MHT CET 2017 :: Physics :: General questions

• Physics - General questions

An ideal gas has pressure 'p'  , volume 'V'  and absolute temperature 'T' .If 'm' is the mass of each molecule  and 'K'  is the Boltzmann constant, then the density of the gas is 

Answer:

Explanation:

For  ideal gas, pV=nRT

 pV= $\frac{m'}{M} RT$ (where, m' is the mass of the gas and M molecular weight)

$p=\frac{m'T}{V}\frac{RT}{M}$

$\therefore$   $p=\frac{\rho RT}{M}$

 where, $\rho$  =$\frac{m'}{V}$ = density of the gas

$\Rightarrow$    $  \rho=\frac{ pM}{RT}=\frac{pM}{NkT}$, where N is Avogadro number

$\Rightarrow$    $  \rho=\frac{ pm}{KT}, $  where    $ m=\frac{M}{N}$= mass of each molecule.

For a rigid diatomic molecule, universal gas constant R= nC_p,  where 'C_p' is the molar specific heat at constant pressure and 'n' is a number. Hence n is  equal to

Answer:

Explanation:

 As we know, for rigid diatomic molecule,

  $\frac{C_{p}}{C_{v}}=\frac{7}{5}$

$\therefore$     ${C_{v}}=\frac{7}{5}C_{p}$     ..........(i)

 also, C_p  -C_v  = R

$\Rightarrow$      ${C_{p}}-\frac{5}{7}C_{p}=R$  [$\because$  drom eq.(i)]

$\Rightarrow$     ${\frac{2}{7}C_{p}}=R$

 $\therefore$      $n={\frac{2}{7}}=0.2857$

When one end of the capillary is dipped in water, the height of the water column is 'h'. The upward force of 105 dyne due to surface tension is balanced by the force due to the weight of water column. The inner circumference  of the capillary is (Surface tension of water = 7 x 10^-2 N/m)

Answer:

Explanation:

 Given, F= 105 dyne= 105 x 10^-5 N

 T= 7 x10^-2  N/m

As we know, circumference of the capillary  x surface tension = upward force

  $2\pi rT=F$

$2\pi r=\frac{F}{T}=\frac{105\times 10^{-5}}{7 \times 10^{-2}}=15 \times 10^{-3}m$

 = 1.5 x 10^-2 m=1.5 cm

  A flywheel at rest is to reach an angular velocity of 24 rad/s in 8 seconds with constant angular acceleration. The total  angle  turned through during this interval is 

Answer:

Explanation:

 Given, $\omega_{0}=0,\omega=24 rad/s$, t= 8 s

 From the relation ,   $\omega=\omega_{0}+\alpha t$

 $\Rightarrow $   $\alpha =\frac{\omega-\omega_{0}}{t}$

Substituting the given values, we get

   $\alpha$  =28/8= 3 rad/s²

 Also, $\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}=0+\frac{1}{2}\times 3 \times (8)^{2}$

 = $\frac{3\times 64}{2}=96$ rad

A lift of mass 'm' is connected to a rope that is moving upward with maximum acceleration 'a' . For maximum safe stress, the elastic limit of the rope is 'T' . The minimum  diameter of the rope is (g= gravitational acceleration)

Answer:

Explanation:

 The maximum tension in the rope =m(g+a)

Stress in the rope ,  $T=\frac{m(g+a)}{\pi r^{2}}$

$\therefore$    $T=\frac{m(g+a)}{\pi r^{2}}=\frac{m(g+a)}{\pi (\frac{d}{2})^{2}}$

 $\Rightarrow$    $ T=\frac{4m(g+a)}{\pi d^{2}}$

$\Rightarrow$    $ d^{2}=\frac{4m(g+a)}{\pi T}$

$\therefore$     d=$\left[\frac{4m(g+a)}{\pi T}\right]^{1/2}$

• Physics - General questions