Physics MCQs for NEET — Practice Questions with Answers

$$\begin{gathered} \mathrm{Particle} \mathrm{velocity}=\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{2\mathrm{\pi av}}{\mathrm{\lambda }}\cos \frac{2\mathrm{\pi }}{\mathrm{\lambda }}\left(\mathrm{vt}-\mathrm{x}\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2\mathrm{\pi a}}{\mathrm{\lambda }}\cos \frac{2\mathrm{\pi }}{\mathrm{\lambda }}\left(\mathrm{vt}-\mathrm{x}\right) \\ \Rightarrow \mathrm{Particle} \mathrm{velocity}=\mathrm{v}\times \frac{2\mathrm{\pi a}}{\mathrm{\lambda }}\cos \frac{2\mathrm{\pi }}{\mathrm{\lambda }}\left(\mathrm{vt}-\mathrm{x}\right)=-\mathrm{v}\frac{\mathrm{dy}}{\mathrm{dx}} \end{gathered}$$

$$\begin{gathered} \left(4\right) \\ \mathrm{Frequency} \mathrm{of} \mathrm{third} \mathrm{overtone} \mathrm{of} \mathrm{closed} \mathrm{pipe}=\frac{7\mathrm{V}}{4{\mathrm{l}}_{\mathrm{c}}} \\ \mathrm{Frequency} \mathrm{of} \mathrm{second} \mathrm{overtone} \mathrm{of} \mathrm{open} \mathrm{pipe}=\frac{3\mathrm{V}}{2{\mathrm{l}}_0} \\ \Rightarrow \frac{7\mathrm{V}}{4{\mathrm{l}}_{\mathrm{c}}} = \frac{3\mathrm{V}}{2{\mathrm{l}}_0} \\ \Rightarrow \frac{{\mathrm{l}}_{\mathrm{c}}}{{\mathrm{l}}_0} = \frac{7}{6} \end{gathered}$$

$$\begin{gathered} Let the frequency of tuning fork=n \\ Then, frequency of sonometer=\frac{1}{2l}\sqrt{\frac{T}{m}} \\ Then, \\ n+5=\frac{1}{2\times 1}\sqrt{\frac{T}{m}} \\ n-5=\frac{1}{2\times 1.05}\sqrt{\frac{T}{m}} \\ \Rightarrow \frac{n+5}{n-5}=\frac{1.05}{1.0}\Rightarrow n=205Hz \end{gathered}$$

$$\begin{gathered} Sound level=10{\log }_{10}\left(\frac{I}{I_0}\right) \\ \Rightarrow 40=10{\log }_{10}\left(\frac{I_1}{I_0}\right) \\ and, 20=10{\log }_{10}\left(\frac{I_2}{I_0}\right) \\ 40-20=10{\log }_{10}\left(\frac{I_1}{I_0}\right)-10{\log }_{10}\left(\frac{I_2}{I_0}\right)=10{\log }_{10}\left(\frac{I_1}{I_2}\right) \\ \frac{I_1}{I_2}=100=\frac{{d_2}^2}{{d_1}^2} \\ {d_2}^2=100m\times 1m \\ {d}_2=10m \end{gathered}$$

(b)Thinking Process

Frequency  in an closed-end tube 

     $\Rightarrow f=\frac{\left(2n-1\right)v}{4l} \left(where, n=1, 2, 3...........\right)$

Also, only odd harmonics exist in a closed-end tube.

Now, given two nearest harmonics are of frequency 220Hz and 260Hz.

So, $\frac{\left(2n-1\right)v}{4l}=220Hz ...\left(i\right)$

Next harmonics occur at,

          $\frac{\left(2n+1\right)v}{4l}=260Hz ...\left(ii\right)$

On subtracting Eq. (i) from Eq (ii). we get 

$\frac{\left\{\left(2n+1\right)-\left(2n-1\right)\right\}v}{4l}=260-220$

$2\left(\frac{v}{4l}\right)=40\Rightarrow \frac{v}{4l}=20Hz$

So, the fundamental frequency of the system=$\frac{v}{4l}=20Hz$

(b) For an open organ pipe

    ${\nu }_n=\frac{n}{21}v, where n=1,2,3...$

For second overtons n=3, $v_{20}=\frac{3}{2L_1}v$

$L_1=$length of open organ pipe

 For closed organ pipe $v_n=\frac{2n+1}{4L}v$

where, n=0, 1,2,3...

lst overtone for closed organ pipe, n=1

$$\begin{gathered} v_{1c}=\frac{3}{4L}v \\ \because v_{2\circ }=v_{1c} \Rightarrow \frac{3v}{2L_1}=\frac{3}{4L}v \\ \Rightarrow L_1=2L \end{gathered}$$

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{L}=75\mathrm{cm}, {\mathrm{f}}_1=420\mathrm{Hz} \mathrm{and} {\mathrm{f}}_2=315\mathrm{Hz} \\ \mathrm{As} \mathrm{two} \mathrm{consecutive} \mathrm{resonant} \mathrm{frequencies} \mathrm{for} \mathrm{a} \mathrm{string} \mathrm{fixed} \mathrm{at} \mathrm{both} \mathrm{ends} \mathrm{will} \mathrm{be}, \\ {\mathrm{f}}_1=\frac{\mathrm{nv}}{2\mathrm{L}} \mathrm{and} {\mathrm{f}}_2=\frac{(\mathrm{n}+1)\mathrm{v}}{2\mathrm{L}} \\ \Rightarrow {\mathrm{f}}_2-{\mathrm{f}}_1=420-315 \\ \Rightarrow \frac{(\mathrm{n}+1)\mathrm{v}}{2\mathrm{L}}-\frac{\mathrm{nv}}{2\mathrm{L}}=105\mathrm{Hz} \\ \Rightarrow \frac{\mathrm{v}}{2\mathrm{L}} =105\mathrm{Hz} \\ \mathrm{Thus}, \mathrm{lowest} \mathrm{resonant} \mathrm{frequency} \mathrm{of} \mathrm{a} \mathrm{string} \mathrm{is} 105\mathrm{Hz}. \end{gathered}$$

When a string is divided into segments, the fundamental frequencies of the segments are inversely proportional to their lengths. The sum of the reciprocals of the fundamental frequencies of the segments is equal to the reciprocal of the original fundamental frequency. Therefore, the correct option is o1: 1/n = 1/n1 + 1/n2 + 1/n3.

For pipe closed at one end 

f_n=n(v/4l)=n(340/4x85x10^-2)=n(100)

Here, n is an odd number ,so for the given condition n can go upto n=11 because n=13 condition will
not be vaild 

n=1,3,5,7,9,11

So, number of possible natural oscillations could be 6.