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1ldo068az | physics | electrostatics | electric-flux-and-gauss-law | <p>A cubical volume is bounded by the surfaces $$\mathrm{x}=0, x=\mathrm{a}, y=0, y=\mathrm{a}, \mathrm{z}=0, z=\mathrm{a}$$. The electric field in the region is given by $$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} x \hat{i}$$. Where $$\mathrm{E}_{0}=4 \times 10^{4} ~\mathrm{NC}^{-1} \mathrm{~m}^{-1}$$. If $$\mathrm{a... | [] | null | 288 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le4a6szq/aa1d5d5b-d21c-486d-a8ee-80b87007b920/a0670e60-ac6b-11ed-8b11-cb59760cfd30/file-1le4a6szr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le4a6szq/aa1d5d5b-d21c-486d-a8ee-80b87007b920/a0670e60-ac6b-11ed-8b11-cb59760cfd30/fi... | integer | jee-main-2023-online-1st-february-evening-shift | 10,791 |
ldqwc25y | physics | electrostatics | electric-flux-and-gauss-law | As shown in figure, a cuboid lies in a region with electric field $E=2 x^{2} \hat{i}-4 y \hat{j}+6 \hat{k} \mathrm{~N} / \mathrm{C}$. The magnitude of charge within the cuboid is $n \in_{0} C$.
<br/><br/>
The value of $n$ is _________ (if dimension of cuboid is $1 \times 2 \times 3 \mathrm{~m}^{3}$ )<br/><br/>
<img src... | [] | null | 12 | Flux through planes parallel to $y-z=2(1)^{2} \times$ Area
<br/><br/>$$
\begin{aligned}
& =2(1)^{2} \times 2 \times 3 \\\\
& =12 \mathrm{Nm}^{2} / \mathrm{C}
\end{aligned}
$$
<br/><br/>Flux through planes parallel to $x-z=-4(2) \times$ Area
<br/><br/>$$
\begin{aligned}
& =-4(2) \times 1 \times 3 \\\\
& =-24 \mathrm{... | integer | jee-main-2023-online-30th-january-evening-shift | 10,793 |
1ldso44r5 | physics | electrostatics | electric-flux-and-gauss-law | <p>In a cuboid of dimension $$2 \mathrm{~L} \times 2 \mathrm{~L} \times \mathrm{L}$$, a charge $$q$$ is placed at the center of the surface '$$\mathrm{S}$$' having area of $$4 \mathrm{~L}^{2}$$. The flux through the opposite surface to '$$\mathrm{S}$$' is given by</p> | [{"identifier": "A", "content": "$$\\frac{q}{2 \\epsilon_{0}}$$"}, {"identifier": "B", "content": "$$\\frac{q}{3 \\epsilon_{0}}$$"}, {"identifier": "C", "content": "$$\\frac{q}{12 \\epsilon_{0}}$$"}, {"identifier": "D", "content": "$$\\frac{q}{6 \\in_{0}}$$"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lei36rxe/59d0fd3a-a039-4cc9-9db6-6ed6319a699d/38c48c20-b403-11ed-bf7e-c52177c53cde/file-1lei36rxf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lei36rxe/59d0fd3a-a039-4cc9-9db6-6ed6319a699d/38c48c20-b403-11ed-bf7e-c52177c53cde/fi... | mcq | jee-main-2023-online-29th-january-morning-shift | 10,794 |
lsamgy2r | physics | electrostatics | electric-flux-and-gauss-law | $C_1$ and $C_2$ are two hollow concentric cubes enclosing charges $2 Q$ and $3 Q$ respectively as shown in figure. The ratio of electric flux passing through $C_1$ and $C_2$ is :<br/><br/>
<img src="data:image/png;base64,UklGRuQQAABXRUJQVlA4INgQAABwKwGdASqwAgADP4HA2WO2MaymofY6QsAwCWlu8p8dnaxiN+e36m/1PsP5/emusGrvrc/337k... | [{"identifier": "A", "content": "$3: 2$"}, {"identifier": "B", "content": "$5: 2$"}, {"identifier": "C", "content": "$2: 5$"}, {"identifier": "D", "content": "$2: 3$"}] | ["C"] | null | <p>The electric flux ($$\Phi$$) through a closed surface, according to Gauss's law, is proportional to the enclosed charge $$Q$$, and is given by $$\Phi = \frac{Q}{\epsilon_0}$$, where $$\epsilon_0$$ is the vacuum permittivity. This is regardless of the shape of the surface, as long as it is closed and it encloses ... | mcq | jee-main-2024-online-1st-february-evening-shift | 10,795 |
jaoe38c1lsflpssv | physics | electrostatics | electric-flux-and-gauss-law | <p>An electric field is given by $$(6 \hat{i}+5 \hat{j}+3 \hat{k}) \mathrm{N} / \mathrm{C}$$. The electric flux through a surface area $$30 \hat{i} \mathrm{~m}^2$$ lying in YZ-plane (in SI unit) is :</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "90"}, {"identifier": "C", "content": "180"}, {"identifier": "D", "content": "150"}] | ["C"] | null | <p>$$\begin{aligned}
& \overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{A}}=30 \hat{\mathrm{i}} \\
& \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\
& \phi=(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(30 \hat{\mat... | mcq | jee-main-2024-online-29th-january-evening-shift | 10,797 |
1lsg6ka1a | physics | electrostatics | electric-flux-and-gauss-law | <p>A particle of charge '$$-q$$' and mass '$$m$$' moves in a circle of radius '$$r$$' around an infinitely long line charge of linear charge density '$$+\lambda$$'. Then time period will be given as :</p>
<p>(Consider $$k$$ as Coulomb's constant)</p> | [{"identifier": "A", "content": "$$T^2=\\frac{4 \\pi^2 m}{2 k \\lambda q} r^3$$\n"}, {"identifier": "B", "content": "$$T=\\frac{1}{2 \\pi r} \\sqrt{\\frac{m}{2 k \\lambda q}}$$\n"}, {"identifier": "C", "content": "$$T=\\frac{1}{2 \\pi} \\sqrt{\\frac{2 k \\lambda q}{m}}$$"}, {"identifier": "D", "content": "$$T=2 \\pi r ... | ["D"] | null | <p>$$\begin{aligned}
& \frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{r}}=\mathrm{m} \omega^2 \mathrm{r} \\
& \omega^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\
& \left(\frac{2 \pi}{\mathrm{T}}\right)^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\
& \mathrm{~T}=2 \pi \mathrm{r} \sqrt{\frac{\m... | mcq | jee-main-2024-online-30th-january-evening-shift | 10,798 |
luxwerbk | physics | electrostatics | electric-flux-and-gauss-law | <p>Five charges $$+q,+5 q,-2 q,+3 q$$ and $$-4 q$$ are situated as shown in the figure. The electric flux due to this configuration through the surface $$S$$ is :</p>
<p><img src="data:image/png;base64,UklGRrgJAABXRUJQVlA4IKwJAABQigCdASoAA2UBP4G+2WY2LyynIVF5EsAwCWlu/Eb4ietQ3f12xfe/H9hlLj1dPr0ljJn//9RRFLS7oh1J1d0OQL3XTB... | [{"identifier": "A", "content": "$$\\frac{q}{\\epsilon_0}$$\n"}, {"identifier": "B", "content": "$$\\frac{3 q}{\\epsilon_0}$$\n"}, {"identifier": "C", "content": "$$\\frac{5 q}{\\epsilon_0}$$\n"}, {"identifier": "D", "content": "$$\\frac{4 q}{\\epsilon_0}$$"}] | ["D"] | null | <p>$$\begin{aligned}
\phi & =\frac{q_{\mathrm{en}}}{\epsilon_0} \\
& =\frac{(5+1-2) q}{\epsilon_0} \\
& =\frac{4 q}{\epsilon_0}
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift | 10,799 |
lv5gsrm9 | physics | electrostatics | electric-flux-and-gauss-law | <p>An electric field, $$\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$$ passes through the surface of $$4 \mathrm{~m}^2$$ area having unit vector $$\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$$. The electric flux for that surface is _________ $$\mathrm{Vm}$$.</p> | [] | null | 12 | <p>The electric flux through a surface is given by the formula:</p>
<p>$$\Phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} = |\overrightarrow{\mathrm{E}}||\overrightarrow{\mathrm{A}}|\cos\theta$$</p>
<p>where $$\overrightarrow{\mathrm{E}}$$ is the electric field, $$\overrightarrow{\mathrm{A}}$$ is th... | integer | jee-main-2024-online-8th-april-morning-shift | 10,802 |
lvc58e7h | physics | electrostatics | electric-flux-and-gauss-law | <p>$$\sigma$$ is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is :</p> | [{"identifier": "A", "content": "$$\\sigma / \\epsilon_o R$$\n"}, {"identifier": "B", "content": "$$\\sigma / \\in_o$$\n"}, {"identifier": "C", "content": "$$\\sigma / 2 \\epsilon_o$$\n"}, {"identifier": "D", "content": "$$\\sigma / 4 \\epsilon_o$$"}] | ["B"] | null | <p>The question is about calculating the electric field at the surface of a thin spherical shell with a uniform surface charge density denoted by $$\sigma$$. To determine the electric field at any point on the surface of the shell, we can use Gauss's law, which is particularly useful for systems with high symmetry like... | mcq | jee-main-2024-online-6th-april-morning-shift | 10,803 |
