MHT CET 2016 :: Physics :: General questions

• Physics - General questions

When open pipe is closed from one end, then third overtone of the closed pipe is higher in frequency by 150 Hz than the second overtone of open pipe. The fundamental frequency of open end pipe will be

Answer:

Explanation:

According to question , fundamental

frequency of organ pipe is given by    $f_{0}=\frac{V_{0}}{2l}$

   [ Symbols have their usual meanings]

 In 2 nd overtone,   $\frac{3}{2}f_{0}=\frac{3V_{0}}{2l}$

 Let   $f=\frac{V_{0}}{4l}$ be the fundamental frequency of closed organ pipe. Third overtone of closed organ pipe at one end is

$f_{3}=\frac{7}{4}\frac{V_{0}}{l}$

 As given,    $\frac{7}{4}\frac{V_{0}}{l}-\frac{3 V_{0}}{2l}=150$

 $\Rightarrow$      $\left(\frac{7}{4}-\frac{6}{4}\right)\frac{V_{0}}{l}=150$

 $\Rightarrow$      $\frac{V_{0}}{4l}=150\Rightarrow \frac{V_{0}}{2l}=300 Hz$

The bob of a simple pendulum  performs SHM with period  T in air and with period T₁ in water, Relation  between T and T₁ is (neglect friction due to water , density of the material  of the bob  is = $\frac{9}{8}\times 10^{3}kg m^{3}$ , density of water =1 gcc^-1)

Answer:

Explanation:

 Time period  of simple pendulum in water is given by

                   $T=2\pi\sqrt{\frac{l}{g_{eff}}}$

[ Symbols have their  usual meanings and   $g_{eff}$ = acceleration due to gravity in water]

 As we know, $g_{eff}$$=g\left(\frac{\sigma-\rho}{\sigma}\right)$

 [ where, $\sigma$ = density of bob.$\rho$ = density of water]

 $=9.8\left(\frac{\frac{9}{8}\times10^{3}-10^{3}}{\frac{9}{8}\times 10^{3}}\right)$

   $=9.8\left(\frac{\frac{9}{8}-1}{\frac{9}{8}}\right)=9.8\left(\frac{1}{8} \times \frac{8}{9}\right)$

 =$\frac{9.8}{9}$

  So, $T_{1}=2\pi \sqrt{\frac{l \times 9}{9.8}}\Rightarrow T_{1}=3T$

 [$\because $  Time period of simple pendulum in air, 

                $T_{}=2\pi \sqrt{\frac{l }{9.8}}$]

Resolving power of telescope increases when

Answer:

Explanation:

 Resolving power of a telescope is given by

 RP= $\frac{d}{1.22 \lambda}$

i.e,   $RP\propto \frac{1}{\lambda}$

 Where $\lambda$ (wavelength of light) decreases, resolving power of a telescope increases

The amount of work done in increasing the voltage  across the plates of the capacitor from 5 V to 10 V is W. The work done in increasing it from 10 V to 15 V will be 

Answer:

Explanation:

 According to question , work done in increasing the voltage from 5 V to 10 V  is given by

$=\frac{1}{2}C(V_{2}^{2}-V_{1}^{2})$

$=\frac{1}{2}\times C\times(10^{2}-5^{2})$

 W$=\frac{1}{2}\times C\times75=\frac{75 C}{2}$  ..........(i)

 Again work done in increasing the plate voltage from 10 V to 15 V is given by

$W_{1}=\frac{1}{2}\times C\times(15^{2}-10^{2})$

 $\Rightarrow $   $W_{1}=\frac{1}{2}\times C\times125=\frac{125 C}{2}$  ......(ii)

 From Eqs.(i) and (ii) , we get

$\frac{W}{W_{1}}=\frac{75 C}{2 \times 125 C}\times2$

$\Rightarrow$     $\frac{W_{1}}{W_{}}=\frac{125}{75}=\frac{5}{3}$

$\Rightarrow$    $W_{1}$  =1.67 W

The value of gravititational acceleration g at a height h above the earth's surface is $\frac{g}{4}$ , then (R=radius of earth)

Answer:

Explanation:

 The value of acceleration due to gravity g at a height h is given by

 $g'= g\left( \frac{R_{e}}{R_{e}+h}\right)^{2}$

[ Symbols have their usual meanings]

when,

 g'= $\frac{g}{4}$   and $R_{e}$= R

 So, $\frac{g}{4}= g\left( \frac{R_{}}{R_{}+h}\right)^{2}$

 $\frac{1}{2}= \frac{R_{}}{R_{}+h}$

 $\Rightarrow$     2R= R+h

  $\Rightarrow$     R= h

• Physics - General questions