Class 11th

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

F= 1000 N

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

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

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

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

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

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 $*$ height

= 1 (1)= 1m³

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

= V ( $\rho$ )= 10³kg

Energy transferred to the surface = 1/2mv²

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

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

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 $*{10}^{-3}*10*{10}^3=10J$

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

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

= $\sqrt[]{2gh}$

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

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

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

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

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.