MHT CET 2019 :: Physics :: General questions

• Physics - General questions

 The luminous  border that surrounds the profile of a mountain just before sun rises behind it, is  an example of 

Answer:

Explanation:

 The luminous border that surrounds the profile of a mountain just before the sun rises behind it is due to the diffraction of light. 

 The stretched string fixed at both ends has 'm' nodes, then the length of the string will be 

Answer:

Explanation:

 For p number of loops in a stretched string, the length is given by

$l=\frac{p\lambda}{2}$   ...........(i)

 As, number of harmonics = number of loops= number of anti-nodes =p  ........(ii)

 Also, number of nodes= number of anti-nodes +1

 Here, number of nodes =m

 Number of anti-nodes=m-1

 from . Eq .(ii)

   p= m-1

 putting this value of p in Eq.(i)  , we get

     $l=\frac{(m-1)\lambda}{2}$

A force  (F)=$-5\hat{i}-7\hat{j}+3\hat{k}$ acting on a partcle  causes a dispalcement (s)=  $3\hat{i}-2\hat{j}+a\hat{k}$  in its own direction. If the work done  is 14J , then  the value of 'a' is 

Answer:

Explanation:

 GIven,   F=$-5\hat{i}-7\hat{j}+3\hat{k}$.

  s=  $3\hat{i}-2\hat{j}+a\hat{k}$

 and W=14 J

 the work done by a force in displaceing a particle  through  a distance is given by 

 W= F.s ........(i)

 Substituting above values in Eq.(i), we get

$14=  (-5\hat{i}-7\hat{j}+3\hat{k}).( 3\hat{i}-2\hat{j}+a\hat{k})$

 $\Rightarrow$    $ 14=  -15+14+3a$

$\Rightarrow $    $ a=\frac{15}{3}=5$

 The equi- conves lens has a focal length 'f' . If the lens is cut along the line perpendicular to the principal  axis and passing through the pole, what will be the focal length of any half part ?

Answer:

Explanation:

 Key idea:     The lens Maker's formula is given as 

$\frac{1}{f}= (\mu_{med}-1)\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)$

 where , f= focal length of lens, R₁= radius of first curved part and R₂= Radius of second curved part 

 As, for equiconvex lens R₁= R₂= R (say)

 So,  $\frac{1}{f}= (\mu_{real}-1)\frac{2}{R_{}}$   ......(I)

Now, if lens  is cut along the line perpendicular to the prinicipal axis as shown in the figure

,

 The new cut  of the lens has, R₁= R and R₂=  $\infty$ . Again by using the lens Maker's formula, focal length of the new part of the lens,

$\frac{1}{f'}= (\mu_{real}-1)\left[\frac{1}{R_{}}-\left(-\frac{1}{\infty}\right)\right]$

$\Rightarrow\frac{1}{f}= (\mu_{real}-1)\left[\frac{1}{R_{}}\right]$   .....(ii)

 So, from the Eqs.(i) and (ii) , we get

  f'=2f

 The magnitude of the magnetic induction  at a point on the axis  at a  large distance (r) from  the centre of a circular coil of  'n' turns and area 'A' carrying current (l)  is given by

Answer:

Explanation:

As we know that the magnetic field on the axis of a circular current carrying loop

$B_{axis}=\frac{\mu_{0}nla^{2}}{2(r^{2}+a^{2})^{3/2}}$..............(i)

where, l= current tjhrough coil,a =radius of a circular loop, r= distance of point from the centre along the axis and n= number of turns in the coil

 area of the coil, A= $\pi  a^{2} $

 $\Rightarrow$     $ a^{2}=\frac{A}{\pi}$   ..........(ii)

and it r>>a then,  $(r^{2}+a^{2})^{3/2}=r^{3}$   ............(iii)

From Eqs.(i) , (ii) and (iii) we get

$B_{}=(\frac{\mu_{0}nl}{2r^{3}})\frac{A}{\pi}\times\frac{2}{2}$

   $\Rightarrow$      $ B_{}=\frac{2\mu_{0}nlA}{4\pi r^{3}}$

So, option (b) is correct

• Physics - General questions