physics ncert solutions class 11th

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

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

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $ρ =$ density of the liquid

PE= dmgx=A $ρ d x g x$

So total PE of the column

= $\int_{0}^{h_{1}} A d x g ρ x$ = $A g ρ \int_{o}^{h_{1}} x d x = A ρ g \frac{h_{1}^{2}}{2}$

But h₁=lsin45

PE=A $ρ$ gl²sin²45/2

Similarly PE of right column = A $ρ$ gl²sin²45/2

Total PE = A $ρ$ gl²sin²45/2+ A $ρ$ gl²sin²45/2

= A $ρ$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A ρ g}{2} [{(l - y)}^{2} + {(l + y)}^{2} - l^{2}$ ]

= $\frac{A ρ g}{2} [2 {(l}^{2} + y^{2})]$ =A $ρ g {(l}^{2} + y^{2})$

Change in KE = ½ mv²

m=A $ρ 2 l$

change in KE= 1/2A $ρ 2 l v^{2}$ =A $ρ l v^{2}$

so from

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

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $ρ =$ density of the liquid

PE= dmgx=A $ρ d x g x$

So total PE of the column

= $\int_{0}^{h_{1}} A d x g ρ x$ = $A g ρ \int_{o}^{h_{1}} x d x = A ρ g \frac{h_{1}^{2}}{2}$

But h₁=lsin45

PE=A $ρ$ gl²sin²45/2

Similarly PE of right column = A $ρ$ gl²sin²45/2

Total PE = A $ρ$ gl²sin²45/2+ A $ρ$ gl²sin²45/2

= A $ρ$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A ρ g}{2} [{(l - y)}^{2} + {(l + y)}^{2} - l^{2}$ ]

= $\frac{A ρ g}{2} [2 {(l}^{2} + y^{2})]$ =A $ρ g {(l}^{2} + y^{2})$

Change in KE = ½ mv²

m=A $ρ 2 l$

change in KE= 1/2A $ρ 2 l v^{2}$ =A $ρ l v^{2}$

so from above eqn

change in PE +change in KE = A $ρ g {(l}^{2} + y^{2})$ +A $ρ l v^{2}$

system being conservative

so change in total energy =0

A $ρ g {(l}^{2} + y^{2}) + A ρ l v^{2}$ =0

Differentiating both with respect to time t

A $ρ g (2 y v) + A ρ l (2 v) a = 0$ =0

Dy/dt =v and dv/dt=a

(gy+la)2A $ρ v$ =0

2A $ρ v = c o n s t a n t$ and 2A $ρ v \neq 0$

La+gy=0

A+(gy/l)=0

d²y/dt²+gy/l=0

d²y/dt²+w²y=0

w= $\sqrt[]{\frac{g}{l}}$

T=2 $π \sqrt[]{\frac{l}{g}}$

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A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. 2 m T A g π ρ = where m is mass of the body and ρ is density of the liquid.

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

Let the log be passed and the vertical displacement at the vertical displacement at the equilibrium position

So mg= buoyant forces = $ρ A x_{o} g$

When it is displaced by further displacement x, the buoyant force is A (x_o+x) $ρ g$

Net restoring force = buoyant forces -weight

=A (x_o+x) $ρ g$ -mg

=A $ρ g x$

As displacement x is downward and restoring force is upward

F_restoring₌-A $ρ g x$ =-kx

So motion is SHM

Acceleration a=F_restoring/m=-kx/m

a=-w²x

w²=k/m

w= $\sqrt[]{\frac{k}{m}}$

T= 2 $π \sqrt[]{\frac{m}{k}} = 2 π \sqrt[]{\frac{m}{A ρ g}}$

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A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand. (a) What is the amplitude of oscillation? (b) Find the frequency of oscillation?

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

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x' is the extension in that case

F= kx'

F=mg

Mg=kx'

X'=mg/k

By dividing x by x'

x/x'= $\frac{2 m g / k}{m g / k} = 2$

so x=2x'

x'=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural l

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

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x' is the extension in that case

F= kx'

F=mg

Mg=kx'

X'=mg/k

By dividing x by x'

x/x'= $\frac{2 m g / k}{m g / k} = 2$

so x=2x'

x'=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural length is equal to amplitude of the oscillation

A=x'=2cm

(b) Time period of the oscillating system depends on mass spring constant given by

T= $2 π \sqrt[]{\frac{m}{k}}$

2mg/k=x

2mg/k=4cm =4 $* 10^{- 2} m$

m/k =4 $* 10^{- 2}$ /2g

k/m=g/2 $* 10^{- 2}$

$v = \frac{1}{T} \sqrt[]{\frac{k}{m}}$

$v = \frac{1}{2 * 3.14} \sqrt[]{\frac{g}{2 * 10^{- 2}}}$

= $\frac{10}{628} * 2.21$ =3.51Hz

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A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s^-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.

(a) Will there be any change in weight of the body, during the oscillation?

(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?

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

Contributor-Level 10

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

(a) The weight of the body changes during oscillations.

(b) Considering the situations in two extreme positions

we can say mg-N= ma

so at the highest point the platform is accelerating downward.

