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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.
| 751. |
A positive charge q is placed in front of conducting solid cube at a distance d from its centre. Find the electric field at the centre of the cube due to the charges appearing on its surface. |
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Answer» Correct Answer - ` q/ (4 pi varepsilon_0d^2)` towards the charge (q) `vec E_q + vec E_("induced") =0 rArr vec E_("induced") =- vec e_q` `vec E_("induced") = q/(4 pi in_0 d^2)` tiwards charge `q`. |
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| 752. |
When the separation between two charges is increased, the electric potential energy of the chargesA. increasesB. decreasesC. remains the sameD. may increase or decrease |
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Answer» Correct Answer - 4 `U = (1)/(4 pi in_(0)) .(q_(1) q_(2))/(r )` `q_(1)` and `q_(2)` may be `+ve or -ve`. |
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| 753. |
When a body is earth connected, electrons from the earth flow into the body. This means the body is `….`A. unchargedB. charged positivelyC. charged negativelyD. an insulator |
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Answer» Correct Answer - A |
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| 754. |
A ball of mass 1 kg carrying a charge `10^(-8)C` moves from a point A at potential 600 V to a point B at zero potential. The change in its kinetic energy isA. `-6xx10^(-6)"erg"`B. `-6xx10^(-6)J`C. `6xx10^(-6)J`D. `6xx10^(-6)"erg"` |
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Answer» Correct Answer - C As work is done by the fiel, KE of the body increased `therefore" "DeltaKE=W=q(V_(A)-V_(B))` `=10^(-8)(600-0)=6xx10^(-6)J` |
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| 755. |
5 J of work is done in moving a positive charge of 0.5C between two points. What is the potential difference between these points ? |
| Answer» Potential difference `= ("work done")/("charge") = (5)/(0.5) = 10V` | |
| 756. |
An electric field is given by `E=(y hati+x hatj)NC^(-1)`. The work done in moving a 1C charge from `r_(A)=(2hati + 2hatj)m" to "r_(B)=(4hati+2hatj)m` isA. 2yB. 3yC. zeroD. infinity |
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Answer» Correct Answer - A Here. Displacement would be, `Deltar=(4-2)hati+(2-2)hatj=2hati` Now, `DeltaV=E.Deltar=(yhati+xhatj).2hati=2y` `therefore" Work done, "W=q(DeltaV)=q.2y` `" "=1xx2y=2yJ" "(because" charge = 1C")` |
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| 757. |
A charge of `+1C` is placed at the centre of a spherical shell of radius 10cm. What will be the work done in moving a charge of `+1 muC` on its surface through a distance of 5 cm? |
| Answer» As the surface of spherical shell will be an equipotential surface, work doen in moving a charge of `+1 M=muC` through any distance on it will be zero. | |
| 758. |
When a `2 mu C` charge is carried from point A to point B, the amount of work done by the electric field is `50 mu J`. What is the potential difference and which point is at a higher potential ? |
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Answer» `V_(B) - V_(A) = (wo rk)/(charg e) = (50 muJ)/(2 muC) = 25 vol t`. Clearly, `V_(B) gt V_(A)`. |
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| 759. |
Two uniformly charged non-conducting hemispherical shells each having uniform charge density `sigma` and radius R from a complete sphere (not stuck together) and surround a concentric spherical conducting shell of radius R/2. If hemispherical parts are in equilibrium then minimum surface charge density of inner conducting shell is :A. `-2 sigma`B. `-sigma//2`C. `-sigma`D. `2 sigma` |
| Answer» Correct Answer - A | |
| 760. |
A capacitor is connected to a battery of 20 V, so that a charge of `100muC` is obtained at the plates. The capacitance of the capacitor isA. `6muF`B. `5muF`C. `9.5muF`D. `10muF` |
