Physics NCERT Exemplar Solutions Class 11th Chapter Six

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

Time of contact =0.001s

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

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 $*{10}^{4}N$

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

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(b) 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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(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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(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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(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

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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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(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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(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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(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