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Work, Energy, and Power physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

A ballon filled with helium rises against gravity increasing its potential energy. The speed of the ballon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.

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Payal Gupta | Contributor-Level 10

3 months ago

This is a long answer type question as classified in NCERT Exemplar

V= volume of ballon

$ρ_{a i r} =$ density of air

$ρ_{H e} =$ density of helium

V( $ρ_{a i r} - ρ_{H e}$ )g= ma= mdv/dt= upthrust

Integrating with respect to t

V( $ρ_{a i r} - ρ_{H e}$ )gt=mv

½ mv²= ½ m v²/m² ( $ρ_{a i r} - ρ_{H e}$ )²g²t²

= ½v²/m ( $ρ_{a i r} - ρ_{H e}$ )²g²t²

= if the ballon rises to height h then s= ut +1/2at²

h=1/2at²= ½ $\frac{V (ρ_{a i r} - ρ_{H e}) g t^{2}}{m}$

so from above equation

1/2mv²= [V( $ρ_{a i r} - ρ_{H e}$ )g][ $\frac{1}{2 m} V (ρ_{a i r} - ρ_{H e}) g t^{2}$ ]

= V( $ρ_{a i r} - ρ_{H e}$ )gh

So ½ mv²+V $ρ_{H e}$ gh= $ρ_{a i r}$ hg

KE_ballon+PE_ballon= change in PE of air

This is a long answer type question as classified in NCERT ExemplarV= volume of ballon <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>=</mo> </math>  density of air <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> <mo>=</mo> </math> density of heliumV( <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> </math> )g= ma= mdv/dt= upthrustIntegrating with respect to tV( <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> </math> )gt=mv½ mv2= ½ m v2/m2 ( <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> </math> )2g2t2= ½v2/m ( <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> </math> )2g2t2= if the ballon rises to height h then s= ut +1/2at2h=1/2at2= ½ <math> <mfrac> <mrow> <mrow> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">V</mi> <mi mathvariant="normal"> </mi> <mo>(</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> <mo>)</mo> <mi mathvariant="normal">g</mi> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </math> so from above equation1/2mv2= [V( <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> </math> )g][ <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>m</mi> </mrow> </mrow> </mfrac> <mi>V</mi> <mi mathvariant="normal"> </mi> <mo>(</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> <mo>)</mo> <mi mathvariant="normal">g</mi> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> ]= V( <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> </math> )ghSo ½ mv2+V <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> </msub> </math> gh= <math> <msub> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> <mi>i</mi> <mi>r</mi> </mrow> </mrow> </msub> </math> hgKEballon+PEballon= change in PE of air

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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:

Vishal Baghel

Using Newton’s formula,

$\Rightarrow (f + d_{1}) (f - d_{2}) = f^{2}$

$\Rightarrow f = \frac{d_{1} d_{2}}{d_{1} - d_{2}} \Rightarrow R = \frac{2 d_{1} d_{2}}{d_{1} - d_{2}}$

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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(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

Payal Gupta

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.

a) So spring will continue compress until the blocks moves with the same velocity.

b) As the surfaces are frictionless momentum of the system will conserved.

c) If the spring is massless whole energy of M1 will be imparted and M1 will be rest .

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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?

(a) The velocity of the bullet will be reduced to half its initial value.

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

(c) The bullet will continue to move along the same parabolic path.

(d) The bullet will move in a different parabolic path.

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Payal Gupta

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

...more

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.

less

A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.

(a) Work done by all forces on man is equal to the rise in potential energy mgL.

(b) Work done by all forces on man is zero.

(c) Work done by the gravitational force on man is mgL.

(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(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.

Hence total work done =-mgL + mgL=0

As the point of application of the contact forces does not move hence work done by reaction forces will be zero.

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

(a) 10.5 N

(b) 21 N

(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

Change in momentum of the ball = m (v-u)= $\frac{3}{20} (- 35 - 35) k g m / s$

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A ballon filled with helium rises against gravity increasing its potential energy. The speed of the ballon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.

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