Tag: physics

Questions Related to physics

In a race, boy A sees another boy B overtaking him at a speed $v$. If they were running in opposite directions, speed of B as seen by A is   

  1. $=v$

  2. $<v$

  3. $>v$

  4. $\geq v$


Correct Option: C
Explanation:

When they are running in same direction, $v = v _B - v _A$

When they are running in opposite directions, $v' = v _B + v _A > v$

A police inspector in a jeep is chasing a pickpocket an a straight road. The jeep going at its maximum speed v (assumed uniform). The pickpocket rides on the motor-cycle of a waiting friend when the jeep is at a distance d away, and the motorcycle starts with a constant acceleration a. The pick pocket will be caught if  

  1. $v\, \geq\, \sqrt{2ad}$

  2. $v^2\, \geq\, \sqrt{2ad}$

  3. $v\, \geq\, \sqrt{3ad}$

  4. $v\, \geq\, \sqrt{2ad^2}$


Correct Option: A
Explanation:

Answer is A.

Suppose the pickpocket is caught at a time t after motorcycle starts. The distance traveled by the motorcycle during this interval is
$s\, =\, \displaystyle \frac{1}{2}at^2$ ...(1)
During this interval the jeep travels a distance
s + d = vt ...(2)
By (1) and (2),
$\displaystyle \frac{1}{2}at^2\, +\, d\, =\, vt$
or,
$t\, =\, \displaystyle \frac{v \pm \sqrt{v^2\, -\, 2ad}}{a}$
The pickpocket will be caught if t is real and positive.
This will be possible if
$v^2\geq\, 2ad\, \quad\, or,\, \quad\, v \geq\, \sqrt{2ad}$
Hence, the pick pocket will be caught if $v^2\geq\, 2ad\, \quad\, or,\, \quad\, v \geq\, \sqrt{2ad}$.

Two sound sources are moving in opposite direction with velocity $v _1$ and $v _2$ $(v _1>v _2)$. Both are moving away from a stationary observer.the frequency of both the source is $900\ Hz$. What is the value of $v _1 - v _2 $  so that the beat frequency observed will be $6\ Hz$  ?


Speed of sound =$300\ ms^{-1}$

  1. $1\ ms^{-1}$

  2. $4\ ms^{-1}$

  3. $3\ ms^{-1}$

  4. $2\ ms^{-1}$


Correct Option: D
Explanation:

$f _1 = 900(\dfrac{300}{300+v _1})$

$\implies f _1= 900(1+ \dfrac{v _1}{300})^{-1}$
$\implies f _1 = 900 - 3v _1$
Similarly 
$f _2 = 900 - 3v _2$
So,
$f _1 -f _2 = 6$
$3(v _1 - v _2)= 6$
$\implies v _1 - v _2 = 2\ ms^{-1}$

Sound waves of equal amplitude wave frequency $(n-1),n _1(nH^1)$. They superinpose to give Beat. The no of Beats poduced per sec cuPll be:

  1. l

  2. 0

  3. 3

  4. 2


Correct Option: D

A sources of sonic oscillations with frequency n= $1700$ Hz and a receiver are located on the same normal to a wall. Both the source and receiver are stationary, and the wall recedes from the source with velocity u= $6.0$ cm/s. Find the beat frequency registred by the receiver. The velocity of sound is equal to $v= 340$ m/s.

  1. $0.2$ Hz

  2. $0.3$ Hz

  3. $0.4$ Hz

  4. $0.6$ Hz


Correct Option: D

Two sound sources (of same frequency ) are placed at distance of 100 meter. An observer, when moving between both sources, hears 44 beats per second. The distance between sound source is now changed to 400 meter then the beats/second heard by observer will be  :

  1. 2

  2. 4

  3. 8

  4. 16


Correct Option: D

Two open organ pipes 80 and 81 cm long found to give 26 beats in 10 sec, when each is sounding its fundamental note. Then the velocity of sound in air is

  1. 337 $m s ^ { - 1 }$

  2. 370 $m s ^ { - 1 }$

  3. 345 $m s ^ { - 1 }$

  4. 350 $m s ^ { - 1 }$


Correct Option: A

Two monochromatic light waves of amplitudes $A$ and $2A$ interfering at a point, have a phase difference of ${60^0}.$ The intensity at that point will be  proportional to :

  1. $3{A^2}$

  2. $5{A^2}$

  3. $7{A^2}$

  4. $9{A^2}$


Correct Option: C

Two sound waves with wavelength $5$m and $5.5$ m respectively. each propoggate in a gas with velocity $300$ m/s. we expect the following number of beats per second

  1. 12

  2. 0

  3. 1

  4. 6


Correct Option: D

Two waves are approaching each other with a velocity of $16\, m/s$ and frequency $n$. the distance between two consecutive nodes is 

  1. $\dfrac{16}{n}$

  2. $\dfrac{8}{n}$

  3. $\dfrac{n}{16}$

  4. $\dfrac{n}{8}$


Correct Option: A