physics ncert solutions class 11th

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) As the body is  falling freely under gravity, the potential energy decreases and kinetic energy increases but total mechanical energy (PE+KE) of the body and earth system will be constant as external force on the system is zero.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) Work done by the frictional force on the cycle and is equal to -200 * 10 =-2000J

As the road is not moving, hence work done by the cycle on the road = zero.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

When the man is squatting on the ground he is tilted somewhat, hence he also has to balance frictional force besides his weight in this case.

R= reactional force = friction +mg

R>mg

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) Force between two protons is same as that of between proton and a positron. As positron is much lighter than proton, it moves away through much larger distance compared to proton.

We know that work done = force (distance). As forces are same in case of proton and positron but distance moved by positron is larger, hence work will be more.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) When electron and proton are moving under influence of their mutual forces the magnetic forces will be perpendicular to their motion hence no work is done by these forces.

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Payal Gupta

Contributor-Level 10

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

Energy given by 1 lit of petrol = 3 * 10 7 J

Efficiency of the car engine=0.5s

Energy used by the car = 0.5 * 3 * 10 7 = 1.5 * 10 7

Total distance travelled s = 15km = 15 * 10 3 m

If f the force of friction= E= f * s

1.5 * 10 7 = f * 15 * 10 3

F= 1.5 * 10 7 15 * 10 3 = 10 3 N

F= 1000 N

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Payal Gupta

Contributor-Level 10

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

Weight of the adult w= mg =600N

Height of each step = h = 0.25m

Total distance travelled = 6km =6000m

Total number of steps = 6000/1= 6000

Total energy utilised in jogging = n * mgh

= 6000 * 600 * 0.25J

= 9 * 10 5 J

Since 10% of intake energy is utilised in jogging

So total energy intake = 10

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Payal Gupta

Contributor-Level 10

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

Mass of the system = 50000kg

Speed of the system v= 36km/h= 10m/s

Compression of the spring x= 1m

KE of the system = 1/2mv2

= ½ (50000) (10)2

= 25000 (100)J= 2.5 * 10 6 J

Since 90 % of KE is lost due to the friction so energy transferred is

E= 1/2kx2= 10% of total KE of the system

= 10 100 * 2.5 * 10 6 J 0r K = 2 * 2.5 * 10 6 10 = 5 * 10 5 N / m

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Payal Gupta

Contributor-Level 10

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

Mass of drop m = 3 * 10 - 5 k g

Terminal velocity = 9m/s

Height = 100cm=1m

Density of water  = 103kg/m3

Area of the surface = 1m2

Volume of the water due to rain V = area * height

= 1 (1)= 1m3

Mas 1 2 * 10 3 * 9 2 = 40.5 * 10 3 J s of water due to rain M= volume (density)

= V ( ρ )= 103kg

Energy transferred to the surface = 1/2mv2

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Payal Gupta

Contributor-Level 10

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

At t=0 suppose bob B is displaced by angle 10 to the right . it is given potential energy E1=E . energy of A, E2=0

When B is released it strikes at A at t=T/4 in the head on elastic collision between B and A comes to rest and A gets velocity of B. therefore E1=0 and E2=E. at A =2T/4, B reaches its extreme right position when KE of A is converted into PE=E2=E . Energy of B, E1=0

At t=3T/4. A reaches its mean position when its PE is converted into KE =E2 =E. it collides elastically with B and transfers whole of its energy to B. thus E2=0 and E1 =E . the entire process is r

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