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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1151. |
A capacitor is an arrangement for storing large amounts of electric charge and hence electric energy in a small space. The electrical capacitance of a capacitor is related to its abillity to store electric charge. We define capacity of a conductor as the ratio of charge Q given to the conductor to the rise in its potential, V i.e., `C = Q//V`. The capacity of an isolted spherical conductor of radius r is `C = 4pi in_(0) r`. In case of a parallel plate capacitor, `C = (in_(0)A)/(d)` where A is area of insulated metal plate and d is distance between the plates. Clearly, capacity depends on size of capacitor. When different capacitors are connected in series, capacity, `C_(s) = (C_(1) C_(2))/(C_(1) + C_(2))` and when capacitors are connected in parallel, `C_(p) = C_(1) + C_(2)` Read the above passage and answer the following questions : (i) From `C = (Q)/(V)`, we find that C can be increased Q or decresing V. Do you agree ? (ii) Capacity of a capacitor is fixed depending on its geometry and the medium used. Is it true ? (iii) Calculate the capacity of a condenser which when connected in series with a conductor of `12 muF` gives us a capacitance of `3 muF`. (iv) What values of life do yo+-earn from this study ? |
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Answer» No, we do agree. This is because capacitance C is fixed, depending on geometry fo condeaser. If we increase Q , V increases but C remains constant and vice-versa. (ii) yes, the statement is true. (iii) Here, `C_(1) = ? C_(2) = 12 muF, C_(s) = 3 muF` From `C_(s) = (C_(1) C_(2))/(C_(1) + C_(2))` `3 = (C_(1) xx 12)/(C_(1) + 12) or 12C_(1) = 3C_(1) + 36 C_(1) = 4muF` (iv) Just as capacity fo a condenser is fixed depending on its geometry, similarly capacity of a person to execuite some work is fixed depending on his buit up. Capacity increases by connecting the capacitors in parallel. exactly in the same way capacity of a group of persons can be increased when each one of them is connected to the same sources , e.g., guide or instractor. This is the implication of this concept in day to day life. |
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| 1152. |
An infinite number of charges each numerically equal to q and of the same sign are placed along the x-axis at `x = 1, x = 2, x = 4, x = 8` and so on. Find electric potential at `x=0`. |
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Answer» Using superpositon principle, we may write electric potential at the origin `(x = 0)` due to various cahrges as `V = (1)/(4pi in_(0)) [(q)/(1) + (q)/(2) + (q)/(4) + (q)/(8) + …….]` `V = (q)/(4pi in_(0)) [ (1)/(1) + (1)/(2) + (1)/(2^(2)) + (1)/(2^(3)) + ......]` As sum of infinite G.P. serious , `S = (a)/(1 - r)` , where a is first term and r is common ration. `V = (q)/(4pi in_(0)) {(1)/((1-1//2))} = (2q)/(4pi in_(0))` |
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| 1153. |
Indenify X in the following nuclear reactions (in the first reaction, n represents a neutron) : (a) `_(1)H^(1) + _(4)Be^(9) rarr X + _(o)n^(1)`, (b) `_(6)C^(12) + _(1)H^(1) rarr X `, (c ) `_(7)N^(15) + _(1)H^(1) rarr _(2)He^(4) + X` |
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Answer» Correct Answer - (a) `_(5)B^(9)`, (b) `_(7)N^(13)`, (c ) `_(6)C^(12)` Using the law of conservation of mass number and charge number, (a) Let `._(Z)X^(A)` As `1+9 = A + a :.` As `A = 9` As `1+4 = Z + 0 :. Z = 5` Hence, `._(Z)X^(A)` is `._(5)5^(9)` is `._(5)B^(9)` (b) Let `._(Z)X^(A)` (b) Let `._(Z)X^(A)` As `12+1 = A :. A = 13` As `6+1 = Z :. Z = 7` Hence, `._(Z)X^(A) ` is `._(7)N^(13)` (c) Let `._(Z)X^(A)` As `15+1 = 4 + A :. A = 12` As `7+1 = 2 + Z :. Z = 6` Hence `._(Z)X^(A)` is `._(6)C^(12)` |
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| 1154. |
A small ball of paper has mass `9xx10^(-5) kg` and carries a charge of `5muC`. When it is held over another charged ball of paper at a distance of 2cm above it, the two balls stay in equilibrium. What is the charge on the second hall ? |
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Answer» Correct Answer - `7.84xx10^(-12)C` `q_(1) = 5 mu C = 5xx10^(-6) C, q_(2) = ?` `r = 2cm = 2xx10^(-2) m` `m= 9xx10^(-5) kg`. `F = (q_(1) q_(2))/(4pi in_(0) r^(2)) = mg` `(9xx10^(9)xx5xx10^(-6) q_(2))/((2xx10^(-2))^(2)) = 9xx10^(-5)xx9.8` `q_(2) = 7.84xx10^(-12)C` |
