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The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is $I$. What is the ratio l/R such that the moment of inertia is minimum?

Answer:

Explanation:

MI of a solid cylinder about its perpendicular bisector of length is

                         $I= m(\frac{l^{2}}{12}+\frac{R^{2}}{4})$

$\Rightarrow I=\frac{mR^{2}}{4}+\frac{ml^{2}}{12}=\frac{m^{2}}{4\pi\rho l}+\frac{ml^{2}}{12}$

                                                                                                                  [ $\therefore \rho\pi R^{2}l=m$ ]

 For l to be maximum,

                                   $\frac{\text{d}I}{\text{dl}}=-\frac{m^{2}}{4\pi\rho}(\frac{1}{l^{2}})+\frac{ml}{6}=0$

                        $\Rightarrow \frac{m^{2}}{4\pi\rho}=\frac{ml^{3}}{6}=0$

Now, putting    $m=\rho \pi R^{2}l$

         $\therefore            l^{3}=\frac{3}{2\pi\rho}.\rho\pi R^{2}l$

    $\frac{l^{2}}{R^{2}}=\frac{3}{2}$

                         $\frac{l}{R}=\sqrt{\frac{3}{2}}$

A body of mass m = 10^-2 kg is moving in a medium and experiences a frictional force F= -kv₂ . Its initial is v₀ =10 ms^-1 . If , after 10 s. its energy is  $\frac{1}{8}mv_{0}^{2}$ , the value of k will be

Answer:

Explanation:

Given , force  F= -kv² 

    $\therefore$   Acceleration, a= $\frac{-k}{m}v^{2}$

  or      $\frac{\text{d}v}{\text{d}t}= \frac{-k}{m}v^{2}\Rightarrow \frac{dv}{v^{2}}=-\frac{k}{m}dt$

  Now with limits , we have

         $\int_{10}^{v} \frac{dv}{v^{2}}= -\frac{k}{m}\int_{0}^{t}dt $

$\Rightarrow \left(- \frac{1}{v}\right)_{10}^{v}= -\frac{k}{m}t\Rightarrow \frac{1}{v} = 0.1+\frac{kt}{m}$

$\Rightarrow v= \frac{1}{0.1+\frac{kt}{m}}= \frac{1}{0.1+1000k}$

 $\Rightarrow \frac{1}{2}\times m\times v^{2}=\frac{1}{8}mv_{0}^{2}$

                 $\Rightarrow v=\frac{v_{0}}{2}=5$

$\Rightarrow  \frac{1}{0.1+1000k}=5\Rightarrow 1=0.5+5000k$

$\Rightarrow  k=\frac{0.5}{5000}\Rightarrow k=10^{-4}kg/m$

C_p and C_v are specific heats at constant pressure and constant volume, respectively. It is observed that C_p - C_v =a for hydrogen gas C_p - C_v =b  for nitrogen gas. The  correct relation between a and b is

Answer:

Explanation:

By mayor's relation, for 1 g mole of a gas.

                      C_P -C_V =R

So, when n gram moles are given,

     $  C_{p}-C_{v}= \frac{R}{n}$

  As per given question,

                    $ a=C_{p}-C_{v}= \frac{R}{2}$ , for H₂     ............(i)

                  $ b=C_{p}-C_{v}= \frac{R}{28}$  , for N₂    ............(ii)

From Eqs. (i) and (ii) , we get

              a=14b

Some energy levels of molecule are shown in the figure. The ratio of the wavelengths r= λ₁/λ₂ is given by 

Answer:

Explanation:

We have   $\lambda =\frac{hc}{\triangle E}$

        So ratio of wavelengths

$\frac{\lambda_{1}}{\lambda_{2}}=\frac{hc/\triangle E_{1}}{hc/\triangle E_{2}}=\frac{\triangle E_{2}}{\triangle E_{1}}= \frac{(\frac{4}{3}E-E)}{2E-E}=\frac{1}{3}$

The following observations were taken for determining surface tension T of water by capillary method. Diameter of capillary, d=1.25 × 10^-2 m rise of water, h=1.45 × 10^-2 m. Using g=9.80 m/s² and the simplified relation  $T= \frac{rhg}{2}\times 10^{3} N/m.$  The possible error in surface tension is closest to

Answer:

Explanation:

By ascent formula, we have surface tension.

                               $T= \frac{rhg}{2}\times 10^{3} N/m.$

                                 $= \frac{dhg}{4}\times 10^{3} N/m.$                 ( $\because r=\frac{d}{2}$ )

                           $\Rightarrow \frac{\triangle T}{T}=\frac{\triangle d}{d}+\frac{\triangle h}{h}$  [ given g is constant]

   So ,percentage

               $= \frac{\triangle T}{T}\times 100=(\frac{\triangle d}{d}+\frac{\triangle h}{h})\times 100$

  $=  (\frac{0.01 \times 10^{-2}}{1.25 \times 10^{-2}}+\frac{0.01 \times 10^{-2}}{1.45 \times 10^{-2}})\times 100$=1.5%

$\therefore \frac{\triangle T}{T}\times 100 $= 1.5%

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