Physics

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

Let there are n number of loops in the string

length corresponding each loop is $\frac{\lambda }{2}$

So we can write L= $\frac{n\lambda }{2}$

$\lambda =\frac{2L}{n}$

v/ $v$ =2L/n

so frequency $v=\frac{n}{2L}$ v= $\frac{n}{2L}\sqrt[]{\frac{T}{\mu }}$

v $\propto n$

so $v_1:v_2:v_3:v_4=n_1:n_2:n_3:n_4$

= 1:2:3:4

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

consider the frequency of tuning fork= 512Hz

(a) L= $\frac{\lambda }{4}$

$\lambda =4L$

V= $v\lambda$ = 512 $*4*17*{10}^{-2}$

= 348.16m/s

(b) we know that v $\propto \sqrt[]{T}$

$\frac{V_{20}}{V_0}$ = $\sqrt[]{\frac{273+20}{273}}=\sqrt[]{\frac{293}{273}}$

$\frac{V_{20}}{V_0}=\sqrt[]{1.073}=1.03$

V_o= $\frac{V_{20}}{1.03}=\frac{348.16}{1.03}=338m/s$

(c) Resonance will be observed at 17cm length of air column only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface because mercury is more denser than water.

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

Given frequency of the wave v=256Hz

          T= $\frac{1}{v}=\frac{1}{256}=3.9*{10}^{3}s$

(a) time taken to pass through mean position is

t=T/4=1/40 =3.9 $*\frac{{10}^{-3}}{4}s=$ 9.8 $*$ 10^-4s

(b) nodes are A, B, C, D, E (having zero displacement)

antinodes are A' and C' (having maximum displacement)

(c) it is clear from diagram A' and C' are forming antinodes have $\lambda$ separation= v/v'=360/256=1.41m

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

We have to observe the displacement and position of different points, then accordingly nature of two waves decided

Points on positions x= 10,20,30,40 never move, always at mean position with respect to time . these are forming nodes which forms a stationary wave

Distance between two successive nodes = $\frac{\lambda }{2}$

$\lambda =2($ node to node distance)

=2 (20-10)

= 2 (10)=20cm

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

As the source is moving towards the observer hence apparent frequency observed is more than the natural frequency

Frequency of whistle $v=400Hz$

Speed of train = v_t= 10m/s

Velocity of sound in air =v= 330m/s

Apparent frequency when source is moving v_app= (v/v-v_t)v

= ( $\frac{330}{330-10})400$

= $\frac{330}{320}*400=412.5Hz$

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

length of pipe

L=20cm =20 $*{10}^{-2}m$

$v_{close}=\frac{v}{4L}=\frac{330}{4*20*{10}^{-2}}$

$v_{close}=330*\frac{100}{80}=412.5Hz$

$\frac{v_{given}}{v_{closed}}=1237.5/412.5$ =3

Hence 3^rd harmonic node of the pipe is resonantly excited by the source of the given frequency.

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

Given that length of wire l=12m

Mass of the wire m=2.10kg

T= 2.06 $*{10}^{4}N$

Speed of transverse wave v= $\sqrt[]{\frac{T}{\mu }}$ = $\sqrt[]{\frac{2.06*{10}^4}{2.1/12}}=\sqrt[]{\frac{2.06*12*{10}^4}{2.1}}=343m/s$

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

Let n₁>n₂ beat frequency $v_b=n_1-n_2$

As we know time per beats = 1/frequency

Time period beats = T_b= 1/ $v_b=1/n_1-n_2$

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

we know the speed of air  v $\propto \sqrt[]{T}$

$\frac{v_t}{v_o}=\sqrt[]{\frac{T_T}{T_o}}=\sqrt[]{\frac{T_T}{273}}$

$\frac{V_T}{V_o}=\frac{3}{1}$

3/1 = $\sqrt[]{\frac{T_T}{T_o}}=\frac{T_T}{273}=9$

T_T=273 $*9=2457K$

=2457-273=2184 ^oC

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

Frequency of vibrations produced by a stretched wire

$v=\frac{n}{2L}\sqrt[]{\frac{T}{\mu }}$

Mass per unit length = mass/length = $\frac{\pi r^{2}l\rho }{l}=\pi r^2\rho$

$v=\frac{n}{2L}\sqrt[]{\frac{T}{\pi r^2\rho }}=v=\sqrt[]{\frac{1}{r^2}}$

$v\propto \frac{1}{r}$ so when radius is tripled v will be 1/3 rd of previous value.