi></mi><mi></mi><mi>k</mi><mi>W</mi></math>

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Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in Fig. 6.10. Initially M₂ is at rest and M₁ is moving toward M₂ with speed v and collides head-on with M₂.

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This is a multiple choice answer as classified in NCERT Exemplar

(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force.

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A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?

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(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

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$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

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=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

So v₁ = v’/ $\sqrt[]{2}$ =v²(v’/2)

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This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv’

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

So v₁ = v’/ $\sqrt[]{2}$ =v²(v’/2)

v₁>v’/2

hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’

(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path, followed will change and the bullet reaches at point B instead of A

(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.

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(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

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A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be

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(c) 1.05 ×104 N

(d) 2.1 × 104 N

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

Time of contact =0.001s

U=126km/h= $\frac{126 \times 1000}{60 \times 60} = \frac{35 m}{s}$

V= -35m/s

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