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The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5 s. In another 10 s it will decrease 10 $\alpha$ times its original magnitude, where $\alpha$ equals,

Answer:

Explanation:

Amplitude   decreases exponentially, In 5 s it remain  0.9 times .Therefore in total  15 s it will remains (0.9) (0.9)(0.9)= 0.729 times its original value

A projectile is given an initial velocity of $(\hat{i}+2j)$  m/s, where   $\hat{i}$  is along the ground and $\hat{j}$ is along the vertical. If g=10 m/s², the equation of its trajectory is

Answer:

Explanation:

intial velocity= (i+2j) m/s

 Magnitude  of initial velocity  $u=\sqrt{(1)^{2}+(2^{2})}=\sqrt{5}m/s$

 Equation of trajectory of projectile is

$y=x\tan\theta -\frac{gx^{2}}{2u^{2}}(1+\tan ^{2}\theta)$

                                    $\left[ \tan\theta=\frac{y}{x}=\frac{2}{1}=2\right]$

 $\therefore$       $  y=x\times2-\frac{10(x)^{2}}{2(\sqrt{5})^{2}}[1+(2)^{2}]$

$=2x-\frac{10(x^{2})}{2\times5}(1+4)$

     = 2x-5x²

Let [ε₀] denote the dimensional formula of the permittivity of vacuum. If M= mass, L= length , T= time and A= electric current then,

Answer:

Explanation:

from COULOMB'S law  

                           $F=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{R^{2}}$

$\epsilon_{0}=\frac{q_{1}q_{2}}{4\pi F R^{2}}$

substituting the units we have ,

 $\epsilon_{0}=\frac{C^{2}}{N.m^{2}}=\frac{[AT]^{2}}{[ML^{}T^{-2}][L^{2}]}=$

                                         $[M^{-1}L^{-3}T^{4}A^{2}]$

The qusetion has statement I and statement II, Of the four choices given after the statements, choices the one that  best describes the two statements.

Statement I: A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as 

$f\left(\frac{1}{2}mv^{2}\right),then f =\left(\frac{m}{M+m}\right)$

Statement II: Maximum energy loss occurs when the particles get stuck together as a result  of the collision

Answer:

Explanation:

Maximum energy loss

  =$\frac{p^{2}}{2m}-\frac{p^{2}}{2(m+M)}\left( \because KE=\frac{p^{2}}{2m}\right)  $

  Before  collision  the mass m and afterr collision the mass m+M

 = $\frac{p^{2}}{2m}\left[\frac{M}{(m+M)}\right]$

 = $\frac{1}{2}mv^{2}\left\{\frac{M}{m+M}\right\}\left(f=\frac{M}{m+M}\right)$

A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the emf induced across the ends of the rod is 

Answer:

Explanation:

$e=\int_{2l}^{3l}(\omega x)Bdx=B\omega\frac{[(3l)^{2}-(2l)^{2}]}{2} =\frac{5Bl^{2}\omega}{2}$

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