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Answer» Correct Answer - B `V=20V, Q=100muC, C=?` `C=Q/V=(100xx10^(-6))/20=5xx10^(-6)F=5muF` |
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| 761. |
Six identical capacitors are joined in parallel, charged to a potential difference of 10 V, separated and then connected in series. Then the potential difference between the free plates isA. 10 VB. 30 VC. 60 VD. `(10/6)V` |
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Answer» Correct Answer - C In parallel potential of each condenser is same `therefore" "V_(1)=V_(2)=V_(3)=V_(4)=V_(5)=V_(6)=10V` Now these condensers are connected is series. `therefore" "V=V_(1)=V_(2)=V_(3)=V_(4)=V_(5)=V_(6)` = 10 + 10 + 10 + 10 + 10 + 10 = 60 V |
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| 762. |
A uniform electric field `vec(E) = -E_(x) hat(i) N//C` for `x lt 0` exists. A right circular cylinder of length `l cm` and radius `r cm` has its centre at the origin and its axis along X-axis. Find out the net outward flux. What is the net charge within the cyclinder ? |
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Answer» Correct Answer - `2pi r^(2) E_(x) (10^(-4)) Nm^(2) C^(-1), 2 pi r^(2) E_(x) in_(0) (10^(-4)) C` Net. Outward flux, `phi_(E) = E_(x) pi ((r )/(100))^(2) + E_(x). pi ((r )/(100))^(2) + 0` `phi_(E) = 2pi r^(2) E_(x) (10^(-4)) Nm^(2) C^(-1)` `q = in_(0) phi_(E) = 2pi r^(2) in_(0) E_(x) (10^(-4)) C` |
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| 763. |
The electric field outside a charged long straight wire is given by `E = (1000)/(r ) V m^(-1)`, and is directed outwards. What is the sign of the charge on the wire ? If two points `A` and `B` are situated such that `r_(A) = 0.2 m` and `r_(B) = 0.4 m`, find the value of `(V_(B) - V_(A))`. |
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Answer» Correct Answer - `+` charge ; `-693.1 V` As the field is directed outwards, charge on the wire must be positive `V_(B) - V_(A) = - int_(A)^(B) vec(E). Vec(dl) = - int_(r = 0.2m)^(r = 0.4m) (1000)/(r ) dr` `V_(B) - V_(A) = -1000 [log_(e) r]_(r = 0.2m)^(r = 0.4m)` `= -1000 [log_(e) 0.4 - log_(e) 0.2]` `= -1000 [log_(e) (0.4)/(0.2)]` `= -1000 log_(e) 2` `V_(B) - V_(A) = -1000xx0.6931` `= -693.1` volt |
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| 764. |
A charged spherical conductor has a surface density of `0.07 C cm^(-2)`. When the charge is increased by `4.4 C`, the surface density changes to `0.084 C cm^(-2)`. Find the initial charge and capacitance of the spherical conductor. |
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Answer» Correct Answer - `5xx10^(3) V` Here, `V = ? q = 5 muC = 5xx10^(-6)C` `C = 1 nF = 10^(-9) F` `V = (q)/(C ) = (5xx10^(-6))/(10^(-9)) = 5xx10^(3) V` |
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| 765. |
125 drops of water each of radius `2 mm` and carrying charge of `1 nC` are made to form a bigger drop. Find the capacitance and potential of the bigger drop. |
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Answer» Correct Answer - `1.1 pF ; 113.6 V` Let `r` and `R` radii of small and bigger drop respectively. Volume of bigger drop `= 1.25xx` volume of small drop `(4)/(3) pi r^(3) = 125xx(4)/(3) pi r^(3)` `R = 5 r = 5xx2xx10^(-3) = 10xx10^(-3)m` Capacitance fo bigger drop `C = 4pi in_(0) R = (10xx10^(-3))/(9xx10^(9)) = 1.1 pF` Charge on bigger drop, `q = 125xx10^(-9)C` `:.` Potential of bigger drop `V = (q)/(C) = (125xx10^(-9))/(1.1xx10^(-12)) = 113.6 V` |
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| 766. |
A charged spherical conductor has a surface density of `0.07 C cm^(-2)`. When the charge is increased by `4.4 C`, the surface density changes to `0.084 C cm^(-2). Find the initial charge and capacitance of the spherical conductor. |
