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What is the minimum energy required to launch a satellite of mass m  from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

Answer:

Explanation:

E= energy of satelite- energy of mass on the surface of plannar

    $=-\frac{GMm}{2r}-\left(-\frac{GMm}{R}\right)$

 Here, r= R+2R= 3R

 Substituting in about the equation we get,

                $E= \frac{5GMm}{6R}$

The diameter pf a plano-convex lens is 6 cm and thickness at the centre is 3mm. If the speed of light in material of lens is $2\times 10^{8}$  m/s, the  focal length of lens is 

Answer:

Explanation:

By pythagoras theorm

$R^{2}=(3)^{2}+(R-0.3)^{2}$

   = R=15 cm

  Refractive index of material of lens

               $\mu=\frac{c}{v}$

Here , c= speed of light in vacum    =3 x10⁸  m/s

 v= speed of light in material of lens  = 2 x 10⁸ m/s

  =   $\frac{3\times 10^{8}}{2 \times 10^{8}}=\frac{3}{2}$

 From lens marker's formula

       $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

 Here, R₁=R and R₂= $\infty$ (For plane surface)

$\frac{1}{f}=(\frac{3}{2}-1)\left(\frac{1}{15}\right)$

  = f= 30 cm

A circular loop of radius n0.3 lies parallel to a much bigger circular loop of radius 20 cm. The centre of the smaller radius 20 cm. The centre of the smaller loop on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the bigger loop, then the flux  linked with smaller loop is 

Answer:

Explanation:

magnetic field at  the centre of smaller loop 

      $B= \frac{\mu_{0}iR_2^2}{2(R_2^2+x^{2})^{3/2}}$

 Area  of smaller loop   $S=\pi R_1^2$

 $\therefore$  Flux  through smaller loop Φ  = BS

 Substituting the values we get, 

$\phi= 9.1\times 10^{-11}Wb$

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 x10³ kg/m³ and 2.2 x 10¹¹ N/m²  respectively?

Answer:

Explanation:

Fundamental frequency of sonometer wire

   $f=\frac{v}{2l}=\frac{1}{2l}\sqrt{\frac{T}{\mu}}=$

                               $\frac{1}{2l}\sqrt{\frac{T}{Ad}}$

 Here,  $\mu$ = mass per unit length of wire,

Also , Young's modulus of elesticity

             $Y=\frac{Tl}{A\triangle l}$

$\Rightarrow\frac{T}{A}=\frac{Y\triangle l}{l}\Rightarrow f=\frac{1}{2l}\sqrt{\frac{Y \triangle l}{ld}}$

l=1.5 m, d=7.7 x 10³ kg/m³

  Y= 2.2 X 10¹¹  N/m²

 After substituting the values we get,

 f≈ 178.2 Hz

Two capacitors C₁  and C₂ are charged to 120 V and 200 V respectively. It is found that by connecting them together the  potential  on each one can be made zero, then

Answer:

Explanation:

Polarity sholud be mentioned in the question . Potential on each of them can be zero , if q_net=0

or  $q_{1}\pm q_{2}=0$

   or   $120C_{1}\pm 200C_{2}=0$

or   $3C_{1}\pm 5C_{2}=0$

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