MHT CET 2020 :: Physics :: General questions

• Physics - General questions

A large open tank containing water has two holes to its wall. A square hole of side a is made at a depth of y and a circular hole of radius r is made at a depth of 16 y from the surface of the water. If an equal amount of water comes out through both the holes per second, then  the relation between r and a will be

Answer:

Explanation:

 Using the principle  of continuity,

              A₁ V₁= A₂ V₂              .............(i)

 where, A₁  is the area of square hole, A₁= a²,

 A₂  is the area of circular  hole,  $A_{2}=\pi r^{2}$,

 v₁ is the velocity at depth y,  $v_{1}=\sqrt{2gy}$

 and v₂  is the velocity  at depth 16y,  $v_{2}=\sqrt{2g(16y)}$

 Substituting values in Eq.(i) , we get

$a^{2}(\sqrt{2gy})=\pi r^{2}(\sqrt{2g(16y)})$

 $a^{2}(\sqrt{2gy})=4\pi r^{2}(\sqrt{2gy})$

$r^{2}=\frac{a^{2}}{4\pi}$

$\therefore$        $ r=\frac{a^{}}{2\sqrt{\pi}}$

The light of wavelength $\lambda$  incident on the surface of metal having work function $\phi$ emits the electrons. The maximum velocity of electrons emitted is 

[c= velocity of light, h= planck's constant, m= mass of electron]

Answer:

Explanation:

 In photoelectric edffect, the maximum kinetic energy possessed  by the particle.

                  $KE_{max}=hv-\phi$

$\frac{1}{2}mv^{2}=\frac{hc}{\lambda}-\phi$           $[\because v=\frac{c}{\lambda}]$

$\Rightarrow $      $ v^{2}=\frac{2(hc-\phi\lambda)}{\lambda m}$

$  v=\left[\frac{2(hc-\phi\lambda)}{\lambda m}\right]^{1/2}$

In a communication system, a repeater is used to extend the range to transmission. It is the combination of 

Answer:

Explanation:

 A repeater is a combination of at least one transmitter, one receiver and one carrier operated relay (COR) configured in a  manner where a signal is received and the transmitter rebroadcasts the received signal on another frequency.

An obstacle is moving towards the source with velocity v. The sound is reflected from the obstacle. If c is the speed  of sound and $\lambda$ is the wavelength, then the wavelength of the reflected wave, $\lambda_{r}$is 

Answer:

Explanation:

 The  frequency of reflected sound wave is 

$f_{r}=f\left(\frac{C+V}{C-V}\right)$

 $\because$ No change  in velocity  occurs due to reflection of sound wave.

Hence,    $\frac{C}{\lambda_{r}}=\frac{C}{\lambda}\left(\frac{C+V}{C-V}\right)\Rightarrow \frac{1}{\lambda_{r}}=\frac{1}{\lambda_{}}\left(\frac{C+V}{C-V}\right)$

$\lambda_{r}=\left(\frac{C-V}{C+V}\right)\lambda$

 A radioactive nucleus emits 4 $\alpha$ -particles and 7 $\beta$-particles in succession. The ratio of number of neutrons  of that of protons is [A= mass number, Z= atomic number]

Answer:

Explanation:

 Let us  assume, a particle X having atomic number Z and mass  number A.

 when an $\alpha $- particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4 . So  , for given case,

$_{Z}X^{A}$   $\underrightarrow{4\alpha-particle}$    $_{Z-8}Y^{A-16}$

 when a  $\beta$- particle is emitted by a nucleus its atomic  number increases by one and mass number remains  unchanged. So, for given case,

$_{Z-8}Y^{A-16}$  $\underrightarrow{7-\beta particle}$  $_{Z-1}Y^{A-16}$

 $\therefore$    Number of neutrons/ number of protons =   $\frac{(A-16)-(Z-1)}{(Z-1)}$

  =  $\frac{A-Z-15}{(Z-1)}$

• Physics - General questions