JEE 2013 :: Main 2013 :: Physics 2013

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A hoop of radius r and mass m rotating with an angular velocity ω₀  is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

Answer:

Explanation:

ω = v/r

From conservation of angular momentum about bottommost  point

  $mr^{2}\omega_{0}=mvr+mr^{2}\times\frac{v}{r}$

$\Rightarrow$    $ v=\frac{\omega_{0}r}{2}$

The shown p-V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat  extracted from the source in a single cycle is 

Answer:

Explanation:

Heat is extracted from the source means heat is given to the system ( or gas) . or Q is positive . This is positive  only along the path ABC

 Heat supplied

 $\therefore$    $ Q_{ABC}=\triangle U_{ABC}+W_{ABC}$

    = nC_V(T_f  -T_i)+ area under  p-V graph

=$n\left(\frac{3}{2}R\right)(T_{c}-T_{A})+2p_{0}V_{0}$

  =$\frac{3}{2}(nRT_{c}-nRT_{A})+2p_{0}V_{0}$

   = $\frac{3}{2}(p_{c}V_{c}-p_{A}V_{A})+2p_{0}V_{0}$

    =  $\frac{3}{2}(4p_{0}V_{0}-p_{0}V_{0})+2p_{0}V_{0}$

    =   $\frac{13}{2}p_{0}V_{0}$

The supply voltage in a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W  bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

Answer:

Explanation:

$P= \frac{V^{2}}{R}$

$R= \frac{120\times120}{60}=240$  Ω

  R_eq = 240+6=246 Ω

$\Rightarrow $    $i_{1}=\frac{V}{R_{eq}}=\frac{120}{246}$

$ v_{1}=\frac{240}{246}\times120=117.073 V$

  $\Rightarrow$   $ i_{2}=\frac{120}{48+6}$

  $v_{2}=\frac{48}{54}\times 120=106.66V$

   v₁-v₂= 10.04 V

A beam of unpolarized light of intensity I₀ is passed through a polaroid A and then through another polaroid  B which is oriented so its principal plane makes an angle  45° relative to that of A. The intensity of the emergent light is

Answer:

Explanation:

 Relation between intensities is

$I_{R}=\left(\frac{I_{0}}{2}\right)\cos^{2}(45^{0}) =\frac{I_{0}}{2}\times\frac{1}{2}=\frac{I_{0}}{4}$

A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 picofarad in parallel witha load resistance 100 kΩ . Find the maximum modulated frequency which coud be dectected by it

Answer:

Explanation:

$\tau= RC= 100\times 10^{3}\times250\times 10^{-12}s$

   = 2.5 x 10^-5 s

  The higher frequency which can be  detected with tolerable  distortion is

$f=\frac{1}{2\pi m_{a}RC}$

    $=\frac{1}{2\pi \times0.6\times2.5\times 10^{-5}}Hz$

   $=\frac{100\times 10^{4}}{25 \times1.2\pi }Hz$

     $=\frac{4}{ 1.2\pi }\times 10^{4}Hz$

   = 10.61 kHz

• Main 2013 - Physics 2013
• Main 2013 - Chemistry 2013
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