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| 1155. |
Two equally charged particles, held `3.2xx10^(-3) m` apart, are released from rest. The initial accelerartion of the first particle is observed to be `7.0 m//s^(2)` and that of the secound to be `9.0 m//s^(2)`. If the mass of the first particle is `6.3xx10^(-7) kg`, what are (a) the mass of the secound particle adn (b) teh magnitude of the charge of each particle ? |
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Answer» Correct Answer - `= 4.9xx10^(-7) kg, q = 7.1xx10^(-11)C` Here, `a_(1) = 7.0 m//s^(2), a_(2) = 9.0 m//s^(2)`, `m_(1) = 6.3xx10^(-7) kg , m_(2) = ?` As `F_(1) = F_(2)` `:. m_(1) a_(1) = m_(2) a_(2)` `m_(2) = (m_(1) a_(1))/(a_(2)) = (6.3xx10^(-7)xx7.0)/(9.0)` `= 4.9xx10^(-7) kg` As `F_(1) = F_(2) = (q_(1) q_(2))/(4pi in_(0) r^(2)) = m_(1) a_(1)` `= 6.3xx10^(-7) xx7.0 = 44.1xx10^(-7)` `:. (9xx10^(9) q^(2))/((3.2xx10^(-3))^(2)) = 44.1xx10^(-7)` `:. q = 7.1xx10^(-11)C` |
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| 1156. |
Uniform electric field of magnitude `100 Vm^-1` in space is directed along the line `y = 3 + x`. Find the potential difference between points `A (3,1) and B (1,3)`.A. ` 100 V`B. ` 200 sqrt 2 V`C. ` 200 V`D. ` 0` |
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Answer» Correct Answer - D ` vec E = 100 cos 45^@ hat I + 100 sin 45^@ hat i` ` vec E= 50 sqrt 2 hat I + 50 sqrt 2 hat j` ` d vec r =- 2 hat I + 2 hat j` ` dV =- vec E. dvec r` ` dV=0` |
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| 1157. |
The SI unit of electric flux is (a) volt metre2 (b) newton per coulomb (c) volt metre (d) joule per coulomb |
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Answer» Correct answer is (c) volt metre |
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| 1158. |
The point charges `+ q, -2q` and `+ q` are placed at point `(x = 0, y = a, z = 0), (x = 0, y = 0, z = 0)` and `(x = a, y = 0, z = 0)`, repectively. The magnitude and direction of the electric dipole moment vector of this charge assembly areA. `sqrt(2qa)` along + y -directionB. `sqrt2qa` along the line joining points `(x = 0, y = 0, z = 0)` and `(x = a, y = a, z = 0)`C. `qa` along the line joining point `(x = 0, y = 0, z = 0)` and `(x = a, y = a, z = 0)`D. `sqrt2qa` along + x-direction |
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Answer» Correct Answer - B |
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| 1159. |
Three charges +5C, +7C and -4C are situated within a closed surface and charges -5C, -7C and +4C are situated outisde the surace what is the T.N.E.I. over the closed surface?A. `-8C`B. 0C. `+8C`D. 10 C |
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Answer» Correct Answer - C T.N.E.I. = `q_(1)+q_(2)+q_(3)` = 5 + 7 - 4 = 8C T.N.E.I. depends upon charge inside the surface only. |
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| 1160. |
State conditions for electrostatic force to be attractive or repulsive. |
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Answer» 1. The force between the two charges will be attractive, if the charges are unlike (one positive and one negative). 2. The force between the two charges will be repulsive, if the charges are similar (both positive or both negative). |
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| 1161. |
The safest way to protect yourself from lighting is to be inside a car. Comment. |
| Answer» The body of the car is metallic. It provides electrostatic shielding to the person in the car, because electric field inside the car is zero. The discharging due to lighting passes to the ground through the metallic body of car. | |
| 1162. |
The safest way to protect yourself from lightening is to be inside a car. Justify. |
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Answer» There is danger of lightning strikes during a thunderstorm. Because trees are taller than people and therefore closer to the clouds above, they are more likely to get hit by lightnings. Similarly, a person standing in open ground is the tallest object and more likely to get hit by a lightning. But car with a metal body is an almost ideal Faraday cage. When a car is struck by lightning, the charge flows on the outside surface of the car to the ground but the electric field inside remains zero. This leaves the passengers inside unharmed. |
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| 1163. |
You are travelling in a car during a thunder storm. In order to protect yourseft from lightening, would you prefer to :A. Remain in the carB. Take shelter under a treeC. Get out and be flat on the groundD. Touch the nearest electrical pole |
| Answer» Correct Answer - A | |
| 1164. |
Assertion: The lightening conductor at the top of a high building has sharp ends. Reason: The surface density of charge at sharp points is very high, resulting in setting up of electric wind.(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.(c) If assertion is true but reason is false.(d) If the assertion and reason both are false.(e) If assertion is false but reason is true. |
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Answer» (a) Both assertion and reason are true and the reason is the correct explanation of the assertion. |
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| 1165. |
Check that the ratio `ke^(2)//Gm_(e) m_(p)` is dimenionless, Look up a table of Physical Constants and determine the value of thisi ratio. What does the ratio signigy ? |
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Answer» `(ke^(2))/(G m_(e) m_(p)) = ([N m^(2) C^(-2)] [C^(2)])/([N m^(2) kg^(-2)] [kg] [kg]) = 1 = (M^(@) L^(@)T^(@))` `:.` The given ratio dimenesionless. using `k = 9xx10^(9) N m^(2) C^(-2), e = 1.6xx10^(-19)C, G = 6.67xx10^(-11) N m^(2) kg^(-2)`, `m_(e) = 9.1xx10^(-31) kg, and m_(p) = 1.66xx10^(-27) kg`, we get , `(ke^(2))/(G m_(e) m_(p)) = 2.29xx10^(39)` This is the ratio of electrostatic force to gravitational force between an electron and a proton. |
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| 1166. |
A circuit shown in the figure consists of a battery of emf `10V` and two capacitance `C_(1)` and `C_(2)` of capacitances `1.0muF` and `2.0muF` respectively. The potential difference `V_(A) - V_(B)` is `5V` A. charge on capacitor `C_(1)` is equal to charge on Capacitor `C_(2)`B. voltage across capacitor `C_(1)` is 5 VC. voltage across capacitor `C_(2)` is 10 VD. energy stored in capacitor `C_(1)` is two times the energy stored in capacitor `C_(2)` |
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Answer» Correct Answer - A::D |
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| 1167. |
In the uniform electric field shown in figure, find : a. `V_A-V_D` b. `V_A-V_C` c. `V_B-V_D` d. `V_C-V_D` |
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Answer» Correct Answer - A::B::C::D `V_A =V_D and V_B=V_C `as the points A and D or B and C are lying on same equipotential surface (`_|_` to electric field lines). Further`V_A or V_D gtV_B or V_C` as electric lines always flow from higher potential to lower potential. `V_A-V_B=V_D-V_C=Ed` `=(20)(1)=20V` |
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| 1168. |
Point `P` lies on the axis of a dipole . If the dipole is rotated by ` 90^@` anti-clock wise , the electric field vector ` vec E` at `P` will rotate by .A. `90^(@)` clockwiseB. `180^(@)` clockwiseC. `90^(@)` anticlockwiseD. `180^(@)` anticlockwise |
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Answer» Correct Answer - A |
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| 1169. |
An electron of mass m and charge q is accelerated from rest in a uniform electric field of strength E. The velocity acquired by it as it travels a distance l isA. `[(2Eql)/m]^(1//2)`B. `[(2Eq)/(ml)]^(1//2)`C. `[(2Em)/(ql)]^(1//2)`D. `[(Eq)/(ml)]^(1//2)` |
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Answer» Correct Answer - A `v=sqrt((2qV)/m)(V=E.l)` `therefore" "v=sqrt((2qEl)/m)` `v=((2qEl)/m)^(1//2)` |
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| 1170. |
The electric field at a point due to an electric dipole, on an axis inclined at an angle `theta ( lt 90^(@))` to the dipole axis, is perpendicular to the dipole axis, if the angle `theta` isA. `tan^(-1) (2)`B. `tan^(-1) ((1)/(2))`C. `tan^(-1) (sqrt2)`D. `tan^(-1) ((1)/(sqrt2))` |
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Answer» Correct Answer - C |
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| 1171. |
