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physics ncert solutions class 11th

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Given below are some functions of x and t to represent the displacement of an elastic wave.

(a) y = 5 cos (4x ) sin (20t)

(b) y = 4 sin (5x – t/2) + 3 cos (5x – t/2)

(c) y = 10 cos [(252 – 250) πt ] cos [(252+250)πt ]

(d) y = 100 cos (100πt + 0.5x )

State which of these represent

(a) A travelling wave along –x direction

(b) A stationary wave

(c) Beats

(d) A travelling wave along +x direction.

Given reasons for your answers.

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Payal Gupta

Contributor-Level 10

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

(a) The equation y = 100 cos (100πt + 0.5x ) is representing a travelling wave along x-direction

(b) The equation y = 5 cos (4x ) sin (20t) represents a stationary wave, because it contains sin cos terms ., combinations of two progressive waves.

(c) The equation y = 10 cos [ (252 – 250) πt ] cos [ (252+250)πt ] involving sum and difference of two near by frequencies 252 and 250 have this equation represents beats formation.

(d) As the equation y = y = 4 sin (5x – t/2) + 3 cos (5x – t/2) involves negative sign with x, have if represents a travelling wave along

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

If c is r.m.s. speed of molecules in a gas and v is the speed of sound waves in the gas, show that c/v is constant and independent of temperature for all diatomic gases.

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Payal Gupta

Contributor-Level 10

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

we know that rms speed of molecules of a gas

C= $\sqrt[]{\frac{3 p}{ρ}} = \sqrt[]{\frac{3 R T}{M}}$

Where M is the molar mass of the gas

Speed of sound wave in gas v= $\sqrt[]{\frac{γ p}{ρ}} = \sqrt[]{\frac{γ R T}{M}}$

On dividing above equation we get

c/v = $\sqrt[]{\frac{3 R T}{M} * \frac{M}{γ R T}}$

c/v = $\sqrt[]{\frac{3}{γ}}$

where $γ$ = adiabatic constant for diatomic gas

$γ = \frac{7}{5}$

c/v=constant

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

The earth has a radius of 6400 km. The inner core of 1000 km radius is solid. Outside it, there is a region from 1000 km to a radius of 3500 km which is in molten state. Then again from 3500 km to 6400 km the earth is solid. Only longitudinal (P) waves can travel inside a liquid. Assume that the P wave has a speed of 8 km s^-1 in solid parts and of 5 km s^-1 in liquid parts of the earth. An earthquake occurs at some place close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth if wave travels along diameter?

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Payal Gupta

Contributor-Level 10

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

Speed of wave in solid =8km/s

Speed of wave in liquid = 5km/s

Required time = [ $\frac{1000 - 0}{8} + \frac{3500 - 1000}{5} + \frac{6400 - 3500}{8}$ ] $* 2$

= [ $\frac{1000}{8} + \frac{2500}{5} + \frac{2900}{8}$ ] $* 2$

= [125+500+362.5] $* 2$ =1975 (diameter = radius $* 2$ )

As we are considering at diametrically opposite point, hence there is a multiplication of 2

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Three

A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground,

(a) The direction of motion of the ball changes every 10 seconds.

(b) Speed of ball changes every 10 seconds.

(c) Average speed of ball over any 20 second interval is fixed.

(d) The acceleration of ball is the same as from the train.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b, c, d) velocity of ball before collision =10+1=11m/s

Speed after collision= 10-1=9m/s

As the speed is changing after travelling 10 m and speed is 1m/s hence, time duration of the changing speed is 10

Since the collision of the ball is perfectly elastic there is no dissipation of energy hence, total momentum and kinetic energy are conserved.

Since the train is moving with constant velocity hence, it will act as inertial frame of reference as that of earth and acceleration will be same in both cases.

As the collision is perfectly elastic so momentum and total energy (K.E and

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Three

A spring with one end attached to a mass and the other to a rigid support is stretched and released.

(a) Magnitude of acceleration, when just released is maximum.

(b) Magnitude of acceleration, when at equilibrium position, is maximum.

