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A parallel plate capacitor of plate area A plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates.If Q, E and W denotes respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system in the process of inserting the slab, then
The correct answer is:
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In the circuit shown in the diagram, E is the e.m.f of the cell, connected to two resistances, each of magnitude R and a capacitor of capacitance C as shown in the diagram. If the switch key K is closed at time t = 0, the growth of potential V across the capacitor will be correctly given by
In the circuit shown in the diagram, E is the e.m.f of the cell, connected to two resistances, each of magnitude R and a capacitor of capacitance C as shown in the diagram. If the switch key K is closed at time t = 0, the growth of potential V across the capacitor will be correctly given by
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Two parallel plate capacitors with area A are connected through a conducting spring of natural length l in series as shown. Plates P and S have fixed positions at separation d. Now the plates are connected by a battery of emf x as shown. If the extension in the spring in equilibrium is equal to the separation between the plates, find the spring constant k
Two parallel plate capacitors with area A are connected through a conducting spring of natural length l in series as shown. Plates P and S have fixed positions at separation d. Now the plates are connected by a battery of emf x as shown. If the extension in the spring in equilibrium is equal to the separation between the plates, find the spring constant k
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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
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n conducting plates are placed face to face. Distance between two consecutive plates is d. Area of plates is 
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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
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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?
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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
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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
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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
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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
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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
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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
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