NiJWQTXa5drcMmG6 | physics | electrostatics | electric-potential-energy-and-electric-potential | A thin spherical conducting shell of radius $$R$$ has a charge $$q.$$ Another charge $$Q$$ is placed at the center of the shell. The electrostatic potential at a point $$P$$ a distance $${R \over 2}$$ from the center of the shell is | [{"identifier": "A", "content": "$${{2Q} \\over {4\\pi {\\varepsilon _0}R}}$$ "}, {"identifier": "B", "content": "$${{2Q} \\over {4\\pi {\\varepsilon _0}R}} - {{2q} \\over {4\\pi {\\varepsilon _0}R}}$$ "}, {"identifier": "C", "content": "$${{2Q} \\over {4\\pi {\\varepsilon _0}R}} + {q \\over {4\\pi {\\varepsilon _0}R}}... | ["C"] | null | Electric potential due to charge $$Q$$ placed at the center of spherical shell at point $$P$$ is
<br><br>$${V_1} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {R/2}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over R}$$
<br><br><img class="question-image" src="https://imagex.cdn.examgoal.net/Fulzm6VzYUnKLFd1t/EiKKTBSgrpk8g... | mcq | aieee-2003 | 10,805 |
Qz2u4D8L4DkJrTXG | physics | electrostatics | electric-potential-energy-and-electric-potential | Two spherical conductors $$B$$ and $$C$$ having equal radii and carrying equal charges on them repel each other with a force $$F$$ when kept apart at some distance. A third spherical conductor having same radius as that $$B$$ but uncharged is brought in contact with $$B,$$ then brought in correct with $$C$$ and finally... | [{"identifier": "A", "content": "$$F/8$$ "}, {"identifier": "B", "content": "$$3$$ $$F/4$$ "}, {"identifier": "C", "content": "$$F/4$$ "}, {"identifier": "D", "content": "$$3$$ $$F/8$$ "}] | ["D"] | null | $$F \propto {{{Q_A}{Q_C}} \over {{x^2}}}$$
<br><br>$$x$$ is distance between the spheres. After first operation charge on $$B$$ is halved i.e $${Q \over 2}.$$
<br><br>and charge on third sphere becomes $${Q \over 2}.$$ Now it is touched to $$C$$, charge then equally
<br><br>distributes them selves to make potential ... | mcq | aieee-2004 | 10,806 |
hcRGNXDoV85s0ZmC | physics | electrostatics | electric-potential-energy-and-electric-potential | A charge particle $$'q'$$ is shot towards another charged particle $$'Q'$$ which is fixed, with a speed $$'v'$$. It approaches $$'Q'$$ upto a closest distance $$r$$ and then returns. If $$q$$ were given a speed of $$'2v'$$ the closest distances of approaches would be | [{"identifier": "A", "content": "$$r/2$$ "}, {"identifier": "B", "content": "$$2r$$ "}, {"identifier": "C", "content": "$$r$$ "}, {"identifier": "D", "content": "$$r/4$$ "}] | ["D"] | null | $${1 \over 2}m{v^2} = {{kQq} \over r}$$
<br><br>$$ \Rightarrow {1 \over 2}m{\left( {2v} \right)^2} = {{kqQ} \over {r'}}$$
<br><br>$$ \Rightarrow r' = {r \over 4}$$ | mcq | aieee-2004 | 10,807 |
Lz3Xvu7apCfZSZ8N | physics | electrostatics | electric-potential-energy-and-electric-potential | Two thin wire rings each having a radius $$R$$ are placed at a distance $$d$$ apart with their axes coinciding. The charges on the two rings are $$+q$$ and $$-q.$$ The potential difference between the centres of the two rings is | [{"identifier": "A", "content": "$${q \\over {2\\pi \\,{ \\in _0}}}\\left[ {{1 \\over R} - {1 \\over {\\sqrt {{R^2} + {d^2}} }}} \\right]$$ "}, {"identifier": "B", "content": "$${{qR} \\over {4\\pi \\,{ \\in _0}\\,{d^2}}}$$ "}, {"identifier": "C", "content": "$${q \\over {4\\pi \\,{ \\in _0}}}\\left[ {{1 \\over R} - {1... | ["A"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/uOJw6w4SfAqNabRTV/N95JCOcDGN2bQYEKqFnqT54ek75cG/OrLdXR5Y7T7WtU6kHxf07S/image.svg" loading="lazy" alt="AIEEE 2005 Physics - Electrostatics Question 205 English Explanation">
<br><br>$${V_A} = {V_{self}} + {V_{due}}$$ to $$(2)$$
<br><br>$$ \Rightarrow {V_A}... | mcq | aieee-2005 | 10,808 |
B6JVgIOFDvXPdmnc | physics | electrostatics | electric-potential-energy-and-electric-potential | Two spherical conductors $$A$$ and $$B$$ of radii $$1$$ $$mm$$ and $$2$$ $$mm$$ are separated by a distance of $$5$$ $$cm$$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres $$A$$ and $... | [{"identifier": "A", "content": "$$4:1$$ "}, {"identifier": "B", "content": "$$1:2$$ "}, {"identifier": "C", "content": "$$2:1$$ "}, {"identifier": "D", "content": "$$1:4$$ "}] | ["C"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/2qqwh79Y5h5q9NopZ/1vI8hKGGYyCiulbpOnRC08WUqHPpq/fZdPzlNUDsRJgcOxBYvLjq/image.svg" loading="lazy" alt="AIEEE 2006 Physics - Electrostatics Question 202 English Explanation">
<br><br>After connection, $${V_1} = {V_2}$$
<br><br>$$ \Rightarrow K{{{Q_1}} \over... | mcq | aieee-2006 | 10,809 |
b9lNWzzXrlRycPv8 | physics | electrostatics | electric-potential-energy-and-electric-potential | Two insulating plates are both uniformly charged in such a way that the potential difference between them is $${V_2} - {V_1} = 20\,V.$$ (i.e., plate $$2$$ is at a higher potential). The plates are separated by $$d=0.1$$ $$m$$ and can be treated as infinitely large. An electron is released from rest on the inner surface... | [{"identifier": "A", "content": "$$2.65 \\times {10^6}\\,m/s$$ "}, {"identifier": "B", "content": "$$7.02 \\times {10^{12}}\\,m/s$$ "}, {"identifier": "C", "content": "$$1.87 \\times {10^6}\\,m/s$$ "}, {"identifier": "D", "content": "$$32 \\times {10^{ - 19}}\\,m/s$$ "}] | ["A"] | null | $$eV = {1 \over 2}m{v^2}$$
<br><br>$$ \Rightarrow v = \sqrt {{{2ev} \over m}} = \sqrt {{{2 \times 1.6 \times {{10}^{ - 19}} \times 20} \over {9.31 \times {{10}^{ - 31}}}}} $$
<br><br>$$ = 2.65 \times {10^6}\,m/s$$ | mcq | aieee-2006 | 10,810 |
d7qzd6QNbjrg9qc7 | physics | electrostatics | electric-potential-energy-and-electric-potential | An electric charge $${10^{ - 3}}\,\,\mu \,C$$ is placed at the origin $$(0,0)$$ of $$X-Y$$ co-ordinate system. Two points $$A$$ and $$B$$ are situated at $$\left( {\sqrt 2 ,\sqrt 2 } \right)$$ and $$\left( {2,0} \right)$$ respectively. The potential difference between the points $$A$$ and $$B$$ will be | [{"identifier": "A", "content": "$$4.5$$ volts "}, {"identifier": "B", "content": "$$9$$ volts"}, {"identifier": "C", "content": "zero "}, {"identifier": "D", "content": "$$2$$ volts"}] | ["C"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/fqys8IgQx5sirqNo5/4fpjKxCE5mFnAcorJQgCo2iExZ2wG/hqHbxL4MkqNWBlsmF5OPBv/image.svg" loading="lazy" alt="AIEEE 2007 Physics - Electrostatics Question 201 English Explanation">
<br><br>The distance of point $$A\left( {\sqrt 2 ,\sqrt 2 } \right)$$ from the or... | mcq | aieee-2007 | 10,811 |
saHNbcQyDNI4w7JN | physics | electrostatics | electric-potential-energy-and-electric-potential | Charges are placed on the vertices of a square as shown. Let $$\overrightarrow E $$ be the electric field and $$V$$ the potential at the center. If the charges on $$A$$ and $$B$$ are interchanged with those on $$D$$ and $$C$$ respectively, then
<img src="data:image/png;base64,UklGRmIGAABXRUJQVlA4IFYGAACwRwCdASrIAUUBPm... | [{"identifier": "A", "content": "$$\\overrightarrow E $$ changes, $$V$$ remains unchanged"}, {"identifier": "B", "content": "$$\\overrightarrow E $$ remains unchanged, $$V$$ changes "}, {"identifier": "C", "content": "both $$\\overrightarrow E $$ and $$V$$ change "}, {"identifier": "D", "content": "$$\\overrightarrow E... | ["A"] | null | As shown in the figure, the resultant electric fields before and after interchanging the charges will have the same magnitude, but opposite directions.
