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Two coaxial solenoids of different radii carry current I  in the same direction. Let F₁ be the magnetic force on the inner solenoid due to the outer one and F₂ be the magnetic force  on the outer solenoid due to the inner one, Then,

Answer:

Explanation:

Consider the two coaxial solenoids. Due to one of the solenoids magnetic field at the centre of the other can be assumed to be constant

 Due to symmetry, forces on upper and lower part of a solenoid  will be equal and opposite and hence resultant is zero , Therefore $F_{1}$ = $F_{2}$=0

In the circuit  shown below, the current in the 1Ω resistor is 

Answer:

Explanation:

Central Idea: Connect point Q  to ground and apply KCL.

   Consider the grounded circuit as shown below.

Applying KCL of point Q we can write incoming current at Q = outgoing current from Q

       $\Rightarrow$      $\frac{V+6}{3}+\frac{V}{1}=\frac{9-V}{5}$

  or      $V\left[\frac{1}{3}+\frac{1}{5}+1\right]=\frac{9}{5}-2$

  or     $V\left[\frac{5+3+15}{15}\right]=\frac{9-10}{5}$

  or        $V\left[\frac{23}{15}\right]=\frac{-1}{5}$

  or      $V=\frac{-3}{23}=-0.13V$

 Thus, current in the 1Ω   resistance  is 0.13 A, from Q to P

When 5V  potential difference is applied across a wire of length 0.1m  the drift speed of electrons is  $2.5\times 10^{-4}ms^{-1}$. If the electron density in the wire is  $8\times 10^{28}m^{-3}$  the resistivity of the material is close to Ωm

Answer:

Explanation:

According to the question

 $v_{d}=2.5\times 10^{-4}m/s$

 $\Rightarrow$         $n=8\times 10^{28}/m^{3}$  

   we know that     $J=nev_{d}$   or   $I=nev_{d}A$

 where,    symbols have their usual  meaning

 $\Rightarrow$     $\frac{V}{R}=nev_{d}A$

      or        $\frac{V}{\frac{\rho L}{A}}=nev_{d}A$

     or     $\frac{V}{{\rho L}}=nev_{d}$

   or     $\rho  =\frac{V}{nev_{d}L}$

      =    $\frac{5}{8\times 10^{28}\times1.6\times 10^{-19}\times2.5 \times 10^{-4}\times0.1}$

          or    $\rho$    =$1.6\times 10^{-5}$  Ωm

A uniformly charged solid sphere of radius R has potential V₀ (measured with respect to ∞) on its surface. For this sphere, the equipotential sufaces,  with potentials  $\frac{3V_{0}}{2},\frac{5V_{0}}{4},\frac{3V_{0}}{4} and \frac{V_{0}}{4}$ have radius R₁, R₂ ,R₃  and R₄  respectively. Then   R₄

Answer:

Explanation:

Potential at the surface of the charged sphere

$V_{0}=\frac{KQ}{R}$  

 $V=\frac{KQ}{r}$ , $r\geq R$

  $= \frac{KQ}{2R^{3}}(3R^{2}-r^{2});r\leq R$

   $V_{centre}=V_{c}=\frac{KQ}{2R^{3}}\times3R^{2}$

    $\frac{3KQ}{2R}=\frac{3V_{0}}{2}$

   $R_{1}=0$

 As potential  decreases for outside points.

Thus, according to the question, we can write

     $V_{R_{2}}=\frac{5V_{0}}{4}=\frac{KQ}{2R^{3}}(3R^{2}-R^{2}_{2})$

     $\frac{5V_{0}}{4}=\frac{V_{0}}{2R^{2}}( 3R^{2}-R_{2}^{2})$

  or      $\frac{5}{2}=3-\left(\frac{R_{2}}{R}\right)^{2}$

          $R_{2}=\frac{R}{\sqrt{2}}$

  Similarly,

      $V_{R_{3}}=\frac{3V_{0}}{4}$

  $\Rightarrow$    $\frac{KQ}{R_{3}}=\frac{3}{4}\times\frac{KQ}{R}$

    or              $R_{3}=\frac{4}{3} R$

                    $V_{R_{4}}=\frac{KQ}{R_{4}} =\frac{V_{0}}{4}$

$\Rightarrow$     $\frac{KQ}{R_{4}}=\frac{1}{4}\times\frac{KQ}{R}$

  OR             R₄   =4R

A train moving on a straight track with speed 20 ms^-1. It is blowing its whistle at the frequency of 1000Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is  close to (speed of sound =320 ms^-1  )

Answer:

Explanation:

 Apparent frequency heard by the person before crossing the train.

    $f_{1}=\left(\frac{c}{c-v_{s}}\right)f_{0}=\left(\frac{20}{320-20}\right)1000$

  Similarily, apparent frequency heard , after crossing the trains

  $f_{2}=\left(\frac{c}{c+v_{s}}\right)f_{0}=\left(\frac{20}{320+20}\right)1000$

                                                     [c=speed of sound]

    $\triangle f= f_{1}-f_{2}=\left(\frac{2cv_{s}}{c^{2}-v_{s}^{2}}\right)f_{0}$

        or      $\frac{\triangle f}{f_{0}}\times100=\left(\frac{2cv_{s}}{c^{2}-v_{s}^{2}}\right)\times 100$

               $=\frac{2\times320\times20}{300\times340}\times100$

                         $=\frac{2\times32\times20}{3\times34}=$   12.54%=12%

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