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4 months ago

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Payal Gupta

Contributor-Level 10

(a) The element V has the highest first ionization enthalpy (? iH1) and positive electron gain enthalpy (? egH) and hence it is the least reactive element. Since inert  gases have positive? egH, therefore, the element-V must be an inert gas. The values of? iH1,  ? iH2 and? egH match that of He.

 

(b) The element II which has the least first ionization enthalpy (? iH1) and a low negative electron gain enthalpy (? egH) is the most reactive metal. The values of ? iH1,  ? iH2 and? egH match that of K (potassium).
               

(c) The element III which has high fi

...more

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, d, e) a) clearly every particle at x will have amplitude =asinkx=fixed

b) for mean position =0

coswt=0

wt= (2n-1) π 2

hence for a fixed of n all particles are having same value of time t= (2n-1) π 2 w

c) amplitude of all the particles are asinkx which is different for different particles at different values of x

d) the energy is a stationary wave is confined between two nodes.

e) particles at different nodes are always at rest.

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) vo=400Hz, v=340m/s

Vm=10m/s

(a) as both source and observer are stationary, hence frequency observed will be same as natural frequency vo=400Hz

(b) the speed of sound v=v+vw= 340+10=350m/s

(c) there will be on effect on frequency because there is no relative motion between source and observer hence c, d are incorrect.

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, d) As we know by y (x, t) = 0.06 sin (2πx/3) cos (120πt)

By comparing the equation with general equation N denotes nodes and A denotes antinodes.

(a) Clearly frequency is common for all the points

(b) Consider all the particles between two nodes they are having same phase at given time

(c) But are having different amplitude of 0.06sin (2 π 3 x ) and because of different amplitudes they are having different energies.

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b, c, d)

during propagation of a plane progressive mechanical wave

(i) clearly the particles O, A and B are having different phase.

(ii) Particles of the wave are having up and SHM.

(iii) For a progressive wave propagating in a fluid

V= B ρ , V= 1 ρ

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c, d) Speed of sound waves in a fluid is given by

V= B ρ

V= 1 ρ , V B so when we increase velocity bulk modulus also increases.

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b, c) y (x, t) = 0.06 sin (2πx/3) cos (120πt)

(a) y (x, t)=asinkxcoswt

(b) w=120 π , f=60hz

(c) k=2 π 3 = 2 π λ , λ = 3 m , f = 60 h z , v =60 (3)=180m/s

(d) since in stationary wave all particles of the medium executes SHM with varying amplitude nodes.

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) y (x, t)=30sin (36t+0.018x+ π 4 )

Y=asin (wt+kx+ )

(a) as the equation involves positive sign with x . hence the wave is travelling from right to left. Hence option a is correct.

(b) W=36

2 π v = 36

v = 36 2 π = 18 π

k = 0.018 = 2 π λ

2 π v v λ = 0.018

w/v=0.018

v=2000cm/s=20m/s

(c) 2 π v = 36

v = 18 π = 5.7Hz

(d) 2 π λ = 0.018 , λ = 2 π /0.0018cm=3.48cm

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) Let frequency of the source is no

Let the speed of sound wave in the medium is v

As observer is stationary

Apparent frequency na= v v - v s n o

v v - v s n o = n a n o

When the train is going away from the observer

Apparent frequency na= v v - v s n o = n a < n o

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) m=2.5kg

μ = mass per unit length

=m/l =2.5/20=125/10=0.125kg/m

V= T μ = 200 0.125

L=v * t

20= 125 2 * 10 5 = 20 * 25 * 5 2 * 10 5

= 20 * 25 0.4 * 10 5 = 1 2

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