Waves — Page 5 | JEE Physics

Q41

Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured in SI units. The tension in the string is :

A 10 N

B 7.5 N

C 5 N

D 12.5 N

Correct Answer

Option D

Solution

y = 0.03 sin(450 t $-$ 9x) v =

{\omega \over k} = {{450} \over 9}

= 50m/s v =

\sqrt {{T \over \mu }} \Rightarrow {T \over \mu }

= 2500 $\Rightarrow$ T = 2500 $\times$ 5 $\times$ 10 $-$ 3 = 12.5 N

Q42

For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5 m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in m) of the are :

A 1, 3, 5, .....

B

{1 \over 1},{1 \over 3},{1 \over 5},

.....

C 1, 2, 3, .....

D

{1 \over 2},{1 \over 4},{1 \over 6},

Correct Answer

Option B

Solution

1.5 = \left( {2{n_1} + 1} \right){\lambda \over 2}

......(1) 5 = n2 $\lambda$ .....(2) (1)

\div

(2)

{{1.5} \over 5} = {{\left( {2{n_1} + 1} \right)} \over {2{n_2}}}

$\Rightarrow$ 3n2 = 10n1 + 5 n1 = 1 ; n2 = 5 $\Rightarrow$ $\lambda$ = 1 n1 = 4 ; n2 = 15 $\Rightarrow$ $\lambda$ = 1/3 n1 = 7 ; n2 = 25 $\Rightarrow$ $\lambda$ = 1/5

Q43

A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to :

A 335 ms–1

B 328 ms–1

C 341 ms–1

D 322 ms–1

Correct Answer

Option B

Solution

In first resonance, length of air column

= {\lambda \over 4}

. So,

{l_1} + e = {\lambda \over 4}

or

11 \times 4 + 4e = \lambda

So, speed of sound is

\Rightarrow v = {f_1}\lambda = 512(44 + 4e)

...... (i) And in second case,

l{'_1} + e = {{\lambda '} \over 4}

or

27 \times 4 + 4e = \lambda '

\Rightarrow v = {f_2}\lambda ' = 256(108 + 4e)

..... (ii) Dividing both Eqs. (i) and (ii), we get

1 = {{512(44 + 4e)} \over {256(108 + 4e)}} \Rightarrow e = 5

cm Substituting value of e in Eq. (i), we get Speed of sound

v = 512(44 + 4e)

= 512(44 + 4 \times 5)

= 512 \times 64

cm s $-$ 1 = 327.68 ms $-$ 1 $\approx$ 328 ms $-$ 1

Q44

The pressure wave, P = 0.01 sin [1000t – 3x] Nm–2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0°C. On some other day, when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 ms–1 . Approximate value of T is

A 12°C

B 15°C

C 4°C

D 11°C

Correct Answer

Option C

Solution

Speed of wave from wave equation

v = - {{\left( {coeffecient{\rm{ }}of{\rm{ }}t} \right)} \over {\left( {coeffecient{\rm{ }}of{\rm{ }}x} \right){\rm{ }}}}

v = - {{1000} \over {( - 3)}} = {{1000} \over 3}

Since speed of wave

\propto \sqrt T

So

= {{1000} \over {{3 \over {336}}}} = \sqrt {{{273} \over T}}

$\Rightarrow$ T = 277.41 K T = 4.41°C

Q45

A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from 440 Hz to 480 Hz, when it gets reflected from the wall. If the speed of sound in air is 345 m/s, then the speed of the car is :

A 36 km/hr

B 54 km/hr

C 24 km/hr

D 18 km/hr

Correct Answer

Option B

Solution

f1 = frequency heard by wall =

\left( {{{{v_s} - 0} \over {{v_s} - {v_c}}}} \right){f_0}

=

{{{{v_s}} \over {{v_s} - {v_c}}} \times 440}

f2 = frequency heard by driver after reflection from wall =

\left( {{{{v_s} + {v_c}} \over {{v_s} - 0}}} \right){f_1}

$\Rightarrow$ 480 =

\left( {{{{v_s} + {v_c}} \over {{v_s}}}} \right){f_1}

$\Rightarrow$ 480 =

\left( {{{{v_s} + {v_c}} \over {{v_s}}}} \right)\left( {{{{v_s}} \over {{v_s} - {v_c}}}} \right) \times 440

