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7 months ago

Class 11th Human Physiology

How many chapters are there in Human Physiology in NEET Syllabus?

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Abhishek Divya

Beginner-Level 5

Human Physiology is fifth unit in NCERT book of class 11th. It comprises total six topics, covering various aspects of digestion, respiration, circulation, excretion, and locomotion.

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7 months ago

Class 11th 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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7 months ago

Class 11th 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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7 months ago

Class 11th 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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7 months ago

Class 11th 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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7 months ago

Class 11th 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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7 months ago

Class 11th 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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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

A raindrop falling from a height h above ground, attains a near terminal velocity when it has fallen through a height (3/4)h. Which of the diagrams shown in Fig. 6.8 correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) When drop falls first velocity increases, hence first KE also increases, after sometime speed is constant this is called terminal velocity, hence KE also become constant PE decreases continuously as the drop is falling continuously . the variation in PE and KE is best represented by b.

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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be

(a) 250π²

(b) 100π²

(c) 5π²

(d) 0

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) Mass m = 5kg

Radius =1m=R

Revolution per minute w= 300rev/min

= 300 ( $2 π$ )rad/min

= $\frac{\frac{300 * 2 * 3.14}{60} r a d}{s} = \frac{10 π r a d}{s}$

Linear speed v= rw

= $\frac{300 * 2 π}{60} = 10 π m / s$

KE= 1/2mv²

= $\frac{1}{2} * 5 * {10 π}^{2} = 250 π$ ²J

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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Six

Which of the diagrams shown in Fig. represents variation of total mechanical energy of a pendulum oscillating in air as function of time?

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) When a pendulum oscillates in air, it will lose energy continuously in overcoming resistance due to air. Therefore total mechanical energy of the pendulum decreases continuously with time. This variation is correctly represented by option c.

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