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Work, Energy, and Power

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2 months ago

Work, Energy, and Power Class 11th

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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Vishal Baghel

Contributor-Level 10

Using Newton's formula,

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

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

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2 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$ -axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$ . If both particles move along the $x$ axis after the collision, the change $Δ λ$ in de-Broglie wavelength of particle $A$ , in terms of its de-Broglie wavelength $(λ_{0})$ before collision is:

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alok kumar singh

Contributor-Level 10

$\Rightarrow {\overset{?}{V}}_{1} = 2 \overset{?}{U_{c m}} - \overset{?}{U_{1}} \Rightarrow 2 [\frac{m / 2 V_{o}}{\frac{m}{2} + \frac{m}{3}}] - V_{o}$

$\Rightarrow \frac{6}{5} V_{o} - V_{o} \Rightarrow \frac{V_{0}}{5}$

$\Rightarrow λ_{o} = \frac{h c}{(\frac{m}{2} V_{o})} λ_{f} = \frac{h c}{(\frac{M}{2} \frac{V_{o}}{5})}$

$\Rightarrow Δ λ = \frac{8 h c}{{m V}_{o}}$

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2 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A wire $A$ , bent in the shape of an arc of a circle, carrying a current of $2 A$ and having radius $2 c m$ and another wire $B$ , also bent in the shape of arc of a circle, carrying a current of $3 A$ and having radius of $4 c m$ , are placed as shown in the figure. The ratio of the magnetic fields due to the wires $A$ and $B$ at the common centre $O$ is:

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alok kumar singh

Contributor-Level 10

$B_{A} = \frac{μ_{0} I θ}{4 π R}$

$\Rightarrow \frac{B_{A}}{B_{B}} = \frac{I_{A} θ_{A} R_{B}}{I_{B} θ_{B} R_{A}}$

$\Rightarrow \frac{2 (\frac{3 π}{2}) (4)}{3 [\frac{5 π}{3}] [2]}$

$\Rightarrow \frac{6}{5}$ .

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2 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A person pushes a box on a rough horizontal platform surface. He applies a force of 200 $N$ over a distance of $15 m$ . Thereafter, he gets progressively tired and his applied force reduces linearly with distance to $100 N$ . The total distance through which the box has been moved is $30 m$ . What is the work done by the person during the total movement of the box?

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alok kumar singh

Contributor-Level 10

Sol. (n) $λ = 5$

$(n, m)$ : Integers

$\begin{array}{r} (\frac{2 m + 1}{2}) λ = \frac{3}{2} \\ \Rightarrow \frac{3 / 2}{5} = \frac{2 m + 1}{2 n} \\ \Rightarrow 3 n = 10 m + 5 \end{array}$

$N, m$ are integers.

So, $m = 1, n = 5, λ = 1$

$\begin{array}{l} m = 4 & n = 15 & λ = \frac{1}{3} \\ m = 7 & n = 25 & λ = \frac{1}{5} \end{array}$

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3 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 11th Chapter Six

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.

(a) While spring is fully compressed all the KE of M₁ is stored as PE of spring.

(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.

(c) If spring is massless, the final state of the M₁ is state of rest.

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

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Payal Gupta

Contributor-Level 10

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 .

d) collision is elastic even if friction is not involved.

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3 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 11th Chapter Six

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 p

...more

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Payal Gupta

Contributor-Level 10

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

...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

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New answer posted

3 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 11th Chapter Six

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.

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Payal Gupta

Contributor-Level 10

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.

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New answer posted

3 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 11th Chapter Six

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

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Time of contact =0.001s

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

V= -35m/s

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

=21/2

F= dp/dt=- $\frac{21 / 2}{0.001} N$ = - 1.05 $* 10^{4} N$

Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.

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New answer posted

3 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 11th Chapter Six

Which of the diagrams in Fig. 6.9 correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity?

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) First velocity of the iron sphere increases and after sometimes becomes constant called terminal velocity. Hence according first KE increases and then becomes constant which is best represented by b.

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New answer posted

3 months ago

Work, Energy, and Power Physics NCERT Exemplar Solutions Class 11th Chapter Six

In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m/s at 45° from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m/s² , the kinetic energy of the shotput when it just reaches the ground will be

(a) 2.5 J

(b) 5.0 J

(c) 52.5 J

(d) 155.0 J

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(d) h =15m, v= 1m/s, m= 10kg, g= 10m/s²

From conservation of mechanical energy

(PE+KE)_initial= (PE+KE)_final

Mgh + $\frac{1}{2} m v^{2}$ =0+KE

KE= mgh + $\frac{1}{2} m v^{2}$

KE= 10 $* 10 * 1.5 + \frac{1}{2} * 10 * 1$

KE= 150+5=155J

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