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Answer» Correct Answer - `22 C ; 5.56xx10^(-12) F` Here, `sigma_(1) = 0.07 C cm^(-2)` and `sigma_(2) = 0.084 C cm^(-2)`. Let `q` be the charge on the spherical conductor of radius `r` cm. `:. Sigma_(1) = (q)/(4pi r^(2)) = 0.07` …(i) and `sigma_(2) = (q + 4.4)/(4pi r^(2)) = 0.084` ….(ii) Dividing (ii) by (i), we get `(q+ 4.4)/(q) = (0.084)/(0.07) = 1.2` or `q = 22 C` From (i), `r = sqrt((q)/(4pi xx 0.07)) = sqrt((22xx7)/(4xx22xx0.07)) = 5cm` `r = 0.05m` `C = 4pi in_(0) r = (0.05)/(9xx10^(6)) = 5.56xx10^(-12) F` |
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| 767. |
The electric field in a region is given by `E=ahati+bhatj`. Hence as and b are constants. Find the net flux passing through a square area of side `I` parallel to y-z plane. |
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Answer» Correct Answer - A::B A square area of side l parallel to y-z plane in vector from can be written as `S=l^2hati` Given, `E=ahati+bhatj` `:.` Electric flux passing through the given area will be `phi=E.S` `=(ahati+bhatj).(l^2hati)` `=al^2` |
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| 768. |
When electrons equal to Avogadro number are transferred from one conductor to another, a potential difference of `10^(6) V` appears between them. Calculate the capacity of the system of two conductors. |
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Answer» Correct Answer - `9.637xx10^(-2) F` Here, `q` = Charge on `6.023xx10^(23)` electrons `= 6.023xx10^(23)xx1.6xx10^(-19) C = 9.637xx10^(4) C` `C = (q)/(V) = (9.637xx10^(4))/(10^(6)) = 9.637xx10^(-2) F` |
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| 769. |
The electric field in a region is given by `E=ahati+bhatj`. Hence as and b are constants. Find the net flux passing through a square area of side `I` parallel to y-z plane.A. `a^(2)l^(2)`B. `al^(2)`C. `b^(2)l^(2)`D. `bl^(2)` |
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Answer» Correct Answer - B A square area of side l parallel t Y-Z plane in vector form can be written as `S=l^(2)hati` Given, `" "E=ahati+bhatj` `therefore" Electric flux passing through the given area will be "` `phi_(E)=E.S=(ahati+bhatj).(l^(2)hati)=al^(2)` |
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| 770. |
When `1.0xx10^(12)` electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Find the capacitance of this two conductor system.A. `2.32xx10^(-4)F`B. `1.6xx10^(-8)F`C. `1.2xx10^(-6)F`D. None of these |
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Answer» Correct Answer - B Number of electrons, `n=1.0xx10^(12)` `therefore" Charge transferred, Q "=n e =1.0xx10^(12)xx1.6xx10^(-19)` `=1.6xx10^(-7)C" "(because e=1.6xx10^(-19)C)` Capacitance between two conductors, `C=(Q)/(V)=(1.6xx10^(-7))/(10)=1.6xx10^(-8)F` |
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| 771. |
The electirc potential at a point `(x, y, z)` is given by `V = -x^(2)y - xz^(3) + 4` The electric field `vecE` at that point isA. `E = (2xy +z^(3))hati + x^(2)hatj +3xz^(2) hatk`B. `E = 2xy hati +(x^(2) +y^(2)) hatj + (3xz - y^(2))hatk`C. `E = z^(3) hati +xyzhatj +z^(2) hatk`D. `E = (2xy - z^(3)) hayi +xy^(2) hatj +3z^(2) xx hatk` |
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Answer» Correct Answer - A |
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| 772. |
Two capacitors having capacitances `C_(1)` and `C_(2)` are charged with 120 V and 200 V batteries respectively. When they are connected in parallel now, it is found that the potential on each one of them is zero. Then,A. `5C_(1)=3C_(2)`B. `3C_(1)=5C_(2)`C. `3C_(1) pm 5C_(2)=0`D. `9C_(1)=4C_(2)` |
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Answer» Correct Answer - C Polarity should be mentioned in the question. Potential on each of them can be zero if, `q_("net")=0`. `q_(1)pmq_(2)=0` `rArr" "120C_(1)pm 200C_(2)=0` `rArr" "3C_(1)pm5C_(2)=0` |
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| 773. |
