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IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

Column I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as $$\lambda$$f. Match each system with statements given in Column II describing the nature and wavelength of the standing waves :

A
(A)$$\to$$(T); (B)$$\to$$(P), (S); (C)$$\to$$(Q), (S); (D)$$\to$$(Q)
B
(A)$$\to$$(P), (T); (B)$$\to$$(P); (C)$$\to$$(Q), (S); (D)$$\to$$(Q)
C
(A)$$\to$$(P); (B)$$\to$$(P), (S); (C)$$\to$$(Q); (D)$$\to$$(Q), (R)
D
(A)$$\to$$(P), (T); (B)$$\to$$(P), (S); (C)$$\to$$(Q), (S); (D)$$\to$$(Q), (R)

Explanation

(1) For a pipe closed at one end, we have

$${{{\lambda _f}} \over 4} = L$$ or $${\lambda _f} = 4L$$

Therefore, the sound waves are longitudinal.

(2) For a pipe open at both ends, we have

$${{{\lambda _f}} \over 2} = L$$ or $${\lambda _f} = 2L$$

Therefore, the sound waves are longitudinal.

(3) For a stretched wire clamped at both ends, we have

$${{{\lambda _f}} \over 2} = L$$

Vibration on the string is transverse.

(4) For a stretched wire clamped at both ends and at mid-point, we have

$${{{\lambda _f}} \over 2} = {L \over 2} \Rightarrow {\lambda _f} = L$$

Therefore, the vibration on the string is transverse.

2

IIT-JEE 2011 Paper 1 Offline

MCQ (Single Correct Answer)
A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is
A
8.50 kHz
B
8.25 kHz
C
7.75 kHz
D
7.50 kHz

Explanation

u = 36 km h$$-$$1 = 10 ms$$-$$1,

$$v$$ = 320 ms$$-$$1, $$\nu $$ = 8 kHz

The sound reflected from the building may be imagined to be coming from the mirror image. The driver is approaching the image-source which is also approaching him with the same speed. Hence the frequency of sound heard by the driver is

$$\nu $$' = $$\nu $$ $$\left( {{{v + u} \over {v - u}}} \right)$$

= 8 kHz $$\times$$ $$\left( {{{320 + 10} \over {320 - 10}}} \right)$$ = 8.5 kHz

3

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)
A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms−1, the mass of the string is
A
5 grams
B
10 grams
C
20 grams
D
40 grams

Explanation

The fundamental mode in a pipe closed at one end and the second harmonic in a string are shown in the figure.

Fundamental frequency of a hollow pipe closed at one end is

$${\upsilon _p} = {v \over {4L}}$$

where, v = speed of sound, L = length of a pipe

Frequency of second harmonic of a string is

$${\upsilon _s} = {2 \over {2l}}\sqrt {{T \over \mu }} = {1 \over l}\sqrt {{T \over \mu }} $$

where, T = tension of the string, m = mass per unit length of the string, l = length of the string

$${\upsilon _s} = {1 \over l}\sqrt {{T \over {{m \over l}}}} = {1 \over l}\sqrt {{{Tl} \over m}} = \sqrt {{T \over {ml}}} $$ ...... (i)

where m is the mass of the string

According to given problem, $${\upsilon _p} = {\upsilon _s}$$

$$\therefore$$ $${v \over {4L}} = \sqrt {{T \over {ml}}} $$ or $$m = {{16{L^2}T} \over {{v^2}l}}$$

Substituting the given values, we get

$$m = {{16 \times {{(0.8)}^2} \times (50)} \over {{{(320)}^2} \times 0.5}} = 0.01$$ kg = 10 gram

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