Physics NCERT Exemplar Solutions Class 11th Chapter Six: Overview, Questions, Preparation

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

(a) work done= increase in PE

= mg (vertical distance travelled)

= mg (s)sin $\theta =1\times 10\times 10sin30$ = 50J

(b) work done against friction = fs

= $\mu N\left(s\right)=\mu mgcos\theta$

= 0.1 $\times 1\times 10\times cos30\times 10=8.66J$

(c) increase in PE =mgh

 =1 $\times 10\times 10\times sin30$

$=100\left(\frac{1}{2}\right)=50J$

(d) according to work energy theorem W= change in KE

= -mgh-fs+FS

= -50-8.66+10 (10)

= 41.34J

(e) force f = FS

= 10 (10)= 100J

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(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Ball 3 is having negligible friction hence, there is no loss of energy.

(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.

(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.

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Let M be the mass of the rocket at any time t and v₁ the velocity of the rocket at the same time t

Let? m = mass of gas ejected in? t time

Relative speed of the gas ejected =u

KE +? t = KE of rocket +KE of gas

= ½ (M-? m) (v+? v)² + ½? m (v-u)²

KE_t= KE of rocket at time t= ½ Mv²

So? K = KE_t+? t -KE_t

= (M? v=? mu)v+1/2? mu²

Since action and reaction forces are equal

M? v/? t=? m/? t|u|

M? v=? m u

So? K= ½? mu²

? K=? W

? W=1/2? mu²

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m =50g = 50 $\times {10}^{3}kg$

Side = 1cm = 0.01m

Speed v = 0.1m/s

Young’s modulus= 2 $\times {10}^{11}N/m$ ²

According to the formula

F/A= Y $\frac{?L}{L}$

And F= K $?L$  where K is the compression in the spring.

K= YA/L = YL

Initial KE= 2 (1/2mv²)= 5 $\times {10}^{-4}J$

Final PE= 2 (1/2)K ( $?L$ )²

$?L=\sqrt[]{\frac{KE}{E}}=\sqrt[]{\frac{KE}{YL}}$  = $\sqrt[]{\frac{5\times {10}^{-4}}{2\times {10}^{11}\times 0.1}}=1.58\times {10}^{-7}m$

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

V= volume of ballon

${\rho }_{air}=$  density of air

${\rho }_{He}=$ density of helium

V( ${\rho }_{air}-{\rho }_{He}$ )g= ma= mdv/dt= upthrust

Integrating with respect to t

V( ${\rho }_{air}-{\rho }_{He}$ )gt=mv

½ mv²= ½ m v²/m² ( ${\rho }_{air}-{\rho }_{He}$ )²g²t²

= ½v²/m ( ${\rho }_{air}-{\rho }_{He}$ )²g²t²

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

h=1/2at²= ½ $\frac{\mathrm{}\mathrm{V}\mathrm{}({\rho }_{air}-{\rho }_{He})\mathrm{g}{t}^2}{m}$

so from above equation

1/2mv²= [V( ${\rho }_{air}-{\rho }_{He}$ )g][ $\frac{1}{2m}V\mathrm{}({\rho }_{air}-{\rho }_{He})\mathrm{g}{t}^2$ ]

= V( ${\rho }_{air}-{\rho }_{He}$ )gh

So ½ mv²+V ${\rho }_{He}$ gh= ${\rho }_{air}$ hg

KE_ballon+PE_ballon= change in PE of air

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

As the block M is at rest and frictional force =Mgsin $\theta$

The force of friction acting between the blocks and incline opposes the tendency of sliding of the block . since block not in motion therefore no work is done. Hence no dissipation of energy.

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When the elevator is descending then electric power is required to prevent it from falling freely under gravity.

Also as the weight inside the elevator increases, its speed of descending increases, therefore there should be a limit on the number of passengers in the elevator to prevent the elevator from descending with a large velocity.

So when we increase the weight it fall with more speed . that is why there is limit in elevator.

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(a) Force is applied on the body to lift it in the upward direction and displacement of the body is also in upward direction. So angle is zero.

Work done = fscos $\theta$ = fs

W= positive

(b) But the gravitational force in downward direction and displacement in upward direction so angle is 180. So work done will be negative. W= fscos 180= -fs

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Force of gravity acts on the car vertically downward while car is moving along horizontal direction. So angle between them is 90. So work done is zero.

