MHT CET 2016 :: Physics :: General questions

• Physics - General questions

A simple pendulum of length l has maximum angular  displacement  $\theta$ . The maximum kinetic energy of the bob of mass m is (g= acceleration  due to gravity)

Answer:

Explanation:

According to question , when bob is at rest, the pendulum has only potential energy

 (PE)=mgl

When bob displaced by smaller  angular  displacement $\theta$ , the pendulum loses PE of bob and it gets converted into kinetic energy (KE).

So, from conservation of energy 

Loss in PE = gain in KE

$\therefore$     KE= mgl-mgl cos $\theta$ = $mgl(1-\cos \theta)$

When the observer moves towards the stationary source with velocity, v₁ the apparent frequency of emitted note is f₁. When the observer moves away from the velocity v₁, the apparent frequency is f₂ . if v is the velocity of sound in air and 

$\frac{f_{1}}{f_{2}}=2, then \frac{v}{v_{1}}=$  ?

Answer:

Explanation:

 According to question ,

 $\frac{f_{1}}{f_{2}}$=2   (given)

[ $f_{1}$ =  apparent frequency when velocity $v_{1}$ is towards the observer.

  $f_{2}$  = apparent frequency when velocity $v_{1}$ is away from the observer]

Now, the apparent frequency of sound  when the observer move towards the source is given by

   $f_{1}=\left(\frac{V}{V-V_{1}}\right)f_{0}$ ............(i)

 [ Symbols have their usual meanings]

 Similarly when observer moves away from the source, apparent frequency is given by

  $f_{2}=\left(\frac{V}{V+V_{1}}\right)f_{0}$  .........(ii)

 From Eqs.(i) and (ii), we get

$\frac{f_{1}}{f_{2}}=\frac{\left(\frac{V}{V-V_{1}}\right)f_{0}}{\left(\frac{V}{V+V_{1}}\right)f_{0}}=\frac{V+V_{1}}{V-V_{1}}$

 $\Rightarrow$      $=\frac{V+V_{1}}{V-V_{1}}=2\Rightarrow2V-2V_{1}=V+V_{1}$

$\Rightarrow$      $V=3V_{1}\Rightarrow \frac{V}{V_{1}}=3$

Two wires having  same length and material are stretched by same force. Their diameters are in the ratio 1:3, the ratio of strain  energy per unit volume for these two wires (smaller to larger diameter) when stretched is 

Answer:

Explanation:

According to question , energy in string per unit volume is given by $\frac{1}{2}$ x stress x strain

$\frac{U}{V}=\frac{1}{2}\times\frac{F}{A}\times\frac{\triangle l}{l}$

[ Symbols have their  usual meanings]

= $\frac{1}{2} \times V \times (strain)^{2}$

$\frac{U_{s}}{S_{L}}=\frac{Y_{s}}{Y_{L}}\times\left(\frac{stress(S)}{Strain(L)}\right)^{2}$

= $\frac{A_{L}}{A_{s}}=\left(\frac{l}{l}\right)^{2}$

 $\Rightarrow$     $\frac{r_{L}^{2}}{r_{s}^{2}}=\frac{3^2}{l^{2}}=9:1$

For a gas $\frac{R}{C_{v}}=0.4$  , where R is the universal gas constant and C_v is molar specific heat at constant volume. The gas is made up of  molecule which are

Answer:

Explanation:

According to question,

    $\frac{R}{C_{v}}=0.4$   

                   [R= universal gas constant and

C_v  = molar specific heat  at constant volume]

As we know, C_p-C_v  = R

So, $\frac{C_{p}-C_{v}}{C_{v}}=0.4\Rightarrow \frac{C_{p}}{C_{v}}=0.4+1=1.4$

 i.e, Y=1.4

 As Y=1.4, the gas is diatomic  in nature

A black rectangular surface of area A emits energy E  per second at 27° C . If length and breadth  are reduced to $\frac{1}{3}$^rd of initial value and temperature is raised  to 327° C . then energy emitted per second becomes

Answer:

Explanation:

According to question ,

   $E= e \sigma A(T^{4}-T_{0}^{4})$

  [ where symbols have their usual meanings]

 when l and b change to $\frac{l}{3}$ and $\frac{b}{3}$ , respectively

 Area becomes $\frac {l}{3} \times \frac{b}{3}=\frac{l b}{3}=\frac {A}{3}$      ($\because$ A=lb)

 Now for two different cases

$\frac{E'}{E}=\frac{A'(227+373)^{4}}{(27+273)^{4}}=\frac{1}{9}(\frac{600}{300})^{4}$

$\therefore$    E'=$\frac{1}{9} \times (2)^{4} \times  E  \Rightarrow  E'= \frac {16 E}{9}$

• Physics - General questions