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*Simulator*

**Previous Years JEE Advanced Questions**

(A) 8.50 kHz (B) 8.25 kHz (C) 7.75 kHz (D) 7.50 kHz

**[JEE 2011]**

**Sol.**(A)

$f^{\prime}=\left(\frac{v}{v-v_{s}}\right)\left(\frac{v+v_{0}}{v}\right) f \Rightarrow f^{\prime}=\left(\frac{320}{320-10}\right)\left(\frac{320+10}{320}\right) \times 8 \Rightarrow f^{\prime} \approx 8.50 \mathrm{kHz}$

**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 $\lambda_{\mathrm{f}}$. Match each system with statements given in

**Column-II**describing the nature and wavelength of the standing waves.

**[JEE 2011]**

**Sol.**((A) $\mathrm{P}, \mathrm{T}(\mathrm{B}) \mathrm{P}, \mathrm{S}(\mathrm{C}) \mathrm{Q}, \mathrm{S}(\mathrm{D}) \mathrm{Q}, \mathrm{R}$)

(A) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open

(B) a low -pressure pulse starts travelling up the pipe, if the other end of the pipe is open

(C) a low pressure pulse starts travelling up the pipe, if the other end of the pipe is closed

(D) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed

**[JEE 2012]**

**Sol.**($\mathrm{B}, \mathrm{D}$)

Information Based (Refer NCERT)

(A) 14.0 cm (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm

**[JEE 2012]**

**Sol.**(B)

In resonance column experiment $\frac{\lambda}{4}=\ell_{1}+e$ so, $\ell_{1}=\frac{\lambda}{4}-e=\frac{v}{4 f}-0.3 d$

$=\frac{336 \times 100 \mathrm{cm}}{4 \times 512}-(0.3)(4 \mathrm{cm})=16.4-1.2=15.2 \mathrm{cm}$

$244 \mathrm{s}^{-1}$ . He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 $\pm 0.005)$ m, the gas in the tube is

(Useful information $: \sqrt{167 \mathrm{RT}}=640 \mathrm{J}^{1 / 2}$ mole $^{-1 / 2} ; \sqrt{140 \mathrm{RT}}=590 \mathrm{J}^{1 / 2}$ mole $^{-1 / 2} .$ The molar masses M in grams are given in the options. Take the values of $\sqrt{\frac{10}{\mathrm{M}}}$ for each gas as given there. )

(A) Neon $\left(\mathrm{M}=20, \sqrt{\frac{10}{20}}=\frac{7}{10}\right)$

(B) Nitrogen $\left(\mathrm{M}=28, \sqrt{\frac{10}{28}}=\frac{3}{5}\right)$

(C) Oxygen $\left(\mathrm{M}=32, \sqrt{\frac{10}{32}}=\frac{9}{16}\right)$

(D) Argon $\left(\mathrm{M}=36, \sqrt{\frac{10}{36}}=\frac{17}{32}\right)$

**[JEE Advanced-2014]**

**Sol.**(D)

(A) The plot below represents schematically the variation of beat frequency with time

(B) The plot below represents schematically the variations of beat frequency with time

(C) The rate of change in beat frequency is maximum when the car passes through Q

(D) $v_{\mathrm{p}}+v_{\mathrm{R}}=2 v_{\mathrm{Q}}$

**[JEE Advanced 2016]**

**Sol.**(A,C,D)

**[JEE Advanced 2017]**

**Sol.**6

Frequency of sound as received by large car approaching the source.

**[JEE Advanced 2018]**

**Sol.**5.00 Hz

(A) The speed of sound determined from this experiment is $332 \mathrm{ms}^{-1}$

(B) The end correction in this experiment is 0.9 cm

(C) The wavelength of the sound wave is 66.4 cm

(D) The resonance at 50.7 cm corresponds to the fundamental harmonic

**[JEE Advanced 2018]**

**Sol.**(A,C)

Let $\mathrm{n}_{1}$ harmonic is corresponding to 50.7 cm & $\mathrm{n}_{2}$ harmonic is corresponding 83.9 cm. since both one consecutive harmonics.

$\therefore$ their difference $=\frac{\lambda}{2}$

$\therefore \frac{\lambda}{2}=(83.9-50.7) \mathrm{cm}$

$\frac{\lambda}{2}=33.2 \mathrm{cm}$

$\lambda=66.4 \mathrm{cm}$

$\therefore \frac{\lambda}{4}=16.6 \mathrm{cm}$

length corresponding to fundamental mode must be close to \frac{\lambda}{4} \& 50.7 cm must be an odd multiple of this length 16.6 × 3 = 49.8 cm. therefore 50.7 is 3^{| \mathrm{rd}}harmonic

If end correction is e, then

\mathrm{e}+50.7=\frac{3 \lambda}{4}

\mathrm{e}=49.8-50.7=-0.9 \mathrm{cm}

\text { speed of sound, } \mathrm{v}=\mathrm{f} \lambda

\therefore \mathrm{v}=500 \times 66.4 \mathrm{cm} / \mathrm{sec}=332.000 \mathrm{m} / \mathrm{s}