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

6.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10^-31 kg, proton mass = 1.67×10^–27 kg, 1 eV = 1.60 ×10^–19 J).

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Payal Gupta | Contributor-Level 10

5 months ago

6.12 Electron mass, me = 9.11*10-31 kg

Proton mass, madhya pradesh = 1.67*10–27 kg

Electron's kinetic energy = 10 keV = 10 $*$ $10^{3}$ $*$ 1.60 *10–19 J = 1.60 *10–15 J

Proton's kinetic energy = 100 keV = 100 $* 10^{3}$ $*$ 1.60 *10–19 J = 1.60 *10–14 J

The electron kinetic energy is given by Eke = (1/2)m $v_{e}^{2}$ where $v_{e}$ is the velocity of electron

$v_{e}$ = $?$ { (2 $*$ Eke )/m} = 5.92 $* 10^{7}$ m/s

The velocity of proton $v_{p}$ = $?$ { (2 $*$ Pke )/m} = 4.37 $* 10^{6}$ m/s

The speed ratio = $v_{e}$ / $v_{p}$ = 13.5

6.12 Electron mass, me = 9.11*10-31 kgProton mass, madhya pradesh = 1.67*10–27 kgElectron's kinetic energy = 10 keV = 10 <math><mo>*</mo></math> <math><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> <math><mo>*</mo></math> 1.60 *10–19 J = 1.60 *10–15 JProton's kinetic energy = 100 keV = 100 <math><mo>*</mo><mi></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> <math><mo>*</mo></math> 1.60 *10–19 J = 1.60 *10–14 JThe electron kinetic energy is given by Eke = (1/2)m <math><msubsup><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msubsup></math> where <math><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></math> is the velocity of electron<math><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></math> = <math><mo>? </mo></math>  { (2 <math><mo>*</mo><mi></mi></math> Eke )/m} = 5.92 <math><mo>*</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>7</mn></mrow></mrow></msup></math> m/sThe velocity of proton <math><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>p</mi></mrow></mrow></msub></math> = <math><mo>? </mo></math>  { (2 <math><mo>*</mo><mi></mi></math> Pke )/m} = 4.37 <math><mo>*</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>6</mn></mrow></mrow></msup></math> m/sThe speed ratio = <math><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub><mi></mi></math> / <math><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>p</mi></mrow></mrow></msub></math> = 13.5

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An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d₁ from C and the distance of the image formed is d₂ from C, the radius of curvature of this mirror is:

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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) The velocity of the bullet will be more than half of its earlier velocity.

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

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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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So v₁ = v’/ $\sqrt[]{2}$ =v²(v’/2)

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(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path

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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

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(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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This is a multiple choice answer as classified in NCERT Exemplar

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