What is the smallest electric force between two charges placed at a distance of `1.0m`? |
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Answer» `F_e=1/(4piepsilon_0).(q_1q_2)/r^2` For `F_e` to be minimum `q_1q_2` should be minimum. We know that `(q_1)_(min)=(q_(2))_(min)=e=1.6xx10^(-19) C ` `(F_(e))_(min)=((9.0xx10^9)(1.6xx10^-19)(1.6xx10^-19))/((1.0)^2)` `=2.304xx10^-28N` |
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| 774. |
Why a third hole in a socket provided for grounding? |
| Answer» All electric appliances may end with some charge due to faulty connections. In such a situation charge will be accumulated on the appliance. When the user touches the appliances he may get as shock. By providing the third hole for grounding al accumulated charge is discharged to the ground and the appliance is safe. | |
| 775. |
There is an electric field E in x-direction. If the work done on moving a charge of `0.2C` through a distance of 2w m along a line making a angle `60^@` with x-axis is 4 J, then what is the value of E?A. `3N//C`B. `4N//C`C. `5N//C`D. `20N//C` |
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Answer» Correct Answer - D |
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| 776. |
There is an electric field E in x-direction. If the work done on moving a charge of `0.2C` through a distance of 2w m along a line making a angle `60^@` with x-axis is 4 J, then what is the value of E?A. `sqrt3N//C`B. `4N//C`C. `5N//C`D. `20N//C` |
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Answer» Correct Answer - D `W=Fscostheta=qEs costheta` `:. E=W/(qscostheta)` `=4/(0.2xx2xxcos60^@)` `=20 N//C` |
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| 777. |
An infinite line of charge `lamda` per unit length is placed along the y-axis. The work done in moving a charge q from `A(a,0)` to `B (2a,0)` isA. `(q lamda)/(2piepsilon_0) `"ln" 2`B. `(qlamda)/(2piepsilon_0)"ln" (1/2)`C. `(qlamda)/4piepsilon_0) "ln" sqrt2`D. `(qlamda)/(4piepsilon_0) "ln"2` |
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Answer» Correct Answer - B `W_(ArarrB)=q(V_B-V_A)` `=q[-int_A^BEdr]=q int_B^A Edr` `=q int_(2a)^a lamda/(2piepsilon_0)dr= (qlamda)/(2piepsilon_0) "in" (1/2)` |
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| 778. |
A particle of mass m having negative charge q moves along an ellipse around a fixed positive charge Q so that its maximum and minimum distances from fixed charge are equal to `r_(1) " and " r_(2)` respectively. Calculate angular momentum L of this particle: |
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Answer» Correct Answer - [`sqrt((mr_(1)r_(2)Qq)/(2piepsi_(0)(r_(1)+r_(2))))`] |
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| 779. |
The electric dipole moment per unit volume of electric dipole isA. polarisationB. diffractionC. interferenceD. reflection |
| Answer» Correct Answer - A | |
| 780. |
Which of the following is example of non polar molecule ?A. `H_(2)`B. HClC. `H_(2)O`D. `HO_(2)` |
| Answer» Correct Answer - A | |
| 781. |
Infinite charges of magnitude `q` each are lying at `x= 1,2,4,8….meter` on `X`-axis. The value of intensity of electric field at point `x=0` due to these charges will beA. `12 xx 10^(9) q NC^(-1)`B. zeroC. `6 xx 10^(9) q NC^(-1)`D. `4 xx 10^(9) q NC^(-1)` |
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Answer» Correct Answer - A |
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| 782. |
A uniform electric field exists in x y plane as shown in Fig. 3.45. Find the potential difference. Between origin O and `A(d,d,0)`. .A. ` Ed (cos theta + sin theta) `B. ` - Ed (sin thera - Cos theta)`C. ` sqrt 2 Ed`D. none of these |
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Answer» Correct Answer - A ` vec E =E cos theta hat I + E sin theta hat j` ` vec (dr) = d hat I +d hat j` ` dV =- vec E. vec (dr) =- Ed cos theta - sin theta` `dV =- Ed (cos theta + sin theta)`. |
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| 783. |
