Tag: physics

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Questions Related to physics

Multiple choice physics motion and types of motion different types of motion basics of motion motion and reference point

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. $v\, \geq\, \sqrt{2ad^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $v^2\geq\, 2ad\, \quad\, or,\, \quad\, v \geq\, \sqrt{2ad}$.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

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. $2\ ms^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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,
$f _1 -f _2 = 6$
$3(v _1 - v _2)= 6$
$\implies v _1 - v _2 = 2\ ms^{-1}$

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The wall acts as a moving reflector. The frequency of the sound reflected by the wall is f' = f * (v + u) / (v - u). The beat frequency is the difference between the reflected frequency and the source frequency, which simplifies to f_beat = f * (2u / (v - u)). Plugging in f=1700, u=0.06 m/s, and v=340 m/s gives 1700 * (0.12 / 339.94), which is approximately 0.6 Hz.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

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 $m s ^ { - 1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The beat frequency is 26 beats / 10 sec = 2.6 Hz. For an open pipe, f = v / (2L). The difference in frequencies is f1 - f2 = (v/2) * (1/L1 - 1/L2) = 2.6. Substituting L1 = 0.80 m and L2 = 0.81 m, we get (v/2) * (0.01 / (0.80 * 0.81)) = 2.6, which solves to v = 336.96 m/s, approximately 337 m/s.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

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. $9{A^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The resultant intensity I is given by I = I1 + I2 + 2 * sqrt(I1 * I2) * cos(phi). Since intensity is proportional to amplitude squared, I1 = k * A^2 and I2 = k * (2A)^2 = 4 * k * A^2. With phi = 60 degrees, cos(60) = 0.5. Thus, I = k * A^2 + 4 * k * A^2 + 2 * sqrt(k * A^2 * 4 * k * A^2) * 0.5 = 5 * k * A^2 + 2 * (2 * k * A^2) * 0.5 = 7 * k * A^2.

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Frequency f = v / lambda. f1 = 300 / 5 = 60 Hz. f2 = 300 / 5.5 = 54.54 Hz. The beat frequency is |f1 - f2| = |60 - 54.54| = 5.46 Hz, which is approximately 6 Hz.