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A particle A of mass m and intial velocity v collides with a particle B of mass $\frac{m}{2}$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λ_A to λ_B after the collision is 

Answer:

Explanation:

For elastic collision

  P_{before collision} =P_{after collsion}

$mv=mv_{A}+\frac{m}{2}v_{B}$

$2v=2v_{A}+2v_{B}$         .........(i)

Now, coefffficent of restitution,

$  e= \frac{v_{B}-v_{A}}{u_{A}-v_{B}}$

Here u_B = 0 ( particle at rest) and for elastic collisione= 1

$\therefore$     $1= \frac{v_{B}-v_{A}}{v}\Rightarrow v =v_{B}-v_{A}$              .............(ii)

From Eq .(i) and Eq .(ii)

                     $v_{A}=\frac{v}{3}$   and $v_{B}=\frac{4v}{3}$

              Hence,  $\frac{\lambda_{A}}{\lambda_{B}}=\frac{(\frac{h}{mV_{A}})}{\frac{h}{\frac{m}{2}V_{B}}}=\frac{V_{B}}{2V_{A}}=\frac{4/3}{2/3}=2$

A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that this density remains the same, the stress in the leg will change by a factor of

Answer:

Explanation:

$Stress=\frac{Weight}{Area}$

     Volume will become (9³) times.

   So weight= volume × density × g will also becomes (9)³ times.

 Area of cross section will become (9)² times.

                  $=\frac{9^{3}\times W_{0}}{9^{2}\times A_{0}}=9 (\frac{W_{0}}{A_{0}})$

     Hence, the stress increases by a factor of 9

A magnetic needle of magnetic moment 6.7 × 10^-2 Am² and moment of inertia 7.5 × 10^-6 kg  m² is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is

Answer:

Explanation:

Time period of oscillations is 

  $T = 2\pi \sqrt{\frac{I}{MB}}$

  $T = 2\pi \sqrt{\frac{7.5\times 10^{-6}}{6.7\times 10^{-2}\times 0.01}}$

     =0.665 s

Hence , time for 10 oscillations is t= 6.65 s

A time-dependent force F =6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 s will be 

Answer:

Explanation:

From Newton's second law, $\frac{\triangle p}{\triangle t}=F$

$\Rightarrow \triangle p=F\triangle t$

 $\therefore$          $p=\int_{}^{}dp =\int_{0}^{1}  F dt$

$\Rightarrow  p= \int_{0}^{1} 6t dt=3kg(\frac{m}{s})$

Also , change in kinetic energy 

 $\triangle k= \frac{\triangle p^{2}}{2m}=\frac{3^{2}}{2\times 1}$

 = 4.5

From work -energy theorm, work done = change in kinectic energy

    So, work done = Δ k= 4.5 J

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15Ω, it shows full-scale deflection. The value of the resistance to be put in series with the galvanometer  to convert it into a voltmeter of range 0-10 V is

Answer:

Explanation:

Suppose a resistance R_s is connected in series with a galvanometer to convert it into a voltmeter.

  $I_{g}(G+R_{s})=V$

$R=\frac{V}{I_{g}}-G$

$\Rightarrow R=1985=1985k$Ω

   or 1985 × 10³Ω

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