MHT CET 2018 :: Physics :: General questions

• Physics - General questions

The ratio of magnetic fields due to a bar magnet at the two axial point P₁ and P₂ which are separated from each other by 10 cm is 25; 2 point P₁ is situated at 10 cm from the centre of the magnet. The magnetic length  of the bar magnet is (point P₁ and P₂  are o2n the same side  of magnet and distance of P₂ from the centre is greater than the distance of P₁  from the centre of the magnet)

Answer:

Explanation:

Let magnetic field at P₁ is B₁ and at P₂ is B₂

$\therefore$    $\frac{B_{1}}{B_{2}}=\frac{\frac{x_{1}}{(x_{1}^{2}-l^{2})^{2}}}{\frac{x_{2}}{(x_{2}^{2}-l^{2})^{2}}}=\frac{(x_{2}^{2}-l^{2})^{2}}{(x_{1}^{2}-l^{2})^{2}}\times\frac{x_{1}}{x_{2}}$

$\Rightarrow$     $\frac{25}{2}=\frac{10}{20}\left[\frac{x_{2}^{2}-l^{2}}{x_{1}^{2}-l^{2}}\right]^{2}$

or   $\left[\frac{x_{2}^{2}-l^{2}}{x_{1}^{2}-l^{2}}\right]=5$

 or $x_{2}^{2}-l^{2}=5x_{1}^{2}-5l^{2}$

 or  $ 4l^{2}=5x_{1}^{2}-5x_{2}^{2}$

  = 5 x (10)²-(20)²

 = 500-400=100

 $\Rightarrow$   $ l^{2}=25$

$\Rightarrow$   l=5 cm

 $\therefore$   2l=10 cm

A pipe closed at one end has a length of 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 1000 Hz are (take, the velocity of sound in air =332 m/s)

Answer:

Explanation:

Fundamental freequency of one  end closed pipe,

 $v=\frac{nv}{4L}$

 Here, n=1

 or   $v_{1}=\frac{332}{4\times 83\times 10^{-2}}$

 or v₁  = 100

As  odd harmonics alone are produced in closed organ  pipe, therefore  possible frequencies are

    3v₁ =300 Hz. 5 v₂ =500 Hz

  7 v₂ = 700 hz , 9 v₂ =900 Hz

 Hence, the number of the possible natural oscillation of air column is 5

The energy of an electron having de-Broglie wavelength $\lambda$ is

(where, h= planck's constant, m= mass of electron)

Answer:

Explanation:

We know that  , 

$\lambda= \frac{h}{\sqrt{2m(KE)}}$

on taking square on both sides,we get

$\lambda^{2}= \frac{h^{2}}{2m(KE)}$

 $KE =  \frac{h^{2}}{2m\lambda^{2}}$

Let  $x=\left[\frac{a^{2}b^{2}}{c}\right]$ be the physical quantity  , if the percentage  error in the measurement of physical quantities a, b and c is 2,3 and 4 per cent respectively, then percentage error in the measurement of x is 

Answer:

Explanation:

Given,

    $x=\frac{a^{2}b^{2}}{c}$

$\frac{\triangle x}{x}=2\frac{\triangle a}{a}+2\frac{\triangle b}{b}+2\frac{\triangle c}{c}$

$\frac{\triangle x}{x}\times 100=2\left(\frac{\triangle a}{a}\times 100\right)+2\left(\frac{\triangle b}{b}\times b\right)+\left(\frac{\triangle c}{c}\times c \right)$

Here, percentage error  in a =2

 percentage error in b=3

percentage error in c=4

   $\left(\frac{\triangle x}{x}\times 100\right)=2\times2+2\times3+4$=4+6+4=14%

A conducting wire has length L₁ and diameter d₁. After stretching the same wire length becomes L₂ and diameter d₂ . The ratio  of resistances before and after stretching is 

Answer:

Explanation:

Resistance of wire is given by R= $\rho\frac{l}{A}$

 where, l is the length of conductor and A is the cross-sectional area.According to the question,

  $R_{1}=\rho\frac{l_{1}}{A_{1}}$  ............(i)

 in the second condition,

 $R_{2}=\rho\frac{l_{2}}{A_{2}}$  ............(ii)

 On dividing  Eq.(i) by Eq.(ii) , we get

 $\frac{R_{1}}{R_{2}}=\frac{l_{1}}{A_{1}}\times\frac{A_{2}}{l_{2}}$

$\frac{R_{1}}{R_{2}}=\frac{A_{2}}{A_{1}}\times\frac{l_{1}}{l_{2}}$

$\frac{R_{1}}{R_{2}}=\frac{\pi(\frac{d_{2}}{2})^{2}}{\pi(\frac{d_{1}}{2})^{2}}\times\frac{l_{1}}{l_{2}}$

$\frac{R_{1}}{R_{2}}=\frac{({d_{2}})^{2}}{({d_{1}})^{2}}\times\frac{l_{1}}{l_{2}}$............(iii)

 Volume of wire  remains constant, Hence , V₁ = V₂

            A₁l₁ = A₂l₂

 $\Rightarrow$   $ \pi \left(\frac{d_{1}}{2}\right)^{2}l_{1}=\pi \left(\frac{d_{2}}{2}\right)^{2}l_{2}$

$\Rightarrow$   $  \frac{\pi d_{1}}{4}^{2}l_{1}=\frac{\pi d_{2}}{4}^{2}l_{2}$

$\frac{l_{1}}{l_{2}}=\frac{(d_{2})^{2}}{(d_{1})^{2}}$    ...............(iv)

Substituting the value of Eq.(iv) in Eq.(iii) , we get

$\frac{R_{1}}{R_{2}}=\frac{(d_{2})^{2}}{(d_{1})^{2}}\times\frac{(d_{2})^{2}}{(d_{1})^{2}}$

 $\frac{R_{1}}{R_{2}}=\frac{(d_{2})^{4}}{(d_{1})^{4}}$

• Physics - General questions