physics ncert solutions class 11th

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 ( $2\pi$ )rad/min

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

Linear speed v= rw

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

KE= 1/2mv²

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

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.

This is a multiple choice answer as classified in NCERT Exemplar

(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

This is a multiple choice answer as classified in NCERT Exemplar

(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

This is a multiple choice answer as classified in NCERT Exemplar

(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 $*\frac{3}{2}*$ 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

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

This is a multiple choice answer as classified in NCERT Exemplar

(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

This is a multiple choice answer as classified in NCERT Exemplar

(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

This is a multiple choice answer as classified in NCERT Exemplar

(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

This is a multiple choice answer as classified in NCERT Exemplar

(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.

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.