<br><br>Also, the potential will be same in both cases as it is a scalar quantity.
<br><br><img class="question-image" src="https://imagex.cdn.examgoal.net/xaRmzFCd0kl... | mcq | aieee-2007 | 10,812 |
EdWWTR8JpqiUdKu9 | physics | electrostatics | electric-potential-energy-and-electric-potential | The potential at a point $$x$$ (measured in $$\mu \,m$$) due to some charges situated on the $$x$$-axis is given by $$V\left( x \right) = 20/\left( {{x^2} - 4} \right)$$ volt
<br/>The electric field $$E$$ at $$x = 4\,\mu \,m$$ is given by | [{"identifier": "A", "content": "$$(10/9)$$ volt / $$\\mu $$ $$m$$ and in the $$ + ve$$ $$x$$ direction"}, {"identifier": "B", "content": "$$\\left( {5/3} \\right)$$ volt/ $$\\mu $$ $$m$$ and in the $$-ve$$ $$x$$ direction"}, {"identifier": "C", "content": "$$\\left( {5/3} \\right)$$ volt/$$\\mu $$ $$m$$ and in the $$+... | ["A"] | null | Here, $$V\left( x \right) = {{20} \over {{x^2} - 4}}volt$$
<br><br>We know that $$E = - {{dV} \over {dx}} = {d \over {dx}}\left( {{{20} \over {{x^2} - 4}}} \right)$$
<br>or, $$E = + {{40x} \over {{{\left( {{x^2} - 4} \right)}^2}}}$$
<br><br>At $$x = 4\mu m,$$
<br><br>$$E = + {{40 \times 4} \over {{{\left( {{4^2} - 4... | mcq | aieee-2007 | 10,813 |
FVnf38AvrFMmUWi6 | physics | electrostatics | electric-potential-energy-and-electric-potential | (This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.)
<p><b>Statement-1 :</b> For a charged particle moving from point $$P$$ to point $$Q$$, the net work done by an electrostatic field on the particle is independent o... | [{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1."}, {"identifier": "C", "content": "Statement- 1 is fa... | ["A"] | null | Statement $$1$$ is true.
<br><br>Statement $$2$$ is true and is the correct explanation of $$(1)$$ | mcq | aieee-2009 | 10,814 |
ecJstlq5xeQtS07S | physics | electrostatics | electric-potential-energy-and-electric-potential | Two points $$P$$ and $$Q$$ are maintained at the potentials of $$10$$ $$V$$ and $$-4$$ $$V$$, respectively. The work done in moving $$100$$ electrons from $$P$$ to $$Q$$ is : | [{"identifier": "A", "content": "$$9.60 \\times {10^{ - 17}}J$$ "}, {"identifier": "B", "content": "$$ - 2.24 \\times {10^{ - 16}}J$$ "}, {"identifier": "C", "content": "$$ 2.24 \\times {10^{ - 16}}J$$ "}, {"identifier": "D", "content": "$$- 9.60 \\times {10^{ - 17}}J$$ "}] | ["C"] | null | $$ {{{W_{PQ}}} \over q} = \left( {{V_Q} - {V_P}} \right)$$
<br><br>$$ \Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right)$$
<br><br>$$ = \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right)$$
<br><br>$$ = + 2.24 \times {10^{ - 16}}J$$ | mcq | aieee-2009 | 10,815 |
rgP80zkYdRStk8mK | physics | electrostatics | electric-potential-energy-and-electric-potential | The electrostatic potential inside a charged spherical ball is given by $$\phi = a{r^2} + b$$ where $$r$$ is the distance from the center and $$a,b$$ are constants. Then the charge density inside the ball is: | [{"identifier": "A", "content": "$$ - 6a{\\varepsilon _0}r$$ "}, {"identifier": "B", "content": "$$ - 24\\pi a{\\varepsilon _0}$$ "}, {"identifier": "C", "content": "$$ - 6a{\\varepsilon _0}$$ "}, {"identifier": "D", "content": "$$ - 24\\pi {\\varepsilon _0}r$$ "}] | ["C"] | null | Electric field
<br><br>$$E = {{d\phi } \over {dr}} = - 2ar\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>By Gauss's theorem
<br><br>$$E = {1 \over {4\theta {\varepsilon _0}{r^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>From $$\left( i \right)$$ and $$\left( ii \right),$$
<br><br>$$q = - 8\pi... | mcq | aieee-2011 | 10,816 |
kZvRddnHfhUblAZ0 | physics | electrostatics | electric-potential-energy-and-electric-potential | This question has statement- $$1$$ and statement- $$2.$$ Of the four choices given after the statements, choose the one that best describe the two statements.