The diagram shows three infinitely long uniform line charges placed on the `X, Y` and ` Z` axis . The word done in moving a unit positive charge from (1,1,1) to (0 , 1,1,) is equal to .A. ` ( lambda ln2 ) / 2 pi varepsilon_(0)`B. `(lambda ln 2 ) // 2 pi varepsilon_(0)`C. `(3lambdal ln 2) //2 pivarepsilon_(0)`D. None |
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Answer» Correct Answer - B Displacement is only along x-axis `Delta =- underset (1) overset (3) (Ò)E_(x) dx` ` E_x ` is field due to dharged wire lying along y-axis `E_(x) =(2K(3lambda))/(x) +(2K(lambda))/(x)rArr E_(x) =(2K(lambda))/(x)xx4` `E_x = (2lambda) (pi in_0 x)` ….(2) From (1) and (2) `Delta V =- underset (1) overset (2) (Ò) (1l)/(phat I_0 x) dx` `Delta V = (2 lambda)/(pi in_(0)) [ln x]_(1)^(2) rArrDelta V =(2 lambda)/(piin_(0)) ln1` `DeltaV=(lambda)/(pi in_(0)) ln2` |
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| 784. |
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. In an electroscope, both leaves are hinged at the top point O. Each leaf has mass m, length l and gets charge q. assuming the charge to be concentrated at ends A and B only, the small angle of deviation `(theta)` between the leaves in static equilibrium, is equal to :A. `((4kq^(2))/(l^(2) mg))^(1//3)`B. `((kq^(2))/(l^(2) mg))^(1//3)`C. `((2kq^(2))/(l^(2) mg))^(1//2)`D. `((64 kq^(2))/(l^(2) mg))^(1//3)` |
| Answer» Correct Answer - A | |
| 785. |
A leaf electroscope is a simple apparatus to detect any charge on a body. It consist of two metal leaves OA and OB, free to rotate about O. Initially both are very slighty separted. When a charged object is touched to the metal knob at the top of the conduction re, charge flows from knob to the top of the conduction rod, charge flows from knob to the leaves through the conducting rod, charge flows from knob to the leaves through the conduction rod. As the leaves are now charged similarly, they start repelling each other and tet separted (deflected by certain angle). The angle of deflection in static equilibrium is an indicator of the amount of charge on the charged boy. If we perform these steps one by one. In which case, the leaves will converge (come closer), as compared to the previous state?A. (i)B. (i) and (iii)C. only (iii)D. In all cases, the leaves will diverge |
| Answer» Correct Answer - C | |
| 786. |
A large nonconducting sheet M is given a uniform charge density. Two unchared small metal rods A and B are placed near the sheet as shown in figure A. M attracts AB. A attracts BC. M attracts BD. B attracts A |
| Answer» Correct Answer - B::D | |
| 787. |
A solid sphere of radius R has a charge Q distributed in its volume with a charge density `rho=kr^a`, where k and a are constants and r is the distance from its centre. If the electric field at `r=(R)/(2)` is `1/8` times that `r=R`, find the value of a. |
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Answer» From gauss theorem `"iont"_s overlineE, overline(dA) = (sum q)/(in_0) rArr E. 4 pi r^2 = 1/in_0 sumq` `E = 1.( 4 pi in_0) (sumq)/r^@` `E_R =1/(4 pi in_0) Q/R_2` …(1) `E_(R//2) = 1/4 pi in_0 (sum q)/(R //2)^2 = 4 1/(4 pi in_0) (sumq)/R^2` ...(2) From (1) and (2) ` E_(R/2)/E_R = (4 sumq)/Q` ...(3) Given that `( E_(R//2)/E_R = 1/8` ...(4) From (3) and (4) ` 1/ 8 = (4 sumq)/Q rArr ( sum q)/Q = 1/(32)` `(int_0^(R//2) rho 4 pi r^2 dr)/(int_0^R rho 34 pi r^2 dr) = 1/( 32)` `( int_0 ^(R//2) Kr^a 4 pi r^2 dr)/(int_0^R Kr^2 4 pi r^2 dr) = 1/( 32)` ltbgt `([r^(a+3)]_0^(R//2))/([r^(a+3)]_0^R) = 1/( 32)` rArr ( 1/2)^(a+3) = (1 /2)^5` ` a + 3 = 5 rArr a =2` . |
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| 788. |
As one penetrates uniformly charged conducting sphere,what happens to the electric field strengthA. increasesB. decreasesC. remains the same as at the surfaceD. is zero at all points |
| Answer» Correct Answer - D | |
| 789. |
