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One end of a horizontal thick copper wire of length 2 L and radius 2R is welded to an end of another horizontal thin copper wire of length L  and radius R. When the arrangement is treated by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is 

Answer:

Explanation:

$\triangle l= \frac{FL}{AY}=\frac{FL}{(\pi r^{2})Y}$

  $\therefore$      $\triangle l\propto \frac{L}{r^{2}}$

 $therefore$    $\frac{\triangle l_{1}}{\triangle l_{2}}=\frac{L/R^{2}}{2L/(2R)^{2}}=2$

The work done on a particle of mass m by a force ,

    $K\left[\frac{x}{(x^{2}+y^{2})^{3/2}}\hat{i}+\frac{y}{(x^{2}+y^{2})^{3/2}}\hat{j}\right]$

 (K is being a constant  of appropriate dimensions), when the particle is taken  from the point (a,0) to the point (0,a) along a circular path of radius a about  the origin in the x-y plane is 

Answer:

Explanation:

$r=OP=x\hat{i}+y\hat{j}$

   $F=\frac{k}{(x^{2}+y^{2})^{3/2}}(x\hat{i}+y\hat{j})$

$=\frac{k}{r^{3}}(r)$

  Since F is along with r or in radial direction. Therefore , work done is zero

The diameter of a cylinder is measured using a vernier callipers with no zero error, It is found that the zero of the vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The vernier scale has 50 division equivalent to 2.45 cm. The 24 ^th division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is 

Answer:

Explanation:

1 mMD= 5.15 cm -5.10 cm=0.05 cm

 $1 VSD=\frac{2.45 cm}{50}=0.049 cm$

$\therefore$  LC=1MSD-1VSD=0.01cm

 Hence , Diameter of cyclinder = (main scale reading)+ (Vernier scale reading )(LC)

  =5.10 +(24) (0.001)=5.124 cm

 The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one -third the size of the object. The wavelength of light inside the lens is 2/3 times the wavelength in free space. The radius of the curved surface of the lens is 

Answer:

Explanation:

$\mu=\frac{\lambda_{air}}{\lambda_{medium}}=\frac{1}{(2/3)}=\frac{3}{2}$

 Further , $|m|=\frac{1}{3}=|\frac{v}{u}|$

    $\therefore$         $|v|=\frac{|u|}{3}$

                 u=-24 m     (real object)

      $\therefore$    v=+8 m (real image)

  Now,   $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}=(\mu-1)(\frac{1}{+R}-\frac{1}{\infty})$

   $\therefore$      $\frac{1}{8}+\frac{1}{24}=(\frac{3}{2}-1)(\frac{1}{R})$

   $\therefore$     R= 3m

A particle of mass m is projected from the ground with an initial speed u₀  at an angle $\alpha$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u₀, The angle that the composite   system  makes with the horizontal immediately after the collision is 

Answer:

Explanation:

From the momentum conservation equation, we have

   P_i=P_j

  $\therefore m(u_{0}\cos\alpha)\hat{i}+m(\sqrt{u_0^2-2gH)}\hat{j}$=(2m)v   .......(i)

   $H=\frac{u_0^2\sin^{2}\alpha}{2g}$      ........(ii)

 From Eqs.(i)   and (ii)

     $v=\frac{u_{0}\cos\alpha}{2}\hat{i}+\frac{u_{0}\cos\alpha}{2}\hat{j}$

  Since both components of v are equal. Therefore it is making 45°  with horizontal.

• advanced 2013 - Physics 2013
• advanced 2013 - Chemistry 2013
• advanced 2013 - Mathematics 2013