<br/>An insulating solid sphere of radius $$R$$ has a uniformly positive charge density $$\rho $$. As a result of this uniform charge distribution there is a fin... | [{"identifier": "A", "content": "Statement- $$1$$ is true, Statement- $$2$$ is true; Statement- $$2$$ is not the correct explanation of Statement- $$1$$."}, {"identifier": "B", "content": "Statement $$1$$ is true, Statement $$2$$ is false."}, {"identifier": "C", "content": "Statement $$1$$ is false, Statement $$2$$ ... | ["C"] | null | The electric field inside a uniformly charged sphere is
<br><br>= $${{\rho .r} \over {3{ \in _0}}}$$
<br><br>The electric potential inside a uniformly charged sphere
<br><br>$$ = {{\rho {R^2}} \over {6{ \in _0}}}\left[ {3 - {{{r^2}} \over {{R^2}}}} \right]$$
<br><br>$$\therefore$$ Potential difference between center a... | mcq | aieee-2012 | 10,817 |
MAgvQPZTOXeSyXPG | physics | electrostatics | electric-potential-energy-and-electric-potential | A charge $$Q$$ is uniformly distributed over a long rod $$AB$$ of length $$L$$ as shown in the figure. The electric potential at the point $$O$$ lying at distance $$L$$ from the end $$A$$ is
<img src="data:image/png;base64,UklGRiQZAABXRUJQVlA4IBgZAABwaQCdASpxAtUAPm00lkikIqIhI125KIANiWlu9kB/QFMA5JvwAuqvrUNX9Cv5T+KvfV/K... | [{"identifier": "A", "content": "$${Q \\over {8\\pi {\\varepsilon _0}L}}$$ "}, {"identifier": "B", "content": "$${{3Q} \\over {4\\pi {\\varepsilon _0}L}}$$ "}, {"identifier": "C", "content": "$${Q \\over {4\\pi {\\varepsilon _0}L\\,\\ln \\,2}}$$ "}, {"identifier": "D", "content": "$${{Q\\ln \\,2} \\over {4\\pi {\\varep... | ["D"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/jKZhr6iOKWUIJcrvo/ILPR11MxwK6IshDFrhKX2r16qjrFN/liAqeOq1ASR4Xb0HPWRznf/image.svg" loading="lazy" alt="JEE Main 2013 (Offline) Physics - Electrostatics Question 185 English Explanation">
<br><br>Electric potential is given by,
<br><br>$$V = \int\limits_L^{... | mcq | jee-main-2013-offline | 10,818 |
mYHueforZ9AVvImQ | physics | electrostatics | electric-potential-energy-and-electric-potential | Assume that an electric field $$\overrightarrow E = 30{x^2}\widehat i$$ exists in space. Then the potential difference $${V_A} - {V_O},$$ where $${V_O}$$ is the potential at the origin and $${V_A}$$ the potential at $$x=2$$ $$m$$ is : | [{"identifier": "A", "content": "$$120$$ $$J/C$$ "}, {"identifier": "B", "content": "$$-120$$ $$J/C$$ "}, {"identifier": "C", "content": "$$-80$$ $$J/C$$ "}, {"identifier": "D", "content": "$$80$$ $$J/C$$ "}] | ["C"] | null | Potential difference between any two points in an electric field is given by,
<br><br>$$dV = - \overrightarrow E .\overrightarrow {dx} $$
<br><br>$$\int\limits_{{V_O}}^{{V_A}} {dV = - \int\limits_0^2 {30{x^2}} } dx$$
<br><br>$${V_A} - {V_O} = \left[ {10{x^3}} \right]_{0}^2 = - 80J/C$$ | mcq | jee-main-2014-offline | 10,819 |
Lh0FdaHGDIWBQYlf | physics | electrostatics | electric-potential-energy-and-electric-potential | A uniformly charged solid sphere of radius $$R$$ has potential $${V_0}$$ (measured with respect to $$\infty $$) on its surface. For this sphere the equipotential surfaces with potentials $${{3{V_0}} \over 2},\,{{5{V_0}} \over 4},\,{{3{V_0}} \over 4}$$ and $${{{V_0}} \over 4}$$ have radius $${R_1},\,\,{R_2},\,\,{R_3}$$ ... | [{"identifier": "A", "content": "$${R_1} = 0$$ and $${R_2} < \\left( {{R_4} - {R_3}} \\right)$$ "}, {"identifier": "B", "content": "$$2R < {R_4}$$ "}, {"identifier": "C", "content": "$${R_1} = 0$$ and $${R_2} > \\left( {{R_4} - {R_3}} \\right)$$ "}, {"identifier": "D", "content": "$${R_1} \\ne 0$$ and $$\\left... | null | null | $$(a,b)$$ We know, $${V_0} = {{Kq} \over R} = Vsurface$$
<br><br>Now, $${V_i} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {R^2}} \right)\,\,\,\,\,$$ [For $$r < R$$]
<br><br>At the center of sphere $$r=0.$$ Here
<br><br>$$V = {3 \over 2}{V_0}$$
<br><br>Now, $${5 \over 4}{{Kq} \over R} = {{Kq} \over {2{R^3}}}\left( {3{R^... | mcqm | jee-main-2015-offline | 10,820 |
h9gwfwlBDbNi9CdftROh6 | physics | electrostatics | electric-potential-energy-and-electric-potential | Within a spherical charge distribution of charge density $$\rho $$(r), N equipotential surfaces of potential V<sub>0</sub>, V<sub>0</sub> + $$\Delta $$V, V<sub>0</sub> + 2$$\Delta $$V, .......... V<sub>0</sub> + N$$\Delta $$V ($$\Delta $$ V > 0), are drawn and have increasing radii r<sub>0</sub>, r<sub>1</sub>, r<su... | [{"identifier": "A", "content": "$$\\rho $$ (r) $$\\alpha $$ r"}, {"identifier": "B", "content": "$$\\rho $$ (r) = constant "}, {"identifier": "C", "content": "$$\\rho $$ (r) $$\\alpha $$ $${1 \\over r}$$ "}, {"identifier": "D", "content": "$$\\rho $$ (r) $$\\alpha $$ $${1 \\over {{r^2}}}$$ "}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265214/exam_images/bdzjl6n2cwc2grx7szo3.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Physics - Electrostatics Question 171 English Explanation">
<br><br>Here... | mcq | jee-main-2016-online-10th-april-morning-slot | 10,822 |
C7lmZNbUrOJKrRhh | physics | electrostatics | electric-potential-energy-and-electric-potential | Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge
densities $$ + \sigma $$, $$ - \sigma $$ and $$ + \sigma $$ respectively. The potential of shell B is : | [{"identifier": "A", "content": "$${\\sigma \\over { \\in {}_0}}\\left[ {{{{b^2} - {c^2}} \\over c} + a} \\right]$$ "}, {"identifier": "B", "content": "$${\\sigma \\over { \\in {}_0}}\\left[ {{{{a^2} - {b^2}} \\over a} + c} \\right]$$ "}, {"identifier": "C", "content": "$${\\sigma \\over { \\in {}_0}}\\left[ {{{{a^2... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263826/exam_images/qr2kas4ypl7ja1klnal1.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Offline) Physics - Electrostatics Question 180 English Explanation">
<br><br>Let charge of shell A, B an... | mcq | jee-main-2018-offline | 10,824 |
mKDPC385uLb9zFngSX3rsa0w2w9jwzam63i | physics | electrostatics | electric-potential-energy-and-electric-potential | In free space, a particle A of charge 1$$\mu $$C is held fixed at a point P. Another particle B of the same charge and
mass 4$$\mu $$g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P
is :
$$\left[ {Take\,{1 \over {4\pi { \in _0}}} = 9 \times {{10}^9}N{m^2}{C^{ - 2}... | [{"identifier": "A", "content": "1.0 m/s"}, {"identifier": "B", "content": "6.32 $$ \\times $$ 10<sup>4</sup> m/s"}, {"identifier": "C", "content": "2.0 $$ \\times $$ 10<sup>3</sup> m/s"}, {"identifier": "D", "content": "1.5 $$ \\times $$ 10<sup>2</sup> m/s"}] | ["B"] | null | q<sub>A</sub> = 1 $$\mu $$c ; q<sub>B</sub> = 1 $$\mu $$c, m<sub>B</sub> = 4 × 10<sup>–9</sup> kg, r<sub>AB</sub> = 10<sup>–3</sup> m<br><br>
$${1 \over 2}{M_B}{V^2} = k{q_A}{q_B}\left\{ {{1 \over {{{10}^{ - 13}}}} - {1 \over {9 \times {{10}^{ - 3}}}}} \right\}$$<br><br>
$${1 \over 2}4 \times {10^{ - 9}}{V^2} = 9 \time... | mcq | jee-main-2019-online-10th-april-evening-slot | 10,825 |
1FyoS11TMmi1xoBbqGsTg | physics | electrostatics | electric-potential-energy-and-electric-potential | A system of three charges are placed as shown
in the figure :<br/><br/>
