Three bodies A, B and C have equal kinetic energies and their masses are $$400\mathrm{g},1.2\mathrm{kg}$$ and $$1.6\mathrm{kg}$$ respectively. The ratio of their linear momenta is :

A body of mass $$50\mathrm{kg}$$ is lifted to a height of $$20\mathrm{m}$$ from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be :

A body is moving unidirectionally under the influence of a constant power source. Its displacement in time t is proportional to :

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be :

A bullet of mass $$50\mathrm{g}$$ is fired with a speed $$100\mathrm{m}/\mathrm{s}$$ on a plywood and emerges with $$40\mathrm{m}/\mathrm{s}$$. The percentage loss of kinetic energy is :

Four particles $$A,B,C,D$$ of mass $$\frac{m}{2},m,2m,4m$$, have same momentum, respectively. The particle with maximum kinetic energy is :

A parallel plate capacitor has a capacitance $\mathrm{C}=200\mathrm{pF}$. It is connected to $230\mathrm{V}$ ac supply with an angular frequency $300\mathrm{rad}/\mathrm{s}$. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

In series LCR circuit, the capacitance is changed from $C$ to $4 C$. To keep the resonance frequency unchanged, the new inductance should be:

Primary side of a transformer is connected to $$230\mathrm{V},50\mathrm{Hz}$$ supply. Turns ratio of primary to secondary winding is $$10:1$$. Load resistance connected to secondary side is $$46\Omega$$. The power consumed in it is :

An AC voltage $$V=20\sin 200\pi t$$ is applied to a series LCR circuit which drives a current $$I=10\sin \left(200\pi t+\frac{\pi }{3}\right)$$. The average power dissipated is:

A capacitor of capacitance $$100\mu \mathrm{F}$$ is charged to a potential of $$12\mathrm{V}$$ and connected to a $$6.4\mathrm{mH}$$ inductor to produce oscillations. The maximum current in the circuit would be :

In an a.c. circuit, voltage and current are given by: $$V=100\sin (100t)V$$ and $$I=100\sin \left(100t+\frac{\pi }{3}\right)\mathrm{mA}$$ respectively. The average power dissipated in one cycle is:

An alternating voltage $$V(t)=220\sin 100\pi t$$ volt is applied to a purely resistive load of $$50\Omega$$. The time taken for the current to rise from half of the peak value to the peak value is:

Primary coil of a transformer is connected to $$220\mathrm{V}$$ ac. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistances shown in figure. The output voltage $$\left(V_0\right)$$ is :

A series L.R circuit connected with an ac source $$E=(25\sin 1000t)V$$ has a power factor of $$\frac{1}{\sqrt{2}}$$. If the source of emf is changed to $$\mathrm{E}=(20\sin 2000\mathrm{t})\mathrm{V}$$, the new power factor of the circuit will be :

A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb :

In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to : A. pure inductor. B. pure capacitor. C. pure resistor. D. combination of an inductor and capacitor. Choose the correct answer from the options given below :

Match List I with List II .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} LIST I LIST II A. Purely capacitive circuit I. B. Purely inductive circuit II. C. LCR series at resonance III. D. LCR series circuit IV. Choose the correct answer from the options given below:

A coil of negligible resistance is connected in series with $$90\Omega$$ resistor across $$120\mathrm{V},60\mathrm{Hz}$$ supply. A voltmeter reads $$36\mathrm{V}$$ across resistance. Inductance of the coil is :

A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now: