ECAT Pre General Science Physics Chapter 12 Electrostatics Online Test With Answers

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ECAT Pre General Science Physics Chapter 12 Electrostatics Online Test

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1 A 10 F capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 volt.The capacitance of second capacitor is 10<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 224);">μ</span>F 20<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 224);">μ</span>F 30<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 224);">μ</span>F 15<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 224);">μ</span>F
2 A certain charge liberates 0.8 g of oxygen. The same charge will liberate. how many g of silver? 108 g 10.8 g 0.8 g 108/0.8 g
3 In a voltmeter the conduction takes place due to Electrons only Holes only Electrons and holes Electrons and ions
4 A conducting wire is drawn to double its length. Final resistivity of the material will be Double of the original one Half of the original one One fourth of the original one Same as original one
5 A piece of fuse wire melts when a current of 15 ampere flows through it. With this current. If it dissipates 22.5 W, the resistance of fuse wire will be Zero 10<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>Ω</b></span> 1<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>Ω</b></span> 0.10<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>Ω</b></span>
6 If 2.2 kilowatt power is transmitted through a 10 ohm line at 22000 volt, the power loss in the form of heat will be 0.1 watt 1 watt 10 watt 100 watt
7 The conductivity of a superconductor is Infinite Very large Very small Zero
8 If 2.2 kilowatt power is transmitted through 1 10 ohm line at 22000 volt, the power loss in the form of heat will be 0.1 watt 1 watt 10 watt 100 watt
9 A 50 volt battery is connected across 10 ohm resistor. The current is 4.5 A. The internal resistance of the battery is Zero 0.5<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>Ω</b></span> 1.1<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>Ω</b></span> 5.0<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>Ω</b></span>
10 A (100 W , 200 W) bulb is connected to a 160 V power supply. The power consumption would be 64 W 80 W 100 W 125 W
11 A wire of radius r has resistance R. If it is stretched to a wire of r/2 radius, then the resistance becomes 2R 4R 16R Zero
12 Two electric bulbs of 200 W and 100 W have same voltage. If R1and R2be their resistance respectively then R<sub>1</sub>= 2R<sub>2</sub> R<sub>2</sub>= 2R<sub>1</sub> R<sub>2 </sub>=<sub> </sub>4R<sub>1</sub> R<sub>1</sub>= 4R<sub>2</sub>
13 A ten ohm electric heater operates on a 110 V line. Calculate the rate at which it develops heat in watts 1310 W 670 W 810 W 1210 W
14 5<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μF</b></span> 10<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μF</b></span> 3<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μF</b></span> 6<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μF</b></span>
15 Taking the earth to be a spherical conductor of diameter 12.8 x 103km. Its capacity will be 711<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μ</b></span>F 611<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μ</b></span>F 811<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μ</b></span>F 511<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μ</b></span>F
16 The nature of capacity of electrostatic capacitor depends on Shape Size Thickness of plates Area
17 A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor Increases Decreases Remain unchanged Becomes infinite
18 The energy required to charge a capacitor of 5μF by connecting D.C. source of 20 KV is 10 KJ 5 KJ 2 KJ 1 KJ
19 When a dielectric material is introduced between the plates of a charged condenser the electric field between the plates Decreases Increases No change May increase or decresase
20 A capacitor of capacity 1μF is charged to 1 KV. The energy stored in J 5 0.5 0.005 50
21 If the distance between the plates of a parallel plate condenser of capacity 10μF is doubled then new capacity will be 5<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μ</b></span>F 20<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μ</b></span>F 10<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μ</b></span>F 15<span style="color: rgb(34, 34, 34); font-family: &quot;Times New Roman&quot;; font-size: 24px; text-align: center; background-color: rgb(255, 255, 248);"><b>μ</b></span>F
22 The capacity of a parallel plat capacitor depends on the Type to metal used Thickness of plates Potential applied across the plates Separation between the plates
23 In a charged capacitor the energy is stored in Both in positive and negative charges Positive charges The edges of the capacitor plates The electric field between the plates
24 A condenser of capacity 50μF is charged to 10 V.The energy stored is 1.25 x 10<sup>-3</sup>J 3.75 x 10<sup>-3</sup>J 2.5 x 10<sup>-3</sup>J 5 x 10<sup>-3</sup>J
25 A metal plate of thickness half the separation between the capacitor plates of capacitance C is inserted. The new capacitance is C C/2 Zero 2C
26 A one microfarad capacitor of a TV is subjected to 4000 V potential difference. The energy stored in capacitor is 8 J 16 J 4 x 10<sup>-3</sup>J 2 x 10<sup>-3</sup>J
27 A medium of dielectric constant 'K' is introduced between the plates of parallel plate condenser. As a result its capcitance Increase k time Decreases k times Decreases 1/K times Remains unchanged
28 Force acting upon a charged particle kept between the plates of a charged condenser if F. IF one of the plates of the condenser is removed, force acting on the same will become Zero F/2 F 2F
29 A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is Charge Q Potential V Capacity Energy U
30 If we increase the distance between two plates of the capacitor, the capacitance will Increase Decrease Remain same First increase then decrease
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