<img src="data:image/png;base64,UklGRoALAABXRUJQVlA4IHQLAACwXACdASrsAtIAPm02l0mkIqKhIDgJMIANiWlu8SAVzBhrCZ+mX8g7Uv7j+Sf7reYr6j+7/lVutV8/92/mX9L/wf5UfCv+u/kXkn8Csk/9h/l36+8U9VX9QPYI9ifkv+J/iv9p/7n90+NT2j/Ufyn1H+qH6ReqN/nf6L7b/5//SeSftA+Av+gf0r/Ofzv+... | [{"identifier": "A", "content": "$${1 \\over {4\\pi {\\varepsilon _0}}}\\left[ { {{{q^2}} \\over d} + {{qQd} \\over {{D^2}}}} \\right]$$"}, {"identifier": "B", "content": "$${1 \\over {4\\pi {\\varepsilon _0}}}\\left[ { - {{{q^2}} \\over d} - {{qQd} \\over {2{D^2}}}} \\right]$$"}, {"identifier": "C", "content": "$${1 \... | ["C"] | null | U<sub>total</sub> = U<sub>self of dipole</sub> + U<sub>interaction</sub><br><br>
= $$ - {{k{q^2}} \over d} - \left( {{{kQ} \over {{D^2}}}} \right)qd$$<br><br>
$$ = - k\left[ {{{{q^2}} \over d} + {{qQd} \over {{D^2}}}} \right]$$ | mcq | jee-main-2019-online-9th-april-morning-slot | 10,827 |
vRdNuRxfkoOLeyuAlYO2g | physics | electrostatics | electric-potential-energy-and-electric-potential | A positive point charge is released from rest at
a distance r<sub>0</sub> from a positive line charge with
uniform density. The speed (v) of the point
charge, as a function of instantaneous distance
r from line charge, is proportional to :-
<img src="data:image/png;base64,UklGRqoGAABXRUJQVlA4IJ4GAABwVwCdASrsAiYBP4G+12a... | [{"identifier": "A", "content": "$$v \\propto \\left( {{r \\over {{r_0}}}} \\right)$$"}, {"identifier": "B", "content": "$$v \\propto \\ln \\left( {{r \\over {{r_0}}}} \\right)$$"}, {"identifier": "C", "content": "$$v \\propto {e^{ + r/{r_0}}}$$"}, {"identifier": "D", "content": "$$v \\propto \\sqrt {\\ln \\left( {{r \... | ["D"] | null | $${1 \over 2}m{V^2} = - q\left( {{V_f} - {V_i}} \right)$$<br><br>
$$E = {\lambda \over {2\pi {\varepsilon _0}r}}$$<br><br>
$$\Delta V = {\lambda \over {2\pi {\varepsilon _0}}}\ln \left( {{{{r_0}} \over r}} \right)$$<br><br>
$${1 \over 2}m{v^2} = {{ - q\lambda } \over {2\pi {\varepsilon _0}}}\ln \left( {{{{r_0}} \ove... | mcq | jee-main-2019-online-8th-april-evening-slot | 10,828 |
wNKzzwUhAdNCElKmWNkcP | physics | electrostatics | electric-potential-energy-and-electric-potential | The electric field in a region is given by
$$\mathop E\limits^ \to = \left( {Ax + B} \right)\mathop i\limits^ \wedge $$
, where E is in NC<sup>–1</sup> and x is in
metres. The values of constants are
A = 20 SI unit and B = 10 SI unit. If the potential
at x = 1 is V<sub>1</sub> and that at x = –5 is V<sub>2</sub>, th... | [{"identifier": "A", "content": "\u2013520 V"}, {"identifier": "B", "content": "180 V"}, {"identifier": "C", "content": "\u201348 V"}, {"identifier": "D", "content": "320 V"}] | ["B"] | null | $$\overrightarrow E = (20x + 10)\widehat i$$<br><br>
$${V_1} - {V_2} = - \int\limits_{ - 5}^1 {\left( {20x + 10} \right)dx} $$<br><br>
$${V_1} - {V_2} = - \left( {10{x^2} + 10x} \right)_{ - 5}^1$$<br><br>
$${V_1} - {V_2} = 10\left( {25 - 5 - 1 - 1} \right)$$<br><br>
$${V_1} - {V_2} = 180\,V$$ | mcq | jee-main-2019-online-8th-april-evening-slot | 10,829 |
9qp3wxSxopj3RNPAuhaTS | physics | electrostatics | electric-potential-energy-and-electric-potential | There is a uniform spherically symmetric surface charge density at a distance R<sub>0</sub> from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R (... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265560/exam_images/p6mnc9avfuvqqs8a41hk.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 12th January Morning Slot Physics - Electrostatics Quest... | ["B"] | null | At any instant 't'
<br><br>Total energy of charge distribution is constant
<br><br>i.e. $${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$$
<br><br>$$ \therefore $$ $${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$$
<br><br>$$ \therefo... | mcq | jee-main-2019-online-12th-january-morning-slot | 10,830 |
uKkiguKXPfwVaPq2QqXeS | physics | electrostatics | electric-potential-energy-and-electric-potential | A solid conducting sphere, having a charge Q,
is surrounded by an uncharged conducting
hollow spherical shell. Let the potential
difference between the surface of the solid
sphere and that of the outer surface of the
hollow shell be V. If the shell is now given a
charge of –4 Q, the new potential difference
between the... | [{"identifier": "A", "content": "V"}, {"identifier": "B", "content": "2V"}, {"identifier": "C", "content": "\u20132V"}, {"identifier": "D", "content": "4V"}] | ["A"] | null | <p>Initially when uncharged shell encloses charge Q, charge distribution due to induction will be as shown,</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l38tc07h/e54ee83d-87e2-4c93-ac59-36910f46e007/3fe88f20-d523-11ec-aeec-6fd1cdec7420/file-1l38tc07n.png?format=png" data-orsrc="https://app-co... | mcq | jee-main-2019-online-8th-april-morning-slot | 10,831 |
fz4z1Wuv1kZ7N8Az0K5vY | physics | electrostatics | electric-potential-energy-and-electric-potential | A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be - | [{"identifier": "A", "content": "$${{Q\\left( {{a^2} + {b^2} + {c^2}} \\right)} \\over {4\\pi {\\varepsilon _0}\\left( {{a^3} + {b^3} + {c^3}} \\right)}}$$"}, {"identifier": "B", "content": "$${Q \\over {4\\pi {\\varepsilon _0}\\left( {a + b + c} \\right)}}$$"}, {"identifier": "C", "content": "$${Q \\over {12\\pi {\\va... | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264540/exam_images/ioij8ebwku9u9alns7km.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Physics - Electrostatics Question 166 English Explanation">
<br><br>Po... | mcq | jee-main-2019-online-10th-january-morning-slot | 10,832 |
0HFsqj2ahMdBtYDEggqus | physics | electrostatics | electric-potential-energy-and-electric-potential | Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be - | [{"identifier": "A", "content": "$${{{Q_2}} \\over {4\\pi {\\varepsilon _0}}}$$"}, {"identifier": "B", "content": "$${{{Q^2}} \\over {2\\sqrt 2 \\pi {\\varepsilon _0}}}$$"}, {"identifier": "C", "content": "$${{{Q_2}} \\over {4\\pi {\\varepsilon _0}}}\\left( {1 + {1 \\over {\\sqrt 3 }}} \\right)$$"}, {"identifier": "D",... | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267351/exam_images/oo3hui5vr6c5bsnpqzbm.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Physics - Electrostatics Question 164 English Explanation">
<br><br>Po... | mcq | jee-main-2019-online-10th-january-evening-slot | 10,833 |
pQ5TDwLxBjpOaZOMnWVzh | physics | electrostatics | electric-potential-energy-and-electric-potential | Three charges Q, + q and + q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is :
<br/><br/><img src="data:image/png;base64,UklGRlIJAABXRUJQVlA4IEYJAABQlwCdASqqAjUCP4HA3GY2MK0nIVc4ssAwCWlu4XaxG/P/fY6eHZP3f/yu8h3... | [{"identifier": "A", "content": "$${{ - q} \\over {1 + \\sqrt 2 }}$$\n"}, {"identifier": "B", "content": "+ q"}, {"identifier": "C", "content": "$$-$$ 2q"}, {"identifier": "D", "content": "$${{ - \\sqrt 2 q} \\over {\\sqrt 2 + 1}}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264232/exam_images/jyhenwdvfabbgmgliars.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Morning Slot Physics - Electrostatics Question 162 English Explanation">
<br><br>U ... | mcq | jee-main-2019-online-11th-january-morning-slot | 10,834 |
LLg3C22oYlnFXM3GsPww7 | physics | electrostatics | electric-potential-energy-and-electric-potential | The given graph shows variation (with distance r form centre) of :
