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An electric dipole has a fixed dipole moment p, which makes angle θ with respect to X-axis. When subjected to an electric field  $E_{1}=E\hat{i}$ , it experiences a torque $T_{1}=\tau\hat{k}$ . When subjected to another electric field $E_{2}=\sqrt{3}E_{1}\hat{j}$  , it experiences a torque T₂ = -T₁ . The angle θ is 

Answer:

Explanation:

 Torque applied on a dipole $\tau$ =pE sinθ where θ = angle between the axis of the dipole and electric field.

     For electric field $E_{1}=E\hat{i}$

          it means the field is directed along positive X direction, so angle between dipole and field will remain θ , therefore torque in this direction.

        $E_{1}=pE_{1}sin\theta$

  In electric field 

      $E_{2}=\sqrt{3}E\hat{j} $, it means field is directed along positive Y -axis, so angle between dipole and field will 90-θ

  Torque  in this direction

       $T_{2}=pE \sin(90-\theta)$

                $=p\sqrt{3}E_{1}\cos\theta$

According to question

      $\tau_{2}=-\tau_{1}\Rightarrow \mid \tau_{2}\mid =\mid\tau_{1}\mid$

  $\therefore   pE_{1}\sin\theta =p\sqrt{3}E_{1}\cos\theta$

$\tan\theta =\sqrt{3}$

 $\Rightarrow$                   $\tan\theta =\tan 60^{0}$

    θ  = 60^°

In Young's double-slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen . The least distance from the common central maximum to the point where the bright  fringes due to both the wavelengths coincide is

Answer:

Explanation:

let n₁th fringe formed due to first wavelength and n₂ th fringe formed due to second wavelength coincide i.e., their distance from common central maxima will be same

    i.e., Y n₁= Y n₂

  $\Rightarrow \frac{n_{1}\lambda_{1}D}{d}=\frac{n_{2}\lambda_{2}D}{d}$

$\Rightarrow \frac{n_{1}}{n_{2}}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{520}{650}=\frac{4}{5}$

Hence, distance of the point of coincidence from the central maxima is

   $y=\frac{n_{1}\lambda_{1}D}{d}=\frac{n_{2}\lambda_{2}D}{d}$

   = $\frac{4\times 650\times 10^{-9}\times15}{0.5\times 10^{-3}}=7.8nm$

A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100g, filled with 170 g of water at room temperature. Subsequently, the temperature  of the system is found to be 75° C . T is (Given, room temperature = 30° C, the specific heat of copper =0.1 cal/ g° C )

Answer:

Explanation:

Heat gained (water+calorimeter) = Heat lost by copper ball

  $\Rightarrow  m_{w}s_{w}\triangle T+m_{c}s_{c}\triangle T =m_{B}s_{B}\triangle T$

$\Rightarrow  170\times 1 \times(75-30) +100\times 0.1 \times (75-30)$

$=100\times 0.1\times (T-75)$

$\therefore$      T = 885° C

In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be

Answer:

Explanation:

In a CE npn transistor amplifier output is 180° out of phase with input.

In amplitude modulation. sinusoidal carrier frequency used is denoted by ω_c   and the signal frequency is denoted by ω_m  . The bandwidth (Δω_m) of the signal is such that Δω_m << ω_c  . Which of the following frequencies is not contained in the modulated wave ?

Answer:

Explanation:

The frequency spectrum of the modulated wave is

Clearly, ω_m is not included in the spectrum.

• Main 2017 - Physics 2017
• Main 2017 - Chemistry 2017
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