In space of horizontal `EF (E = (mg)//q)` exist as shown in figure and a mass m is released at the end of a light rod. If mass m is releases from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position .A. ` sqrt ((g)/(l))`B. ` sqrt ((2g)/(l))`C. `sqrt ((3g)/(l))`D. `sqrt ((5g)/(l))` |
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Answer» Correct Answer - B From conservation of energy work done by electrifield + work done by gravitational field ` = 1/2 I omega^2` ` q. E. l sin 45^@ _ Mg l (1 - cos 45(@)= 1/2 ml^2 omega^2` `q(mg)/q l(sqrt 2) + mgl (1 - 1/(sqrt 2)) = 1/2 ml^2 omega^2` ` mgl = 1/2 ml^2 omega^2 rArr omega = sqrt (2 g)/l` . |
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| 790. |
A 5 pF capacitor is connected in series with a 10 μF capacitor and the combination is connected across a 9 V battery. The potential differences across the capacitors are in the ratio (A) 4 : 1 (B) 3 : 1 (C) 2 : 1 (D) 1 : 1 |
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Answer» Correct option is (C) 2 : 1 |
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| 791. |
A 2 μF capacitor, charged to a p.d. of 200 V, is connected across an uncharged capacitor. If the common p.d. is 20 V, the capacitance of the second capacitor is (A) 18 μF (B) 20 μF (C) 36 μF (D) 40 μF |
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Answer» Correct option is (A) 18 μF |
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| 792. |
A 5 μF capacitor is charged to a p.d. of 10 V. If it is further charged, so that its p.d. increases to 20 V, the electric energy stored in it increases by (A) 450 μJ(B) 500 μJ (C) 750 μJ (D) 900 μJ |
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Answer» Correct option is (C) 750 μJ |
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| 793. |
The energy density in the region between the plates of a charged parallel-plate air capacitor is given by the expression(A) \(\cfrac 12\)ε0E2(B) \(\cfrac 12\)ε0E (C) \(\cfrac{E^2}{2ε_0}\)(D) \(\cfrac{σ ^2}{ɛ_o }\) |
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Answer» Correct option is (A) \(\cfrac 12\)ε0E2 |
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| 794. |
A parallel plate capacitor having a plate separation of `2mm` is charged by connecting it to a `300v` supply. The energy density isA. `0.01 J//m^(3)`B. `0.1 J//m^(3)`C. `1 J//m^(3)`D. `10 J//m^(3)` |
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Answer» Correct Answer - B `u=1/2in_(0)KE^(2)` `=(8.85xx10^(-12))/2xx1xx(300/(2xx10^(-3)))^(2)` `=0.1J//m^(3)` |
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| 795. |
If a `10muF` capacitor is to have an energy constant of 1 joule, it must be placed across a potential difference ofA. 900 VB. 750 VC. 447.2 VD. 200 V |
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Answer» Correct Answer - C `E=1/2CV^(2)` `therefore" "V=sqrt((2E)/C)=sqrt((2xx1)/(10xx10^(-6)))=447.2V` |
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| 796. |
Can a body have charge and still be at zero potentialA. yes, alwaysB. yes, but not alwaysC. neverD. depends upon the nature of charge |
| Answer» Correct Answer - B | |
| 797. |
Can a body have electric potential and still be unchargedA. yes, alwaysB. yes, but not alwaysC. neverD. depends upon the type of potential |
| Answer» Correct Answer - B | |
| 798. |
If an electron moves from rest from a point at which potential is `50` volt to another point at which potential is `70` volt, then its kinetic energy in the final state will beA. jouleB. ergC. electron-voltD. newton |
| Answer» Correct Answer - C | |
| 799. |
Two equal and opposite charges separated by a finite distance isA. electric dipoleB. electric torqueC. electric dipole momentD. none of these |
| Answer» Correct Answer - A | |
| 800. |
The product of positive charge and distance between the two charges isA. electric dipoleB. electric dipole momentC. electric fluxD. electric intensity |
| Answer» Correct Answer - B | |