<br/><br/><img src="data:image/png;base64,UklGRsYKAABXRUJQVlA4ILoKAADQjQCdASoAA5QBP4HA2mY2MK0nITLpAsAwCWlu4XKRG/P5829oGP7Zy9nuJQf6OAsN/Df+b14s9N/+BV3SdCo4dONqbU171/qvU+bJxtTam1NqbU2mhJ/gmOKkfBMcVI+CMR5yY4qR8ExxUj4Jji3pP8ExxUj4JjipHwTBpaPgmOKkfBMcVI+CY5... | [{"identifier": "A", "content": "Electric field of a uniformly charged sphere "}, {"identifier": "B", "content": "Electric field of a uniformly charged spherical shell"}, {"identifier": "C", "content": "Potential of a uniformly charged sphere"}, {"identifier": "D", "content": "Potential of a uniformly charged spherical... | ["D"] | null | As the field inside the uniformly charged hollow
sphere or spherical shell is zero, so the potential inside it is
constant, whereas outside it varies inversely with distance. | mcq | jee-main-2019-online-11th-january-morning-slot | 10,835 |
b8a3iDdpbA5nR3XhHEjgy2xukeu1p6p5 | physics | electrostatics | electric-potential-energy-and-electric-potential | A two point charges 4q and -q are fixed
on the x-axis at x = $$ - {d \over 2}$$
and x = $${d \over 2}$$
respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will :
<img src="data:image/png;base64,UklGRoADAABXRUJQVlA4IHQDAADwFgC... | [{"identifier": "A", "content": "increase by $${{3{q^2}} \\over {4\\pi {\\varepsilon _0}d}}$$"}, {"identifier": "B", "content": "increase by $${{2{q^2}} \\over {3\\pi {\\varepsilon _0}d}}$$"}, {"identifier": "C", "content": "decrease by $${{{q^2}} \\over {4\\pi {\\varepsilon _0}d}}$$"}, {"identifier": "D", "content": "... | ["D"] | null | $$\Delta U = {1 \over {4\pi {\varepsilon _0}}}.{{4q.q} \over {(3d/2)}} - {1 \over {4\pi {\varepsilon _0}}}.{{4q.q} \over {(d/2)}}$$<br><br>$$ = {{4{q^2}} \over {4\pi {\varepsilon _0}}}\left( {{2 \over d}} \right)\left( { - {2 \over 3}} \right)$$<br><br>= decrease by $${{4{q^2}} \over {3\pi {\varepsilon _0}d}}$$ | mcq | jee-main-2020-online-4th-september-morning-slot | 10,837 |
1KHLCaZUjsFMDcrPLOjgy2xukf3uxetk | physics | electrostatics | electric-potential-energy-and-electric-potential | Concentric metallic hollow spheres of radii R and 4R hold charges Q<sub>1</sub>
and Q<sub>2</sub>
respectively. Given that
surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is : | [{"identifier": "A", "content": "$${{3{Q_2}} \\over {4\\pi {\\varepsilon _0}R}}$$"}, {"identifier": "B", "content": "$${{3{Q_1}} \\over {4\\pi {\\varepsilon _0}R}}$$"}, {"identifier": "C", "content": "$${{3{Q_1}} \\over {16\\pi {\\varepsilon _0}R}}$$"}, {"identifier": "D", "content": "$${{{Q_2}} \\over {4\\pi {\\vareps... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265460/exam_images/aeu7ki4snictajaatgwf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Evening Slot Physics - Electrostatics Question 132 English Explanation">
<br>$$\sigm... | mcq | jee-main-2020-online-3rd-september-evening-slot | 10,840 |
k6Ws4MBMQrRu7jCPomjgy2xukfg749nb | physics | electrostatics | electric-potential-energy-and-electric-potential | A solid sphere of radius R carries a charge
Q + q distributed uniformly over its volume. A
very small point like piece of it of mass m gets
detached from the bottom of the sphere and
falls down vertically under gravity. This piece
carries charge q. If it acquires a speed v when
it has fallen through a vertical height y... | [{"identifier": "A", "content": "v<sup>2</sup> = $$y\\left[ {{{qQ} \\over {4\\pi {\\varepsilon _0}R\\left( {R + y} \\right)m}} + g} \\right]$$"}, {"identifier": "B", "content": "v<sup>2</sup> = $$2y\\left[ {{{qQR} \\over {4\\pi {\\varepsilon _0}{{\\left( {R + y} \\right)}^3}m}} + g} \\right]$$"}, {"identifier": "C", "c... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263285/exam_images/zvkzprbjiwvpe9vuifdc.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Morning Slot Physics - Electrostatics Question 130 English Explanation">
<br><br>App... | mcq | jee-main-2020-online-5th-september-morning-slot | 10,841 |
KkfjLO3WF7JAsho77Djgy2xukfl25hbl | physics | electrostatics | electric-potential-energy-and-electric-potential | Ten charges are placed on the circumference
of a circle of radius R with constant angular
separation between successive charges.
Alternate charges 1, 3, 5, 7, 9 have charge (+q)
each, while 2, 4, 6, 8, 10 have charge (–q) each.
The potential V and the electric field E at the
centre of the circle are respectively.
<br/>... | [{"identifier": "A", "content": "V = 0; E = 0"}, {"identifier": "B", "content": "$$V = {{10q} \\over {4\\pi {\\varepsilon _0}R}}$$; $$E = {{10q} \\over {4\\pi {\\varepsilon _0}{R^2}}}$$"}, {"identifier": "C", "content": "$$V = {{10q} \\over {4\\pi {\\varepsilon _0}R}}$$; E = 0"}, {"identifier": "D", "content": "V = 0; ... | ["A"] | null | Net charge = 5q - 5q = 0
<br><br>Potential of centre = V = $${{K\sum q } \over r}$$
<br><br>V<sub>C</sub> = $${{K\left( 0 \right)} \over r}$$ = 0
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264883/exam_images/jzyrlwuzyu2q8gsea9oj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 au... | mcq | jee-main-2020-online-5th-september-evening-slot | 10,842 |
ZYQO4VcRx4Q77uWLSf1klrz7xj7 | physics | electrostatics | electric-potential-energy-and-electric-potential | 512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ________ V. | [] | null | 128 | Let charge on each drop = q<br><br>radius = r<br><br>$$v = {{kq} \over r}$$<br><br>$$ \Rightarrow $$ $$2 = {{kq} \over r}$$<br><br>radius of bigger<br><br>$${4 \over 3}\pi {R^3} = 512 \times {4 \over 3}\pi {r^3}$$<br><br>$$R = 8r$$<br><br>$$ \therefore $$ $$v = {{k(512)q} \over R} = {{512} \over 8}{{kq} \over r} = {{51... | integer | jee-main-2021-online-25th-february-morning-slot | 10,843 |
25Z1kLsnhI1VvCJXLf1klunqo6e | physics | electrostatics | electric-potential-energy-and-electric-potential | 27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is __________ times that of a smaller drop. | [] | null | 243 | $$(27)\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}$$<br><br>R = 3r<br><br>Potential energy of smaller drop :<br><br>$${U_1} = {3 \over 5}{{k{q^2}} \over r}$$<br><br>Potential energy of bigger drop :<br><br>$$U = {3 \over 5}{{k{Q^2}} \over R}$$<br><br>$$U = {3 \over 5}{{k{{(27q)}^2}} \over R}$$<br><br>$$... | integer | jee-main-2021-online-26th-february-evening-slot | 10,844 |
1ktbnxlg7 | physics | electrostatics | electric-potential-energy-and-electric-potential | The two thin coaxial rings, each of radius 'a' and having charges +Q and $$-$$Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is : | [{"identifier": "A", "content": "$${Q \\over {2\\pi {\\varepsilon _0}}}\\left[ {{1 \\over a} + {1 \\over {\\sqrt {{s^2} + {a^2}} }}} \\right]$$"}, {"identifier": "B", "content": "$${Q \\over {4\\pi {\\varepsilon _0}}}\\left[ {{1 \\over a} + {1 \\over {\\sqrt {{s^2} + {a^2}} }}} \\right]$$"}, {"identifier": "C", "conten... | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264682/exam_images/e64mwnxui8pfey2pftd6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Evening Shift Physics - Electrostatics Question 100 English Explanation"> <br><br>$${V... | mcq | jee-main-2021-online-26th-august-evening-shift | 10,845 |