W= fscos90=0

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No total mechanical energy of the body falling freely under gravity is not conserved because a small part of its energy is utilised against resistive force of air, which is non conservative force. Gain in KE< loss in PE.

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No, work done is not always zero in circular path it is zero only when all the forces are zero.

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Total linear momentum of the system of two balls is always conserved. While balls are in contact, there may be deformation which means elastic PE which came from part of KE .But they are in contact while collision so according to above case kinetic energy will not conserved.

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

m =100kg

Height h= 10m

Power= rate of work done =change of energy /time= mgh/t

= $\frac{100\times 9.8\times 10}{20}=5\times 98=490W$

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Average work done by a human heart per beat =0.5J

Total work done during 72 beats= 72 $\times 0.5$ =36J

Power = work done/time

=36/60=0.6W

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When charge particle moves in a uniform magnetic field the path of particle is circular . so when it moves in circular path its radius Is also constant . so its kinetic energy also constant.

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According to work energy theorem = change in kinetic energy = work done

Kinetic energy of the body = force (displacement)

As kinetic energy is same so both will move with same velocities.

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

At B the velocity of B is vertically downward, therefore when string is cut at B then it fall downwards.

At C velocity along horizontally right, so when we cut it at C then it will move to right. But under the action of gravity its path becomes parabola.

At X when we cut it moves tangentially in forward direction. So under of the action of gravity it also follows parabola path.

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

mechanical energy =KE+PE

E_o = KE + V (x)

KE= E₀ -V (x)

At A x=0 v (x)=E_o

KE= E_o-E_o =0

atB, V (x)o

so KE>0

at C and D, V (x)= 0

KE is maximum at FV (x)= E_o

Hence KE= 0

As KE= 1/2mv²

Therefore at A and F where KE =0, v=0

At C and D KE is maximum therefore v is maximum.

At B  KE is positive but not maximum but it has some value.

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

Let v₁ and v₂ are velocities of two balls after collision

According to conservation of momentum

2mv_o = mv₁+ mv₂

2v_o= v₁+v₂

and e= v₂-v₁/2v_o

v₂=v₁+2v_oe

2v₁=2v_o-2ev_o

V₁=v_o (1-e) since e<1 so ball will move after collision.

b)by principle of conservation of linear momentum

P=P₁+P₂

For inelastic collision some KE is lost hence $\frac{p^2}{2m}$ > $\frac{p_1^2}{2m}\frac{p_2^2}{2m}$

P²>p₁²+p₂²

Thus P, P₁ and P₂ are related as shown in fig

P²>p₁²+p₂² this condition only holds when angle is 90.

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

TE=KE+PE

E=V+K

For region A given V>E 

so ‘K=E-V

V>E

 So E-V <0

Hence K<0 this is not possible.

For region B given V

So E-V>0

This is only possible because total energy is greater than PE

For region C given K>E

 So K-E>0

So PE =V= E-K <0

Which is possible because PE can be negative.

For region D given V>K

This is possible because for the system PE may be greater than KE

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

When the ball A reaches bottom point its velocity in horizontal direction

(a) two balls have same mass and the collision between them is elastic therefore ball A transfers its entire linear momentum to ball B. hence ball A will come to rest after collision and does not rise at all.

(b) speed at B = speed with which A hits the ball B

= $\sqrt[]{2gh}$

$=\sqrt[]{2\times 9.8\times 1}$

$=\sqrt[]{19.6}=4.42m/s$

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

mass of the rain drop = 1g=1 $\times {10}^{-3}kg$

Height of falling h= 1km = 10³m and g = 10m/s² and sped of drop =50m/s

(a) Loss of PE of the drop =mgh= 1 $\times {10}^{-3}\times 10\times {10}^3=10J$

(b) Gain in KE of the drop = 1/2mv²= ½ $\times 1\times {10}^{-3}\times {50}^2$ =1.250J

(c) No gain in KE is not equal to the loss in its Pe, because a part of PE is utilised in doing work against the viscous drag in air.

