MHT CET 2019 :: Physics :: General questions

• Physics - General questions

P and Q are two  non- zero vectors  inclined to each other at an angle '$\theta$'.'p. and 'q' are unit vectors along P and Q respectively. The   component  of Q in the direction of Q will ve 

Answer:

Explanation:

 Let two vectors P and Q are represented by  graph as below 

 Here, Q_p  is a vector in the direction of P. Then , from the right angle triangle , we get

$\cos\theta=\frac{Q_{p}}{Q}$     ......(i)

$\Rightarrow$     $ Q_{p}= Q \cos\theta$

 Also,   $\cos\theta=\frac{P.Q}{P Q}\Rightarrow \frac{Q_{p}}{Q}=\frac{P.Q}{P Q}$

$\Rightarrow $   $Q_{p}=\frac{P.Q}{P }$          ......(ii)

 As given that , $\hat{P}$ is the unit vector along P. then

  $\hat{P}=\frac{\underline{P}}{P}$   .......(ii)

 Putting the value of P from Eq.(iii) to Eq. (ii) , we get

  $Q_{p}=\hat{P}.Q$

 A lift is tied with thick iron ropes having mass 'M'. The maximum acceleration of the lift is 'a' m/s² and the maximum safe stress is 'S' N/m². The minimum  diameter  of the rope is 

Answer:

Explanation:

The maximum stress produced in a rope is given by 

$\sigma_{max}=\frac{Force}{Area}=\frac{Mg}{\pi r^{2}}$

As the lift is accelerationg with acceleration a, then

$\sigma_{max}=\frac{M(g\pm a)}{\pi r^{2}}$

$\Rightarrow $   $r^{2}=\frac{M(g\pm a)}{\pi S}$    [ Given , $\sigma_{max}$ = S]

$ \frac{d^{2}}{4}=\frac{M(g\pm a)}{\pi S}$           $\left[\because r=\frac{d}{2}\right]$

$d= \sqrt{\frac{4M(g\pm a)}{\pi S}}$

 As acceleration is maximum i.e, g'= g+a, so

$d= \sqrt{\frac{4M(g\pm a)}{\pi S}}$

 The range of an ammeter of resistance 'G' can be increased from 'l'  to 'nl'  by connecting

Answer:

Explanation:

 The range of an ammeter can be increased by connecting  a shunt or parallel resistance to it,

 whose value is given by 

$R=\left(\frac{l_{g}}{l-l_{g}}\right) G\Omega$  .......(i)

 To increase the range from l to nl , the current  passing through it is l= nl_g

 Using this value in  Eq.(i), we get

$R=\left(\frac{l_{g}}{nl_{g}-l_{g}}\right) G$

 =  $\frac{G}{n-1}\Omega$

Two light balls are suspended as shown in the figure. When the stream of air passes through  the space between them, the distance between the balls will

Answer:

Explanation:

 When a stream of air passes through the space between the balls, the pressure reduces between them as compared to the atmospheric pressure on either side. So, according to Bernoulli's theorem, the pressure energy should be constant and is compensated by the movement of balls toward each other. So, the distance between the balls will decrease.

Find the wrong statement from the following about the equation of stationary wave given by Y=0.04  cos ($\pi$x) sin (50 $\pi$ t) m where t is in second. Then for the stationary wave.

Answer:

Explanation:

 Key idea .  The displacement  of a wave in term of time period is given by

$y= A\sin\left[ 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right]$

 This equation in terms of speed of wave (v) becomes   $y= A\sin\left[ \frac{2\pi}{\lambda}(vt-x)\right]$

 The given equation of wave is 

$y= 0.04 \cos (\pi x) \sin (50 \pi t)$

  $y= 0.02 \sin(50 \pi t+ \pi x)+0.02 \sin (50 \pi t- \pi x)$

[$\because$     $ 2 \sin A\cos B= \sin (A+B) .\cos (A-B)]$

 Thus, the given wave is the combination of two waves.

 $y_{1}=0.02 \sin (50 \pi t+\pi x)$     ( in -ve x- direction)

and   $y_{2}=0.02 \sin (50 \pi t-\pi x)$         (in +ve x direction )

 Comparing them with the general equation  of wave  $a\sin (\omega t+kx)$, we get

 Amplitude  , a= 0.02 m

 Time period , T=  $\frac{2 \pi}{50 \pi}= \frac{1}{25}=0.04 s$

 Wave length ,  $\lambda= \frac{2 \pi}{\pi}=2 m$

 velocity ,  v=  $ \frac{50 \pi\times \lambda}{2 \pi}=\frac{100}{2}=50 ms^{-1}$

     So, option (a) is wrong 

• Physics - General questions