(c) Speed is maximum when mass is at equilibrium position.

(d) Magnitude of displacement is always maximum whenever speed is minimum.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, c) as we know restoring force is F=-kx

Potential energy of the spring = 1/2kx²

The restoring force is central hence, when particle released it will execute SHM about equilibrium position

Also acceleration= f/m=-kx/m

But at equilibrium position when x=o, a also 0 and at equilibrium position potential energy converted into kinetic energy so speed will also maximum that time.

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Three

For the one-dimensional motion, described by x = t–sint

(a) x (t) > 0 for all t > 0.

(b) v (t) > 0 for all t > 0.

(c) a (t) > 0 for all t > 0.

(d) v (t) lies between 0 and 2.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) x= t-sint

Velocity v = dx/dt=

d/dt (t-sint)=

=1-cost

When cost =1, velocity v=0

V_max=1-cost_min = 1- (-1)=2

V_{min =}1- (cost)_max= 1-1=0

Hence v lies between 0 and 2

Acceleration a=dv/dt=-sint

When v=0 then cost =1

V_max= 1- (-1)=2, V_min=1-1=0

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Three

A graph of x versus t is shown in Fig Choose correct alternatives from below. (a) The particle was released from rest at t = 0. (b) At B, the acceleration a > 0. (c) At C, the velocity and the acceleration vanish. (d) Average velocity for the motion between A and D is positive. (e) The speed at D exceeds that at E.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, c, d) at point A graph will parallel to time axis hence, v=dx/dt=0. As the starting point is A hence, we can say that the particle is starting from rest.

At C, the graph changes slope, hence velocity also changes. As graph at C is almost parallel to time axis hence, we can say that velocity vanishes.

Velocity is 0. So particle at rest that time. As slope of D will be greater than E . so velocity will be greater at D.

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Three

The variation of quantity A with quantity B, plotted in Fig. 3.2 describes the motion of a particle in a straight line.

(a) Quantity B may represent time.

(b) Quantity A is velocity if motion is uniform.

(c) Quantity A is displacement if motion is uniform.

(d) Quantity A is velocity if motion is uniformly accelerated.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, c, d) when we are calculating velocity of a displacement time graph we have to take slope similarly we have to take slope of velocity time graph to calculate acceleration . when slope is constant motion will be uniform.

When we are representing motion by a graph it may be displacement time, velocity time or acceleration time. Hence B may represent time . for uniform motion velocity time graph should be a straight line parallel to time axis.

if Quantity B may represent time. Also Quantity A is displacement if motion is uniform. Quantity A is velocity if motion is uniform

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Three

At a metro station, a girl walks up a stationary escalator in time t₁ . If she remains stationary on the escalator, then the escalator take her up in time t₂. The time taken by her to walk up on the moving escalator will be

(a) (t₁ + t₂)/2

(b) t₁ t₂/(t₁– t₁)

(c) t₁ t₁/( t₁+ t₁)

(d) t₁–t₂

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) in this situation we have to find net velocity with respect to the earth that will be equal to velocity of the girl plus velocity of escalator.

Let displacement is L the

Velocity of girl v_g= L/t₁

Velocity of escalator v_e= L/t₂

net velocity = velocity of escalator +velocity of girl

V_g=l/t₁ and V_e=l/t₂

So net velocity = L/t₁+ L/t₂

So t= $\frac{t_{1} t_{2}}{t_{1} + t_{2}}$

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4 months ago

physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Three

The displacement of a particle is given by x = (t – 2)² where x is in metre and t in seconds. The distance covered by the particle in first 4 seconds is

(a) 4 m

(b) 8 m

(c) 12 m

(d) 16 m

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) x = (t-2)²

V= dx/dt= 2 (t-2)

And acceleration a = 2

When t =0 v= -4m/s

When t= 2s v= 0m/s

When t= 4s v= 4m/s

Distance travelled = area of graph =area of OAC+area of ABD= $\frac{4 * 2}{2} + \frac{1}{2} * 2 * 4$ =8m

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