**Q1.**The wavelength of two notes in air are 36/195 m and 36/193m. each note produces 10 beats per second separately with a third note of fixed frequency. The velocity of sound in air in m/s is

Solution

Beat frequency=v_1~v_2 Let the frequency of third note be n. Then, 195v/36-v=10 …(i) And v-193v/36=10 …(ii) Adding Eqs. (i) and (ii) v/18=20 ⟹ v=360 ms^(-1)

Beat frequency=v_1~v_2 Let the frequency of third note be n. Then, 195v/36-v=10 …(i) And v-193v/36=10 …(ii) Adding Eqs. (i) and (ii) v/18=20 ⟹ v=360 ms^(-1)

**Q2.**Law of superposition is applicable to only

Solution

B

B

**Q3.**A car moving with a velocity of 36 km^(-1) crosses a siren of frequency 500 Hz. The apparent frequency of siren after passing it will be

Solution

Given that, velocity of the car =36000m/3600s=10ms^(-1) v_emitted=500 Hz v_sound=330 ms^(-1) We know that =v_emitted×((v_sound-v_obs ))/c_sound Hence, observed frequency v_obs=500×(330-10)/330 v_obs=485Hz

Given that, velocity of the car =36000m/3600s=10ms^(-1) v_emitted=500 Hz v_sound=330 ms^(-1) We know that =v_emitted×((v_sound-v_obs ))/c_sound Hence, observed frequency v_obs=500×(330-10)/330 v_obs=485Hz

**Q4.**Who wrote the famous Book “Constitution of India - Defaced and Defied”?

Solution

Minimum audible frequency =20 Hz ⇒v/4l=20⇒l=336/(4×20)=4.2 m

Minimum audible frequency =20 Hz ⇒v/4l=20⇒l=336/(4×20)=4.2 m

**Q5.**A whistle revolves in a circle with an angular speed of 20 rad/sec using a string of length 50 cm. If the frequency of sound from the whistle is 385 Hz, then what is the minimum frequency heard by an observer, which is far away from the centre in the same plane? (v=340 m/s)

Solution

Minimum frequency will be heard, when whistle moves away from the listener. n_min=n(v/(v+v_s )) where v=rÏ‰=0.5×20=10 m/s ⇒n_min=385(340/(340+10))=374Hz

Minimum frequency will be heard, when whistle moves away from the listener. n_min=n(v/(v+v_s )) where v=rÏ‰=0.5×20=10 m/s ⇒n_min=385(340/(340+10))=374Hz

**Q6.**The velocity of sound I air is 330ms^(-1). The rms velocity of air molecules (Î³=1.4) is approximately equal to

Solution

C_(rms=v) √(3/Î³)=330×√(3/1.4)=471.4 ms^(-1)

C_(rms=v) √(3/Î³)=330×√(3/1.4)=471.4 ms^(-1)

**Q7.**For the stationary wave y=4 sin〖(Ï€x/15) cos〖(96Ï€t)〗 〗, the distance between a node and the next antinode is

Solution

Comparing given equation with standard equation y=2a sin〖2Ï€x/Î»〗 cos〖2Ï€vt/Î»〗 gives us 2Ï€/Î»=Ï€/15⇒Î»=30 Distance between nearest node and antinodes =Î»/4=30/4=7.5

Comparing given equation with standard equation y=2a sin〖2Ï€x/Î»〗 cos〖2Ï€vt/Î»〗 gives us 2Ï€/Î»=Ï€/15⇒Î»=30 Distance between nearest node and antinodes =Î»/4=30/4=7.5

**Q8.**A source is moving towards an observer with a speed of 20 m/s and having frequency of 240 Hz. The observer is now moving towards the source with a speed of 20 m/s. Apparent frequency herad by observer, if velocity of sound is 340 m/s, is

Solution

n^'=n((v+v_O)/(v-v_S ))=240((340+20)/(340-20))=270 Hz

n^'=n((v+v_O)/(v-v_S ))=240((340+20)/(340-20))=270 Hz

**Q9.**A tuning fork of frequency 340 Hz is vibrated just above the tube of 120 cm height. Water is poured slowly in the tube, what is the minimum height of water necessary for the resonance?

Solution

Using relation v=vÎ» Or Î»=v/v=340/340-1m If length of resonance columns are l_1,l_2 and l_3, then l_1=Î»/4=1/4 m=25 cm (for first resonance) l_2=3Î»/4=75cm (for second resonance) l_3=5Î»/4=125 cm for third resonance This case of third resonance is impossible because total length of the tube is 120 cm So, minimum height of water =120-75=45cm

Using relation v=vÎ» Or Î»=v/v=340/340-1m If length of resonance columns are l_1,l_2 and l_3, then l_1=Î»/4=1/4 m=25 cm (for first resonance) l_2=3Î»/4=75cm (for second resonance) l_3=5Î»/4=125 cm for third resonance This case of third resonance is impossible because total length of the tube is 120 cm So, minimum height of water =120-75=45cm

**Q10.**Two waves are represented by y_1=4 sin404Ï€t and y_2=3 sin400Ï€t. Then

Solution

Given ∶y_1=4 sin〖404Ï€t,y_2=3 sin400Ï€t 〗 ∴Ï‰_1=404Ï€,Ï‰_2=400Ï€,A_1=4,A_2=3 Ï‰_1=2Ï€v_1⇒404Ï€=2Ï€v_1⇒v_1=202 Hz Ï‰_2=2Ï€v_2⇒400Ï€=2Ï€v_2⇒v_2=200 Hz Beat frequency =v_1-v_2=202-200=2 Hz I_max/I_min =((A_1+A_2)/(A_1-A_2 ))^2=((4+3)/(4-3))^2=(7/1)^2=49/1

Given ∶y_1=4 sin〖404Ï€t,y_2=3 sin400Ï€t 〗 ∴Ï‰_1=404Ï€,Ï‰_2=400Ï€,A_1=4,A_2=3 Ï‰_1=2Ï€v_1⇒404Ï€=2Ï€v_1⇒v_1=202 Hz Ï‰_2=2Ï€v_2⇒400Ï€=2Ï€v_2⇒v_2=200 Hz Beat frequency =v_1-v_2=202-200=2 Hz I_max/I_min =((A_1+A_2)/(A_1-A_2 ))^2=((4+3)/(4-3))^2=(7/1)^2=49/1