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Easy
Question
A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dieletric constant K 2 = .The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed v, the time constant as a function of time t is
The correct answer is:
Related Questions to study
Physics-
For the circuit shown, which of the following statement is true
For the circuit shown, which of the following statement is true
Physics-General
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n conducting plates are placed face to face. Distance between two consecutive plates is d. Area of plates is 
dielectric slab of dielectric constant k is inserted between the first and second plates and the assembly is charged by a battery of emf x. Find the charge stored in the assembly
n conducting plates are placed face to face. Distance between two consecutive plates is d. Area of plates is dielectric slab of dielectric constant k is inserted between the first and second plates and the assembly is charged by a battery of emf x. Find the charge stored in the assembly
Physics-General
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A source is moving across a circle given by the equation 
in with constant speed 
clockwise sense A detector is stationary at the point ( 2R, 0 ) w.r.t. the centre of the circle. The frequency emitted by the source is 
The coordinates of the source when the detector detects mimimum frequency is
A source is moving across a circle given by the equation in with constant speed clockwise sense A detector is stationary at the point ( 2R, 0 ) w.r.t. the centre of the circle. The frequency emitted by the source is The coordinates of the source when the detector detects mimimum frequency is
Physics-General
Physics-
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Two pulses are emitted by sources at S and | S . What is the time lag by which observer observe them?
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Two pulses are emitted by sources at S and | S . What is the time lag by which observer observe them?
Physics-General
Physics-
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Find the time 
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Find the time
Physics-General
Physics-
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Find the time 
Observe O is ahead by L from source S which are moving along same line with velocities V0 and VS respectively. The speed of sound is V. The source emits a wave pulse that reaches the obsever in time t1
At time t=T, the source reaches at | S . It is obvious that the observer will not be at O this time. The source emits a wavepulse at this time to reach the observer in time t2, which is measured from t=0
Find the time
Physics-General
Physics-
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
The maximum intensity produced at D is given by
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
The maximum intensity produced at D is given by
Physics-General
Physics-
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
If a minima is formed at the detector then, the magnitude of wavelength of the wave produced is given by
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
If a minima is formed at the detector then, the magnitude of wavelength of the wave produced is given by
Physics-General
Physics-
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
If a maxima is formed at a detector then, the magnitude of wavelength of the wave produced is given by
A narrow tube is bent in the form of a circle of radius R, as shown in the figure. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed
If a maxima is formed at a detector then, the magnitude of wavelength of the wave produced is given by
Physics-General
Physics-
Superposition of waves results in maximum and minimum of intensities such as in case of standing waves. This phenomenon is called as interference. Another type of superposition result in interference in time which is called as beats. In this case waves are analyzed at a fixed point as a function of time. If the two waves are of nearby same frequency are superimposed, at a particular point, intensity of combined waves gives a periodic peak and fall. This phenomenon is beats. If w1 and w2 are the frequencies of two waves then by superimposed y = y1 + y2 , we get at x = 0, 
Thus amplitude frequency is small and fluctuates slowly. A beat i.e., a maximum of intensity occurs, also intensity depends on square of amplitude. The beat frequency is given by 
Number of beats per second is called as beat frequency. A normal ear can detect only upto 10 Hz of frequency because of persistence of ear The frequency of beats produced in air when two sources of sound are activated, one emitting wavelength 32 cm, other 32.2 cm is 
Superposition of waves results in maximum and minimum of intensities such as in case of standing waves. This phenomenon is called as interference. Another type of superposition result in interference in time which is called as beats. In this case waves are analyzed at a fixed point as a function of time. If the two waves are of nearby same frequency are superimposed, at a particular point, intensity of combined waves gives a periodic peak and fall. This phenomenon is beats. If w1 and w2 are the frequencies of two waves then by superimposed y = y1 + y2 , we get at x = 0, Thus amplitude frequency is small and fluctuates slowly. A beat i.e., a maximum of intensity occurs, also intensity depends on square of amplitude. The beat frequency is given by Number of beats per second is called as beat frequency. A normal ear can detect only upto 10 Hz of frequency because of persistence of ear The frequency of beats produced in air when two sources of sound are activated, one emitting wavelength 32 cm, other 32.2 cm is
Physics-General
Physics-
A traveling wave on stretched string can be understood by the function y = f(x -vt). Here v is the wave speed ‘x’ is co-ordinate of point and ‘y’ is its instantaneous displacement. To describe the wave completely, we must specify the function f. If the wave moves in negative x-direction y (x, t) = f(x + vt) and if it moves in positive x-direction y (x, t) = f(x - vt). The general relation for a traveling wave must satisfy the relation 
, if plane wave exists. The particle velocity and wave velocity are related by Vpa = - (slope) (wave velocity ). Answer the following questionsConsider the snapshot of a wave traveling in positive x-direction
A traveling wave on stretched string can be understood by the function y = f(x -vt). Here v is the wave speed ‘x’ is co-ordinate of point and ‘y’ is its instantaneous displacement. To describe the wave completely, we must specify the function f. If the wave moves in negative x-direction y (x, t) = f(x + vt) and if it moves in positive x-direction y (x, t) = f(x - vt). The general relation for a traveling wave must satisfy the relation , if plane wave exists. The particle velocity and wave velocity are related by Vpa = - (slope) (wave velocity ). Answer the following questionsConsider the snapshot of a wave traveling in positive x-direction
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Radio waves coming at angle a to vertical are received by a ladder after reflection from a nearby water surface and also directly. What can be height of antenna from water surface so that it records a maximum intensity (a maxima) (wavelength = l )
Radio waves coming at angle a to vertical are received by a ladder after reflection from a nearby water surface and also directly. What can be height of antenna from water surface so that it records a maximum intensity (a maxima) (wavelength = l )
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Physics-
Two coherent narrow slits emitting wave length l in the same phase are placed parallel to each other at a small separation of 2 l , the sound is detected by moving a detector on the screen S at a distance D (>> l ) from the slit S1 as shown in figure.Find the distance x such that the intensity at P is equal to the intensity at O
Two coherent narrow slits emitting wave length l in the same phase are placed parallel to each other at a small separation of 2 l , the sound is detected by moving a detector on the screen S at a distance D (>> l ) from the slit S1 as shown in figure.Find the distance x such that the intensity at P is equal to the intensity at O
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Physics-
Sound from two coherent sources S1 and S2 are sent in phase and detected at point P equidistant from both the sources. Speed of sound in normal air is V0 , but in some part in path S1 , there is a zone of hot air having temperature 4 times, the normal temperature, and width d. What should be minimum frequency of sound, so that minima can be found at P?
Sound from two coherent sources S1 and S2 are sent in phase and detected at point P equidistant from both the sources. Speed of sound in normal air is V0 , but in some part in path S1 , there is a zone of hot air having temperature 4 times, the normal temperature, and width d. What should be minimum frequency of sound, so that minima can be found at P?
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Physics-
Figure shows two snapshots of medium particles within a time interval of 1/60 s. Find the possible time periods of the wave
Figure shows two snapshots of medium particles within a time interval of 1/60 s. Find the possible time periods of the wave
Physics-General