If `10^(10)` electrons are acquired by a body every second, the time required for the body to get a total charge of C will beA. 2hB. 2 daysC. 2 yrD. 20 yr |
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Answer» Correct Answer - D |
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| 1172. |
In which orientation,a diole placed, in a uniform field is in (i) stable (ii) unstable equilibrioum? |
| Answer» If `theta` is angle between `vec(P)` and `vec(E)`, then `theta = 0^(@)` fpr stable equilibrium, `theta = 180^(@)` for unstable equibrium. | |
| 1173. |
An electric dipole is held at an angle `theta` in a uniform electric field E. Will there be any (i) net translating force (ii) torque acting on in ? Explain. |
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Answer» (i) There will be no net translating force. (ii) Torque on the dipole `: tau = pE sin theta`. It will align the dipole along the filed. Once the dipole in aligned, `tau = 0`. For detail, refer to `Art.1 (b) .17` |
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| 1174. |
Two charges `q_(1)` and `q_(2)` are placed `30 cm` apart, as shown in the figure. A third charge `q_(3)` is moved along the arc of a circle of radius `40 cm` from `C` to `D`. The change in the potential energy o fthe system is `(q_(3))/(4pi epsilon_(0))k`., where `k` is A. `8q_(2)`B. `8q_(1)`C. `6q_(2)`D. `6q_(1)` |
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Answer» Correct Answer - A |
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| 1175. |
A test charge q is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in Fig. (i) Calculate the potential difference between A and C (ii) At what point [of A and C] is the electric potential more and why? |
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Answer» (i) p.d. does not depend upon the path along which the test charge moves. `:. E = - (dV)/(dx) = (-(V_(C) - V_(A)))/(2-6) = (V_(C) - V_(A))/(4)` or `V_(C) - V_(A) = 4E`. Therefore, `V_(C) - V_(A)` (ii) Direction of electric field is in the direction of decreasing potential. So, `V_(C) gt V_(A)` |
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| 1176. |
Two charges `q_(1)` and `q_(2)` are placed `30 cm` apart, as shown in the figure. A third charge `q_(3)` is moved along the arc of a circle of radius `40 cm` from `C` to `D`. The change in the potential energy o fthe system is `(q_(3))/(4pi epsilon_(0))k`., where `k` is A. `8 q_(2)`B. `8 q_(1)`C. `6 q_(2)`D. `8q_(1)` |
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Answer» Correct Answer - 1 `U_(1) = (1)/(4 pi in_(0)) [ (q_(1) q_(3))/(0.4) + (q_(2) q_(3))/(0.5)]` `U_(2) = (1)/(4 pi in_(0)) [(q_(1) q_(3))/(0.4) + (q_(2) q_(3))/(0.1)]` `U_(2) - U_(1) = (1)/(4 pi in_(0)) q_(2) q_(3) [ (1)/(0.1) - (1)/(0.5)]` `= (q_(3))/(4 pi in_(0)) (8q_(2)) = (q_(3))/( 4 pi in_(0)) k` `k = 8 q_(2)` |
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| 1177. |
A test charge `q_(0)` is moved without acceleration from A to C over the path ABC as shown in Fig. Calcualate potential difference beetween A and C. |
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Answer» As is clear from Fig. `vec(AB) + vec(BC) = vec(AC)` As `E = - (dV)/(dr) = (-(V_(C) - V_(A)))/(d) = (V_(A) - V_(C))/(d)` `:. V_(A) - V_(C) =E.d.` |
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| 1178. |
Four dipoles each of magnitudes of charges `+-e` are placed inside a sphere. The total flux of E coming out of the sphere isA. zeroB. `(4e)/epsilon_0`C. `(8e)/epsilon_0`D. none of these |
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Answer» Correct Answer - A `phi=q_("in")/epsilon_0` `q_("in")=0` `:. phi =0` |
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| 1179. |
Let `V_(0)` be the potential at the origin in an electric field `vecE=E_(x)hati+E_(y)hatj`. The potential at the point `(x,y)` isA. `V_(0)-xE_(x)-yE_(y)`B. `V_(0)+xE_(x)+yE_(y)`C. `xE_(x)+yE_(y)-V_(0)`D. `(sqrt(x^(2)+y^(2)))sqrt(E_(x)^(2)+E_(y)^(2))-V_(0)` |
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Answer» Correct Answer - A |
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| 1180. |
What are polar molecules? Explain with examples. |
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Answer» 1. Polar molecules are the molecules in which the centre of positive charge and the negative charge is naturally separated. 2. Molecules of water, ammonia, sulphur dioxide, sodium chloride etc. have an inherent separation of centres of positive and negative charges. Such molecules are called polar molecules. |