N=mg-ma

a=w²A

N=mg-mw²A

A= amplitude of motion m=50kg v=2m/s

w=2 $π v = 4 π r a d / s$

A= 5cm = 5 $* 10^{- 2} m$

N= 50 $* 9.8 - 50 * {(4 π)}^{2} * 5 * 10^{- 2} = 95.5 N$

When it is accelerating towards mean position that is vertically upwards

N-mg=ma=Mw²A

N=mg+mw²A

N=m (g+w²A)

N= 50 [9.8+ ${(4 π)}^{2} * 5 * 10^{- 2}$ ]

N= 884N

Machine reads the normal reaction

Maximum weight =884N

Minimum weight=95.5N

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For two vectors A and B, A B A B + = − is always true when (a) A B = ≠ 0 (b) A⊥ B (c) A B = ≠ 0 and A and B are parallel or anti parallel (d) when either A or B is zero.

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

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- |A+B|=|A-B|

$\sqrt[]{|{A |}^{2} + |{B |}^{2} + 2| A| | B | c o s θ} = \sqrt[]{|{A |}^{2} + |{B |}^{2} - 2| A| | B | c o s θ}$

$|{A |}^{2} + |{B |}^{2} + 2| A| | B | c o s θ$ = $|{A |}^{2} + |{B |}^{2} - 2| A| | B | c o s θ$

4|A|B|cos $θ$ =0

|A|²+|B|²cos $θ$ =0

A=0 or B=0 so $θ = 90$ . so A perpendicular B

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For a particle performing uniform circular motion, choose the correct statement(s) from the following:

(a) Magnitude of particle velocity (speed) remains constant.

(b) Particle velocity remains directed perpendicular to radius vector.

(c) Direction of acceleration keeps changing as particle moves.

(d) Angular momentum is constant in magnitude but direction keeps changing.

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

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- (i) speed will constant throughout

(ii) velocity will be tangential in the direction of motion

(iii) centripetal acceleration will be a= v²/r, will always be towards centre of the circular path.

(iv) angular momentum is constant in magnitude and direction out of the plane perpendicularly as well.

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Following are four different relations about displacement, velocity and acceleration for the motion of a particle in general. Choose the incorrect one (s) :

(a) V_av= $\frac{1}{2} [v (t_{1}) + v (t_{2})]$

(b) V_av= $\frac{r {(t}_{2}) - r {(t}_{1})}{t_{2} - t_{1}}$

(c) V_av= $\frac{1}{2} [v (t_{1}) - v (t_{2}) (t_{2} - t_{1})]$

$(d) V a v = \frac{v {(t}_{2}) - v {(t}_{1})}{t_{2} - t_{1}}$

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

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c

Explanation – as we know average acceleration is a_av= $\frac{? v}{? t} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

But when acceleration is not uniform V_av is not equal to v₁+v₂/2

So we can write $? v = \frac{? r}{? t}$

$? r = r_{2} - r_{1}$ = v₂-v₁ (t₂-t₁)

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A particle slides down a frictionless parabolic (y + x² ) track (A – B – C) starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then

(a) KE at P = KE at B

(b) Height at P = height at A

(c) Total energy at P = total energy at A

(d) Time of travel from A to B = time of travel from B to P.

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

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation– as the given track y=x² is a frictionless track thus total energy will be same throughout the journey.

Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P

Hence (KE)_B= (KE)_P

Total energy at A = PE= total energy at B = KE= total energy at P

= PE+KE

Potential energy at A is converted to KE and PE at P hence

(PE)P< (PE)A

Hence (height)P= (height)A

As height of p < height of A

Hence path length AB > path length BP

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Two particles are projected in air with speed vo at angles θ1 and θ2 (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices

(a) Angle of projection : q1 > q2

(b) Time of flight : T1 > T2

(c) Horizontal range : R1 > R2

(d) Total energy : U1 > U2 .

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

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a,b,c

Explanation – H= $\frac{u^{2} {s i n}^{2} θ}{2 g}$

H₁=V_o²sin^{2
$θ$}₁/2g , H₂=V_o²sin^{2
$θ$}₂/2g

H₁>H₂

V_o²sin^{2
$θ$}₁/2g= V_o²sin^{2
$θ$}₂/2g

Sin^{2
$θ$}₁>sin^{2
$θ$}₂

Sin^{2
$θ$}₁ – sin² $θ$ ₂>0

(Sin $θ$ ₁ – sin $θ$ ₂)( Sin $θ$ ₁ + sin $θ$ ₂)>0

Sin $θ$ ₁>sin $θ$ ₂ or ₁ >₂

T= $\frac{2 u s i n θ}{g} = \frac{2 v_{o} s i n θ}{g}$

T₁= $\frac{2 v_{o} s i n ϑ_{1}}{g}$ , T₂= $\frac{2 v_{o} s i n ϑ_{2}}{g}$

T₁> T₂

R= $\frac{u^{2} s i n 2 θ}{g} = \frac{v_{o}^{2} s i n 2 θ}{g}$

Sin $θ$ ₁>sin $θ$ ₂

Sin2 $θ$ ₁> sin2 $θ$ ₂

$\frac{R_{1}}{R_{2}} = \frac{S i n 2 θ_{1}}{s i n 2 θ_{2}} 1$

R₁>R₂

Total energy for the first particle

U₁=K.E+P.E=1/2m_{1
$v_{o}^{2}$}

U₂= K.E+P.E= 1/2m_{2
$v_{o}^{2}$}

Total energy for the second particle

So m₁= m₂ then U₁=U₂

So m₁>m₂ then U₁>U₂

So m₁2 then U1

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It is found that |A+B|=|A|. this necessarily implies.

(a) B=0

(b) A,B are antiparallel

(c) A,B are perpendicular

(d) A.B<0

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