1l54vg1ub | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>If the electric potential at any point (x, y, z) m in space is given by V = 3x<sup>2</sup> volt. The electric field at the point (1, 0, 3) m will be :</p> | [{"identifier": "A", "content": "3 Vm<sup>$$-$$1</sup>, directed along positive x-axis."}, {"identifier": "B", "content": "3 Vm<sup>$$-$$1</sup>, directed along negative x-axis. "}, {"identifier": "C", "content": "6 Vm<sup>$$-$$1</sup>, directed along positive x-axis."}, {"identifier": "D", "content": "6 Vm<sup>$$-$$1<... | ["D"] | null | <p>$$\overrightarrow E = -{{dV} \over {dx}}\widehat i$$</p>
<p>$$\overrightarrow E = - 6x\widehat i$$</p>
<p>So, $$\overrightarrow E $$ at (1, 0, 3) is</p>
<p>$$\overrightarrow E = - 6\widehat i$$ V/m</p> | mcq | jee-main-2022-online-29th-june-evening-shift | 10,846 |
1l59q7pw2 | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>27 identical drops are charged at 22V each. They combine to form a bigger drop. The potential of the bigger drop will be _____________ V.</p> | [] | null | 198 | <p>Let the charge on one drop is q and its radius is r.</p>
<p>So for one drop $$V = {{kq} \over r}$$</p>
<p>For 27 drops merged new charge will be Q = 27 q and new radius is R = 3r</p>
<p>So new potential is</p>
<p>$$V' = {{kQ} \over R} = 9{{kq} \over r} = 9 \times 22$$ V</p>
<p>= 198 V</p> | integer | jee-main-2022-online-25th-june-evening-shift | 10,848 |
1l5w3ic10 | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>Eight similar drops of mercury are maintained at 12 V each. All these spherical drops combine into a single big drop. The potential energy of bigger drop will be ____________ E. Where E is the potential energy of a single smaller drop.</p> | [] | null | 32 | From law of conservation of charge
<br/><br/>
$$
\begin{aligned}
&q_{i}=\mathrm{q}_{\mathrm{f}} \Rightarrow 8 \times\left(4 \pi \mathrm{E}_{0} \mathrm{R}\right) \times 12=\left(4 \pi E R^{1}\right) \times \mathrm{V}_{f} \\\\
&\Rightarrow 96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} \mathrm{R}^{1} \\\\
&\text { And, } 8 \times... | integer | jee-main-2022-online-30th-june-morning-shift | 10,849 |
1l6p4ymxs | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>Given below are two statements.</p>
<p>Statement I : Electric potential is constant within and at the surface of each conductor.</p>
<p>Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.</p>
<p>In the light of the above statements, choose t... | [{"identifier": "A", "content": "Both Statement I and Statement II are correct"}, {"identifier": "B", "content": "Both Statement I and Statement II are incorrect"}, {"identifier": "C", "content": "Statement I is correct but Statement II is incorrect"}, {"identifier": "D", "content": "Statement I is incorrect but Statem... | ["A"] | null | <p>Since $${\overrightarrow E _{net}} = \overrightarrow 0 $$ in the bulk of a conductor</p>
<p>$$\Rightarrow$$ Potential would be constant.</p>
<p>$$\Rightarrow$$ Statement I is correct</p>
<p>Since a conductor's surface is equipotential, $$\overrightarrow E $$ just outside is perpendicular to the surface.</p> | mcq | jee-main-2022-online-29th-july-morning-shift | 10,850 |
ldo6p32l | physics | electrostatics | electric-potential-energy-and-electric-potential | Considering a group of positive charges, which of the following statements is correct ? | [{"identifier": "A", "content": "Net potential of the system cannot be zero at a point but net electric field can be zero at that point"}, {"identifier": "B", "content": "Net potential of the system at a point can be zero but net electric field can't be zero at that point."}, {"identifier": "C", "content": "Both the ne... | ["A"] | null | $V=\frac{\sum K Q_{i}}{r_{i}}$
<br/><br/>Here, $Q_{i}$ and $r_{i}$ are positive.
<br/><br/>$\therefore V > 0$
<br/><br/>The correct statement is:
<br/><br/>(A) Net potential of the system cannot be zero at a point but net electric field can be zero at that point.
<br/><br/><b>Explanation:</b>
<br/><br/>In a group o... | mcq | jee-main-2023-online-31st-january-evening-shift | 10,851 |
1ldpk49kn | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>Which of the following correctly represents the variation of electric potential $$(\mathrm{V})$$ of a charged spherical conductor of radius $$(\mathrm{R})$$ with radial distance $$(\mathrm{r})$$ from the center?</p> | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldq01kcp/894f2002-ca45-4858-8d9a-1399ec8814fd/fa4cdaa0-a490-11ed-8eb4-ef270ed2fedb/file-1ldq01kcq.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1ldq01kcp/894f2002-ca45-4858-8d9a-1399ec8814fd/fa4... | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leh7vykb/8ec66bb1-70c0-404b-a3fe-3e227d9cea4d/d0fd51a0-b388-11ed-b684-7d62e194d524/file-1leh7vykc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leh7vykb/8ec66bb1-70c0-404b-a3fe-3e227d9cea4d/d0fd51a0-b388-11ed-b684-7d62e194d524/fi... | mcq | jee-main-2023-online-31st-january-morning-shift | 10,852 |
1ldsbwybp | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>For a charged spherical ball, electrostatic potential inside the ball varies with $$r$$ as $$\mathrm{V}=2ar^2+b$$.</p>
<p>Here, $$a$$ and $$b$$ are constant and r is the distance from the center. The volume charge density inside the ball is $$-\lambda a\varepsilon$$. The value of $$\lambda$$ is ____________.</p>
<p>... | [] | null | 12 | <p>$$V = 2a{r^2} + b$$</p>
<p>$$ \Rightarrow E = - {{dV} \over {dr}} = - 4ar$$</p>
<p>$$ \Rightarrow {1 \over {4\pi \varepsilon }}{Q \over {{r^2}}} = - 4ar$$</p>
<p>$$ \Rightarrow {Q \over {{4 \over 3}\pi {r^3}}} = 3 \times \varepsilon \times ( - 4a) = - 12a\varepsilon $$</p>
<p>$$ \Rightarrow \lambda = 12$$</p> | integer | jee-main-2023-online-29th-january-evening-shift | 10,853 |
1ldwrigpd | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>The electric potential at the centre of two concentric half rings of radii R$$_1$$ and R$$_2$$, having same linear charge density $$\lambda$$ is :</p>
<p><img src="data:image/png;base64,UklGRuQQAABXRUJQVlA4INgQAACwzwCdASoAA6MBP4G+2WQ2MCymorGp2sAwCWlu7rOSxo2dwxox58fQfn9fv1y9k1v5wc3d7hf3Dvq+pbC/Vu9B4yV///VQeX3BkIGCovg... | [{"identifier": "A", "content": "$$\\frac{\\lambda}{2\\in_0}$$"}, {"identifier": "B", "content": "$$\\frac{\\lambda}{\\in_0}$$"}, {"identifier": "C", "content": "$$\\frac{2\\lambda}{\\in_0}$$"}, {"identifier": "D", "content": "$$\\frac{\\lambda}{4\\in_0}$$"}] | ["A"] | null | $V_{1}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{\lambda\left(\pi R_{1}\right)}{R_{1}}$
<br/><br/>
$$
\begin{aligned}
V_{2}= & \frac{1}{4 \pi \varepsilon_{0}} \times \frac{\lambda\left(\pi R_{2}\right)}{R_{2}} \\\\
V_{\text {net }} & =V_{1}+V_{2} \\\\
& =2 \times \frac{1}{4 \pi \varepsilon_{0}} \pi \lambda \\\\
& =\... | mcq | jee-main-2023-online-24th-january-evening-shift | 10,854 |
1lgrjoft7 | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>64 identical drops each charged upto potential of $$10 ~\mathrm{mV}$$ are combined to form a bigger drop. The potential of the bigger drop will be __________ $$\mathrm{mV}$$.</p> | [] | null | 160 | The problem involves 64 identical drops, each charged up to a potential of $$10 \mathrm{~mV}$$, that are combined to form a bigger drop. We are asked to find the potential of the bigger drop.