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| 1181. |
What are non-polar molecules? Explain with examples. |
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Answer» i. Non-polar molecules are the molecules in which the centre of positive charge and the negative charge is one and the same. They do not have a permanent electric dipole. When an external electric field is applied to such molecules, the centre of positive and negative charge are displaced and a dipole is induced. ii. Molecules such as H2, CI2, CO2, CH4, etc., have their positive and negative charges effectively centred at the same point and are called nonpolar molecules. |
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| 1182. |
Which of the following is example of polar molecule ?A. `H_(2)O`B. `O_(2)`C. `H_(2)`D. `CO_(2)` |
| Answer» Correct Answer - A | |
| 1183. |
Non polar molecules have donot permanentA. forceB. pressureC. electric dipole momentD. torque |
| Answer» Correct Answer - C | |
| 1184. |
A molecule in which centre of gravity of positive nuclei and revolving electrons coincide isA. polar moleculeB. non polar moleculeC. bipolar moleculeD. unipolar molecule |
| Answer» Correct Answer - B | |
| 1185. |
Polar molecules have permanentA. resistanceB. currentC. electric dipole momentD. magnetic flux |
| Answer» Correct Answer - C | |
| 1186. |
A molecule in which centre of gravity of positive nuclei and revolving electrons do not coincide isA. polar moleculeB. non polar moleculeC. bipolar moleculeD. unipolar molecule |
| Answer» Correct Answer - A | |
| 1187. |
If the dielectric constant and dielectirc strength be denoted by `K` and `x` respectively, then a meterial suitable for use as a dielectric in a capacitor must haveA. high K and high XB. high K and low XC. low K and high XD. low K and low X |
| Answer» Correct Answer - A | |
| 1188. |
Let `[in_(0)]` denote the dimensional formula of the permittivity of vacuum. If `M = mass, L = length, T = time and A = elctric current`, then :A. ` [varepsilon_0] = M ^(-1) L^2 A^(-2)]`B. ` [varepsilon_0] = M ^(-1) L^2 A^(-2)]`C. ` [varepsilon_0] = M ^(-1) L^3 A^(-2)]`D. ` [varepsilon_0] = M ^(-1) L^3 A^(-2)]` |
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Answer» Correct Answer - D `(Energy)/(Volume 0 1/2 in_0 = 1/2 in_0 [F/q]^2` ` (ML^2 T^(-2))/L^2 = in_0 [(MLT^(-2)/(AT)]^2` `in_0 rarr M^(-1) L^(-3) T^4 A^2 1`. |
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| 1189. |
The dimension of `((1)/(2))epsilon_(0)E^(2)` (`epsilon_(0)` : permittivity of free space, E electric fieldA. ` MLT(-1)`B. ` ML^2 T^(-2)`C. ` MLT^(-2)`D. `MLT^-1)T^(-2)` |
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Answer» Correct Answer - A::B::C::D (a) `U=1/2 in_0 E^2 J//m^3` is energy stored per unit volume in capaciteo `[U] =(ML^(-1) T^(-2))` |
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| 1190. |
Find out the electrostalic force between two point charges placed in air (each of +1 C) if they are separately by 1 m. |
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Answer» `F_(e)=(kq_(1)q_(2))/r^(2)=(9xx10^(9)xx1xx1)/1^(2)=9xx10^(9)N` From the above result, we can say that 1 C charge is too large to realize. In nature, charge is usually of the order of `muC` |
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| 1191. |
charge conservation is always valid. Is it also true for mass ? |
| Answer» No, mass conservation is not always. In some nuclear reactions, some mass is lost nd it is converted into energy. | |
| 1192. |
What are the differences between charging by induction and charging by conduction ? |
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Answer» Major differences between two methods of charging are as follows : (i) In induction, two bodies are close to each other but do not touch each other while in conduction they touch each other. (Or they are connected by a metallic wire) (ii) In induction, total charge of a body remains unchanged while in conduction it changes. (iii) In induction, induced charge is always opposite in nature to that of source charge while in conduction charge on two bodies finally is of same nature. |
