JEE 2014 :: Main 2014 :: Physics 2014

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 A particle moves with simple harmonic motion in a straight line, In first   $\tau$   sec. after starting from rest it travels a distance a and in next $\tau$   sec. it travels 2a, in same direction, then 

Answer:

Explanation:

 In   SHM , a particle  starts from rest, we have

i.e,   $x=A\cos \omega t,$     at     t=0,x=A

when $ t=\tau$     then x=A-a   .......(i)

when  $t=2\tau$   , then x=A-3a     .......(ii)

 On comparing Eqs. (i)  and (ii), we get

 $A-a=A\cos \omega \tau$

    $A-3a=A\cos 2\omega \tau$

 As   $\cos2 \omega t=2\cos^{2}\tau-1$

$\Rightarrow$           $\frac{A-3a}{A}=2\left(\frac{A-a}{A}\right)^{2}-1$

$\Rightarrow$           $\frac{A-3a}{A}=\frac{2A^{2}+2a^{2}-4Aa-A^{2}}{A^{2}}$

  $A^{2}-3aA=A^{2}+2a^{2}-4Aa$

$a^{2}=2aA$

 A=2a

  Now,   $A-a=A\cos\omega\tau$

$\Rightarrow$      $\cos\omega\tau=1/2$

$\frac{2\pi}{T}\tau=\frac{\pi}{3}$

   $\Rightarrow $          $ T=6\pi$

An open glass tube is immersed in mercury in such away that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional  46 cm. What will be length  of the air column  above the mercury  in the tube now  ? (Atmospheric pressure =76 cm of Hg)

Answer:

Explanation:

 In this question, the system is accelerating horizontally i.e, no component of acceleration in vertical direction. Hence, the pressure in the vertical direction will remain  unaffected

   i.e,    p₁  =p₀+ρgh

 Again, we have to use the concept that the pressure in the same level will be same.

For air trapped in tube, p₁V₁=p₂V₂

 $p_{1}=p_{atm}=\rho g76$

  V₁= A.8

  [A= area of cross-section]

$p_{2}=p_{atm}-\rho g(54-x)$

 $=\rho g(22+x)$

  V₂ =A.x

$\rho g 76\times 8A=\rho g(22+x)Ax$

    $x^{2}+22x-78\times 8=0$

  $\Rightarrow$                 $x=16 cm$

 One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperature at A, B and C are 400K, 800K, and 600K. respectively, Choose the correct statement 

Answer:

Explanation:

According  to first law  of thermodynamics, we get

(i) Change in internal energy from A to B. i.e, $\triangle U_{AB}$

        $\triangle U_{AB}=nC_{V}(T_{B}-T_{A})$

  $=1\times \frac{5R}{2}(800-400)=1000R$

  (ii) Change  in internal energy from B to C

   $\triangle U_{BC}=nC_{V}(T_{C}-T_{B})$

  $=1\times \frac{5R}{2}(600-800)=-500R$

  (iii)  $\triangle U_{isothermal}=0$

(iv)   Change in internal energy from C to A i.e,   $\triangle U_{CA}$

    $\triangle U_{CA}=nC_{V}(T_{A}-T_{C})$

    =   $1\times\frac{5R}{2}(400-600)=-500R$

 Three rodes of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross-section of each rod is 4 cm². End of copper rod is maintained at 100° C whereas ends of brass and steel are kept at 0° C. Lengths of the copper, brass and steel rods are 46,13 and 12 cm respectively, The rods are thermally insulated from surroundings except at ends. Thermal conductivities are 0.92 ,0.26 and 0.12 in CGS units, respectively. Rate of heat flow through copper rod is 

Answer:

Explanation:

 In thermal conduction , it is found that in  steady state the heat current is direcly proportional to the area  of cross-section A which is proportional to the change in  temperature (T₁-T₂)

 Then,    $\frac{\triangle Q}{\triangle t}=\frac{KA(T_{1}-T_{2})}{x}$

  According to thermal conductivity , we get

i.e,    $\frac{dQ_{1}}{dt}=\frac{dQ_{2}}{dt}+\frac{dQ_{3}}{dt}$

$\frac{0.92(100-T)}{46}=\frac{0.26(T-0)}{13}+\frac{0.12(T-0)}{12}$

$\Rightarrow $       $T=40^{0}C$

  $\therefore $                $\frac{dQ_{1}}{dt}=\frac{0.92\times4(100-40)}{40}$

   =4.8 cal/s

 There is a circular tube in a vertical plane. Two liquids that do not mix and of densities  d₁ and d2  are filled in the tube. Each liquid subtends  90⁰ C  angle at centre. Radius joining  their interface  makes an angle   $\alpha$  with vertical, Ration d₁/d₂  is 

Answer:

Explanation:

 Equating  pressure at A, we get

   $R\sin\alpha d_{2}+R\cos\alpha d_{2}+R(1-\cos \alpha)d_{1}$

   =   $R(1-\sin\alpha)d_{1}$

$(\sin \alpha+\cos\alpha)d_{2}=d_{1}(\cos\alpha-\sin \alpha)$

  $\frac{d_{2}}{d_{1}}=\frac{1+\tan \alpha}{1-\tan \alpha}$

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