MHT CET 2019 :: Physics :: General questions

• Physics - General questions

In the hydrogen emission spectrum, for any series, the principal quantum number is n. Corresponding maximum wavelength  $\lambda$  is (R= Rydberg's constant )

Answer:

Explanation:

 In hydrogen emission spectrum , wavelength is given by

$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For maximum wavelength of any principal quantum number n,

n₁ =n and n₂= n+1

$\therefore$  $\frac{1}{\lambda_{max}}=R\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)$

 $=R\left[\frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}}\right]$

$=R\frac{(n^{2}+2n+1-n^{2})}{n^{2}(n+1)^{2}}$

 $\frac{1}{\lambda_{max}}=\frac{R^{2}(2n+1)}{n^{2}(n+1)^{2}}$

$\lambda_{max}=\frac{n^{2}(n+1)^{2}}{R^{}(2n+1)}$

 A hole is drilled halfway to the centre of the earth. A body weighs 300N on the surface of the earth. How much will, it weigh at the bottom of the hole?

Answer:

Explanation:

 Given, distance of bottom of hole from the surface of earth (d)= half the radius of earth =  $\frac{R_{e}}{2}$

If  g be the value of gravitational acceleration on the surface of earth , then weight of body

   mg= 300N

 If 'g' be the gravitational  acceleration at the bottom of hole , then

$g'=g\left(1-\frac{d}{R_{e}}\right)=g\left(1-\frac{\frac{R_{e}}{2}}{R_{e}}\right)\Rightarrow g'=\frac{g}{2}$

$\therefore$     Weight of the body on the bottom of hole,

$mg'=\frac{mg}{2}=\frac{300}{2}=150 N$

The damped SHM , the SI unit of damping constant is 

Answer:

Explanation:

In damped SHM, damping force is  proportional to the velocity of the oscillator

i.e,     $F_{d}\propto v=F_{d}=bv$

 where, b is a damping constant

 $\therefore$ Damping constant , b$=\frac{F_{d}}{v}$

 SI unit of damoing constant

 b= SI unit of force / SI unit of velocity $=\frac{kg.m.s^{-2}}{m.s^{-1}}$

 = kg. s^-1= kg/s

 The vectors (A+B) and (A-B)  are at right angles to each other. This is possible under the condition

Answer:

Explanation:

 Vectors (A+B) and (A-B) are at right angle to each other , therefore

 (A+B).(A-B)=0

 A.A+B.A-A.B-B.B=0

 |A|+  BA cos $\theta$- AB cos $\theta$ -|B|²=0

   |A|²=|B| ²

 Hence , |A|= | B|

 A rod l m long is acted upon by couples as shown in the figure. The moment of the couple is $\tau$ Nm. If the force at each end of the rod, then the magnitude of each force is 

$(\sin 30^{0}=\cos 60^{0}=0.5)$

Answer:

Explanation:

Length of rod =l

 Moment of couple,

    $\tau=Fl\sin \theta$

$\therefore$    $F= \frac{\tau}{l\sin\theta}=\frac{\tau}{l\sin 30^{0}}=\frac{\tau}{l\times\frac{1}{2}}$

 Force, F= $\frac{2 \tau}{l}$

• Physics - General questions