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A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying  at distance L from the end A is

Answer:

Explanation:

$V=\int_{L}^{2L} \frac{kdQ}{x}$

$=\int_{L}^{2L} \frac{k\left(\frac{Q}{L}\right)dx}{x}$

$=\frac{Q}{4 \pi \epsilon_{0}L}\int_{L}^{2L} \left(\frac{1}{x}\right)dx$

$=\frac{Q}{4 \pi \epsilon_{0}L}[\log_{e} x]_L^{2L}$

$=\frac{Q}{4 \pi \epsilon_{0}L}[\log_{e} 2L-\log_{e} L]$

$=\frac{Q}{4 \pi \epsilon_{0}L}ln(2)$

Two short bar magnets of length 1cm each have magnetic moments 1.20 Am²  and 1.00 Am² respectively. They are placed on a horizontal table parallel to each other with their  N poles pointing towards the south. They have a common magnetic equator and are separated by a  distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the midpoint O of the  line joining  their centre s is close to (Horizontal component of the earth's magnetic induction is 3.6 x 10^-5 Wb/m²)

Answer:

Explanation:

B_net  = B₁+B₂ +B_H

 $B_{net}=\frac{\mu_{0}}{4\pi}\frac{(M_{1}+M_{2})}{r^{3}}+B_{H}$

  $=\frac{10^{-7}(1.2+1)}{(0.1)^{3}}+3.6\times 10^{-5}$

= $2.56\times 10^{-4} Wb/m^{2}$

Two charges, each equal to q are kept at x=-a  and x=a  on the x-axis. A particle of mass m and charge   q₀= q/2 is placed at the origin. If charges q₀  is given a small displacement y(y << a) along the y-axis,  the net force acting on the particle  is proportional to 

Answer:

Explanation:

$F_{net}=2F \cos\theta$

$F_{net}=\frac{2kq(\frac{q}{2})}{(\sqrt{y^{2}+a^{2})^{2}}}.\frac{y}{(\sqrt{y^{2}+a^{2})}}$

$F_{net}=\frac{2kq(\frac{q}{2})y}{(y^{2}+a^{2})^{3/2}}.\Rightarrow\frac{kq^{2}y}{a^{3}}\propto y$

In a hydrogen-like atom electron makes the transition from an energy level with a quantum number n to another with a quantum number (n-1). If n>>1,  the frequency  of radiation emitted is proportional  to 

Answer:

Explanation:

$   \triangle E=hv$

$v= \frac{\triangle E}{h}=k\left[\frac{1}{(n-1)^{2}}-\frac{1}{n^{2}}\right]$

$=\frac{k2n}{n^{2}(n-1)^{2}}$

$=\frac{2k}{n^{3}}\propto\frac{1}{n^{3}}$

Assume that a drop of liquid evaporates by decrease in its surface energy so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, the density of liquid  is $\rho$   and L is its talent heat of vaporization

Answer:

Explanation:

Decrease in surface energy= hear required  in vaporization

 $\therefore$      $ T(ds)= L(dm)$

$\therefore$     $ T(2)(4\pi r)dr= L(4\pi r^{2}dr)\rho$

 $\therefore   $     $ r= \frac{2T}{\rho L}$

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