$\Rightarrow$ 480 =

\left( {{{{v_s} + {v_c}} \over {{v_s} - {v_c}}}} \right) \times 440

$\Rightarrow$ 480 =

\left( {{{345 + {v_c}} \over {345 - {v_c}}}} \right) \times 440

$\Rightarrow$ 12 =

\left( {{{345 + {v_c}} \over {345 - {v_c}}}} \right) \times 11

$\Rightarrow$ 11 × 345 + 11vc = 12 × 345 – 12vc $\Rightarrow$ vc =

{{345} \over {23}}

m/s =

{{345} \over {23}} \times {{18} \over 5}

= 54 km/hr

Q46

In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air is 330 m/s, the tuning fork frequency is :

A 2200 Hz

B 3300 Hz

C 1100 Hz

D 550 Hz

Correct Answer

Option A

Solution

{l_1} = l - 17

{l_2} = l - 24.5

We know,

v = 2f({l_1} - {l_2})

$\Rightarrow$

330 = 2 \times f \times [(f \times [(l - 17) - (l - 24.5)] \times 10^{ - 2}

$\Rightarrow$

165 = f \times 7.5 \times {10^{ - 2}}

$\Rightarrow$

f = {{165 \times 1000} \over {7.5}}

$\Rightarrow$

f = 2200\,Hz

Q47

The driver of a bus approaching a big wall notices that the frequency of his bus's horn changes from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is 330 ms–1.

A 91 kmh–1

B 81 kmh–1

C 61 kmh–1

D 71 kmh–1

Correct Answer

Option A

Solution

Frequency received by wall,

{f_w} = \left( {{{330} \over {330 - v}}} \right){f_0}

Frequency after reflection,

f' = \left( {{{330 + v} \over {330}}} \right){f_w}

= \left( {{{330 + v} \over {330}}} \right) \times \left( {{{330} \over {330 - v}}} \right){f_0}

$\Rightarrow$

490 = \left( {{{330 + v} \over {330 - v}}} \right)420

$\therefore$ v = 25.2 m/s = 91 km/h

Q48

Two identical strings X and Z made of same material have tension TX and TZ in them. If their fundamental frequencies are 450 Hz and 300 Hz, respectively, then the ratio TX/TZ is

A 2.25

B 0.44

C 1.25

D 1.5

Correct Answer

Option A

Solution

f =

{1 \over {2l}}\sqrt {{T \over \mu }}

For identical string

l

and $\mu$ will be same f $\propto$

\sqrt T

$\therefore$

{{450} \over {300}} = \sqrt {{{{T_x}} \over {{T_y}}}}

$\Rightarrow$

{{{{T_x}} \over {{T_y}}} = {9 \over 4}}

= 2.25

Q49

A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 × 104 N. When the tension is changed to T, the velocity changed to v/2. The value of T is close to :

A 30.5 × 104 N

B 2.50 × 104 N

C 10.2 × 102 N

D 5.15 × 103 N

Correct Answer

Option D

Solution

v = \sqrt {{T \over \mu }}

$\therefore$

{{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}}

v1 = v, v2 =

{v \over 2}

$\Rightarrow$

{v \over {{v \over 2}}} =

\sqrt {{{2.06 \times {{10}^4}} \over {{T_2}}}}

$\Rightarrow$ T2 =

{{{2.06 \times {{10}^4}} \over 4}}

= 5.15 × 103 N

Q50

Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm2) is 90 ms-1. If the Young's modulus of wire is 16

\times

1011 Nm-2, the extension of wire over its natural length is :

A 0.03 mm

B 0.04 mm

C 0.02 mm

D 0.01 mm

Correct Answer

Option A

Solution

Velocity of the wave, v =

\sqrt {{T \over \mu }}

$\Rightarrow$ T = v2 $\mu$ We know, Youngs modulus, Y =

{{{F \over A}} \over {{{\Delta l} \over l}}}

=

{{{T \over A}} \over {{{\Delta l} \over l}}}

[As here F = T] $\Rightarrow$

{Y{{\Delta l} \over l}}

=

{{T \over A}}

=

{{{{v^2}\mu } \over A}}

$\Rightarrow$

\Delta

l =

{{{{v^2}\mu l} \over {AY}}}

=

{{90 \times 90 \times {{60 \times {{10}^{ - 3}}} \over {60 \times {{10}^{ - 2}}}} \times 60 \times {{10}^{ - 2}}} \over {1 \times {{10}^{ - 6}} \times 16 \times {{10}^{11}}}}

= 3 $\times$ 10-5 m = 0.03 mm