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 repeated.

b)the plot of energy with time 0

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Mass of drop m = 3 $\times {10}^{-5}kg$

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 $\times$ height

= 1 (1)= 1m³

Mas $\frac{1}{2}\times {10}^3\times 9^2=40.5\times {10}^{3}J$ s of water due to rain M= volume (density)

= V ( $\rho$ )= 10³kg

Energy transferred to the surface = 1/2mv²

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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 $\times {10}^{6}J$

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

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

= $\frac{10}{100}\times 2.5\times {10}^{6}J$ 0r K = $\frac{2\times 2.5\times {10}^6}{10}=5\times {10}^{5}N/m$

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 $\times$ mgh

= 6000 $\times$ 600 $\times$ 0.25J

= 9 $\times {10}^5$ J

Since 10% of intake energy is utilised in jogging

So total energy intake = 10

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

Energy given by 1 lit of petrol = 3 $\times {10}^{7}J$

Efficiency of the car engine=0.5s

Energy used by the car = 0.5 $\times 3\times {10}^7=1.5\times {10}^7$

Total distance travelled s = 15km = 15 $\times {10}^{3}m$

If f the force of friction= E= f $\times s$

1.5 $\times {10}^7$ = f $\times 15\times {10}^3$

F= $\frac{1.5\times {10}^7}{15\times {10}^3}={10}^{3}N$

F= 1000 N

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

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

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(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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(c) when we are considering the two bodies as system the total external force on the system will be zero . hence total linear momentum of the system remains conserved.

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(c) From conservation of mechanical energy

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

V= $\sqrt[]{2gh}$

Hence speed is same for both stones .

a₁=acceleration along inclined plane = gsin $\theta$ ₁

Similarly for stone a₂=gsin $\theta$ ₂ as $\theta$ ₂> $\theta$ ₁, hence a₂>a₁

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(b) Total energy is E=KE+PE……………1

When particle is at x =x_m i.e at extreme position, returns back. Hence at x=x_m and x=0 KE=0

From eqn1

E= PE+0 = PE= V (x_m)=1/2kx_m²

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(b) When two bodies of equal masses collides elastically, their velocities are interchanged.

When ball 1 collides with ball 2 then velocity of ball1, v₁ becomes zero and velocity of ball 2, v₂ becomes v.

V₁=0, V₂=v

When ball 2 collides will ball 3 v₂ =0, v₃=v

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(b) v=ax^3/2

m=0.5kg, a= 5m/s²

a= dv/dt= vdv/dt= ax^3/2d/dx (ax^3/2)

= ax^3/2a $\times \frac{3}{2}\times$ x^1/2= 3/2a²x²

Force = ma= m $\frac{3}{2}$ a²x²

Work done = ${\int }_0^{2}fdx={\int }_0^3\frac{3}{2}m{a}^2{x}^2$ dx

= $\frac{3}{2}$ ma² (x³/3)₀²

= $\frac{1}{2}m{a}^2\times 8=\frac{1}{2}\times \left(0.5\right)\times 25\times 8$ =50J

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(b) Power =constant

P+dw/dt=fds/dt

Fds=fdscos0

P=fds/dt=constant

Fv=constant

[MLT^-2] [LT^-1]= constant

[L²T^-3]= constant

L $\propto T$ ^3/2

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(d) When the earth is the closest to the sun, speed of the earth is maximum, hence KE is maximum. When the earth is the farthest from the sun speed is minimum hence, KE is minimum but never zero and negative.

This variation is correctly represented by option d

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(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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(a) Mass m = 5kg

Radius =1m=R

Revolution per minute  w= 300rev/min

= 300 ( $2\pi$ )rad/min

= $\frac{\frac{300\times 2\times 3.14}{60}rad}{s}=\frac{10\pi rad}{s}$

Linear speed v= rw

= $\frac{300\times 2\pi }{60}=10\pi m/s$

KE= 1/2mv²

= $\frac{1}{2}\times 5\times {10\pi }^2=250\pi$ ²J

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(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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(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 $\times 10\times 1.5+\frac{1}{2}\times 10\times 1$

KE= 150+5=155J

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(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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(c) m =150g =3/20kg

Time of contact =0.001s

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

V= -35m/s

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

=21/2

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

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

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

U_i + K_i = U_f + K_f

0+1/2mv²= mgh + 1/2mv’

$\frac{{\left(v\text{'}\right)}^2}{2}=\frac{v^2}{2}=-gh$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^2-2gh}$ ……….1

Let speed after emerging be v₁ then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^2}{2}-gh}$ ………….2

From eqn 1 and 2

$\frac{v\text{'}}{v_1}=\frac{\sqrt[]{v^2-2gh}}{\sqrt[]{v^2-2gh/\sqrt[]{2}}}=\sqrt[]{2}$

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.

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