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| 1193. |
A capacitor of capacitance `2muF` is charged to a potential difference of `5V`. Now, the charging battery is disconected and the capacitor is connected in parallel to a resistor of `5Omega` and another unknown resistor of resistance `R` as shown in figure. If the total heat produced in `5Omega` resistance is `10muJ` then the unknown resistance `R` is equal to A. `10 Omega`B. `15 Omega`C. `(10//3)Omega`D. `7.5 Omega` |
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Answer» Correct Answer - C |
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| 1194. |
If at a certain stage during the charging of a capacitor of capacitance C. the charge and potential difference are q and e, the work dW required to transfer an additional amount of charge dq is (A) vdq(B) \(\cfrac{dq}v\)(C) \(\cfrac{vdq}C\)(D) \(\cfrac{q^2}{2C}\) |
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Answer» Correct option is (A) vdq |
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| 1195. |
Explain concept of charging by induction. |
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Answer» 1. If an uncharged conductor is brought near a charged body, (not in physical contact) the nearer side of the conductor develops opposite charge to that on the charged body and the far side of the conductor develops charge similar to that on the charged body. This is called induction. 2. This happens because the electrons in a conductor are free and can move easily in presence of charged body. 3. A charged body attracts or repels electrons in a conductor depending on whether the charge on the body is positive or negative respectively. 4. Positive and negative charges are redistributed and are accumulated at the ends of the conductor near and away from the charged body. 5. In induction, there is no transfer of charges between the charged body and the conductor. So when the charged body is moved away from the conductor, the charges in the conductor are free again. |
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| 1196. |
Explain concept of charging by conduction. |
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Answer» 1. When certain dissimilar substances, like fur and amber or comb and dry hair, are rubbed against each other, electrons get transferred to the other substance making them charged. 2. The substance receiving electrons develops a negative charge while the other is left with an equal amount of positive charge. 3. This can be called charging by conduction as charges are transferred from one body to another. |
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| 1197. |
Does in charging the mass of a body change? |
| Answer» Yes, as charging a body means addition or removal of electrons and electron has a mass. | |
| 1198. |
Does the attraction between the comb and the piece of papers last for longer period of time? |
| Answer» No, because the comb loses its net charge after some time. The exess charge of the comb transfers to earth through our body after some time. | |
| 1199. |
Can two similarly charged bodies attract each other? |
| Answer» Yes, when the charge on one body `(q_1)` is much greater than that on the other `(q_2)` and they are close enough to each other so that force of attraction between `q_1` and induced charge on the other exceeds the force of repulsion between `q_1 and q_2`. However, two similar point charges can never attraction each other because no induction will take place here. | |
| 1200. |
A leaf electroscope is a simple apparatus to detect any charge on a body. It consists of two metal leaves OA and OB, free to rotate about O. Initially both are very slightly separated. When a charged object is touched to the metal knob at the top of the conducting rod, charge flows knob to the leaves through the conducting rod. As the leaves are now charges similarly, they start repelling each other and get separated, (deflected by certain angle). The angle of deflection in static equilibrium is an indicator of the amount of charge on the charged body. When a + 20 C rod is touched to the knob, the deflection of leaves was `5^(@)`, and when an identical rod of -40 C is touched, the direction was found to be `9^(@)`. If an identical rod of +30 C is touched, then the deflection may be : |
| Answer» Correct Answer - C | |