<br/><br/>
The potential of each drop is given by:<br/><br/>
$$V = \frac{Kq}{r}$$<br/><br/>
where $$K$$ is the Coulomb constant,... | integer | jee-main-2023-online-12th-april-morning-shift | 10,855 |
1lgyfsuqi | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>Three concentric spherical metallic shells X, Y and Z of radius a, b and c respectively [a < b < c] have surface charge densities $$\sigma,-\sigma$$ and $$\sigma$$ respectively. The shells X and Z are at same potential. If the radii of X & Y are 2 cm and 3 cm, respectively. The radius of shell Z is _______... | [] | null | 5 | <p>Given three concentric spherical shells X, Y, and Z with radii a, b, and c respectively, and with surface charge densities ( $\sigma$ ), ( $-\sigma$ ), and ( $\sigma$ ) respectively, we know that the potential at the surface of a sphere due to a uniform surface charge is given by:</p>
<p>$ V = \frac{1}{4\pi\epsilon_... | integer | jee-main-2023-online-10th-april-morning-shift | 10,856 |
1lgyqr0ia | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>Electric potential at a point '$$\mathrm{P}$$' due to a point charge of $$5 \times 10^{-9} \mathrm{C}$$ is $$50 \mathrm{~V}$$. The distance of '$$\mathrm{P}$$' from the point charge is:</p>
<p>(Assume, $$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}$$ )</p> | [{"identifier": "A", "content": "0.9 cm"}, {"identifier": "B", "content": "90 cm"}, {"identifier": "C", "content": "3 cm"}, {"identifier": "D", "content": "9 cm"}] | ["B"] | null | <p>The electric potential (V) at a distance (r) from a point charge (Q) is given by the formula:</p>
<p>$ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} $</p>
<p>In this case, we know (V) and (Q), and we're asked to solve for (r). We can rearrange the formula to solve for (r):</p>
<p>$ r = \frac{1}{4\pi\epsilon_0} \frac{... | mcq | jee-main-2023-online-8th-april-evening-shift | 10,857 |
jaoe38c1lscpyqpm | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>The electric potential at the surface of an atomic nucleus $$(z=50)$$ of radius $$9 \times 10^{-13} \mathrm{~cm}$$ is __________ $$\times 10^6 \mathrm{~V}$$.</p> | [] | null | 8 | <p>$$\begin{aligned}
& \text { Potential }=\frac{\mathrm{kQ}}{\mathrm{R}}=\frac{\mathrm{k} . \mathrm{Ze}}{\mathrm{R}} \\
& =\frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}} \\
& =8 \times 10^6 \mathrm{~V}
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-evening-shift | 10,860 |
luz2umur | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>At the centre of a half ring of radius $$\mathrm{R}=10 \mathrm{~cm}$$ and linear charge density $$4 \mathrm{~nC} \mathrm{~m}^{-1}$$, the potential is $$x \pi \mathrm{V}$$. The value of $$x$$ is _________.</p> | [] | null | 36 | <p>$$\begin{aligned}
V & =\frac{K Q}{R} \\
& =\frac{9 \times 10^9 \times 4 \times 10^{-9} \pi R}{R} \\
& =36 \pi
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-morning-shift | 10,862 |
lv0vyvcr | physics | electrostatics | electric-potential-energy-and-electric-potential | <p>An infinitely long positively charged straight thread has a linear charge density $$\lambda \mathrm{~Cm}^{-1}$$. An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1luzqmfde/bacaa397-88aa-4410-af0d-88c01670842f/a7169320-fa7b-11ee-9ce8-f9d90546778f/file-1luzqmfdf.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1luzqmfde/bacaa397-88aa-4410-af0d-88c01670842f/a71... | ["D"] | null | <p>The electron revolves in a circle, so its kinetic energy remains same.</p>
<p>So option (2) best represent the given situation.</p> | mcq | jee-main-2024-online-4th-april-morning-shift | 10,863 |
8MTviplPuzKIkeMH | physics | geometrical-optics | lenses | A thin glass (refractive index $$1.5$$) lens has optical power of $$-5$$ $$D$$ in air. Its optical power in a liquid medium with refractive index $$1.6$$ will be | [{"identifier": "A", "content": "$$-1$$ $$D$$ "}, {"identifier": "B", "content": "$$1$$ $$D$$ "}, {"identifier": "C", "content": "$$-25$$ $$D$$ "}, {"identifier": "D", "content": "$$25$$ $$D$$"}] | ["B"] | null | $${1 \over {{f_a}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$${1 \over {{f_m}}} = \left( {{{{\mu _g}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
<br><br>$${1 \over {{f_m}... | mcq | aieee-2005 | 10,865 |
N93RksJ30NYA8ysz | physics | geometrical-optics | lenses | Two lenses of power $$-15$$ $$D$$ and $$+5$$ $$D$$ are in contact with each other. The focal length of the combination is | [{"identifier": "A", "content": "$$ + 10\\,cm$$ "}, {"identifier": "B", "content": "$$ - 20\\,cm$$"}, {"identifier": "C", "content": "$$ - 10\\,cm$$"}, {"identifier": "D", "content": "$$ + 20\\,cm$$"}] | ["C"] | null | Power of combination is given by
<br><br>$$P = {P_1} + {P_2} = \left( { - 15 + 5} \right)D$$ $$\,\,\,\,\,\,\,\,\, = - 10D.$$
<br><br>Now, $$P = {1 \over f} \Rightarrow f = {1 \over P} = {1 \over { - 10}}$$ metre
<br><br>$$\therefore$$ $$f = - \left( {{1 \over {10}} \times 100} \right)cm = - 10\,cm.$$ | mcq | aieee-2007 | 10,866 |
jNhKtkSIRYIy9woC | physics | geometrical-optics | lenses | In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance $$u$$ and the image distance $$v,$$ from the lens, is plotted using the same scale for the ... | [{"identifier": "A", "content": "$$\\left( {{f \\over 2},{f \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {f,f} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( {4f,4f} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {2f,2f} \\right)$$ "}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265852/exam_images/c4pi5km2sotdchk0vdjn.webp" loading="lazy" alt="AIEEE 2009 Physics - Geometrical Optics Question 184 English Explanation">
<br><br>Here $$u = - 2f,v = 2f$$
<br><br>As $$|u|$$ increases, $$v$$ decreases for $$|u| ... | mcq | aieee-2009 | 10,868 |
h6o7Ed95gvDtiKy2 | physics | geometrical-optics | lenses | An object $$2.4$$ $$m$$ in front of a lens forms a sharp image on a film $$12$$ $$cm$$ behind the lens. A glass plate $$1$$ $$cm$$ thick, of refractive index $$1.50$$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of fi... | [{"identifier": "A", "content": "$$7.2$$ $$m$$ "}, {"identifier": "B", "content": "$$24$$ $$m$$ "}, {"identifier": "C", "content": "$$3.2$$ $$m$$ "}, {"identifier": "D", "content": "$$5.6$$ $$m$$ "}] | ["D"] | null | The focal length of the lens
<br><br>$${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$$
<br><br>$$ = {{20 + 1} \over {240}} = {{21} \over {240}}$$
<br><br>$$f = {{240} \over {21}}cm$$
<br><br>Shift $$ = t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \r... | mcq | aieee-2012 | 10,869 |
QBKQfus1LKHDHtNk | physics | geometrical-optics | lenses | Diameter of a plano-convex lens is $$6$$ $$cm$$ and thickness at the center is $$3mm$$. If speed of light in material of lens is $$2 \times {10^8}\,m/s,$$ the focal length of the lens is | [{"identifier": "A", "content": "$$15$$ $$cm$$ "}, {"identifier": "B", "content": "$$20$$ $$cm$$ "}, {"identifier": "C", "content": "$$30$$ $$cm$$ "}, {"identifier": "D", "content": "$$10$$ $$cm$$ "}] | ["C"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/PX9LeYP8KWRWDFYVw/7s4aVYynlCAzCLApMDxTss02CcRuY/YJrGi0uEP7jzJtcydwLIqj/image.svg" loading="lazy" alt="JEE Main 2013 (Offline) Physics - Geometrical Optics Question 178 English Explanation">
<br><br>$$\therefore$$ $$n = {{Velocity\,\,of\,\,light\,\,in\,\,v... | mcq | jee-main-2013-offline | 10,870 |
xNIG8WpORNYinWn2 | physics | geometrical-optics | lenses | A thin convex lens made from crown glass $$\left( {\mu = {3 \over 2}} \right)$$ has focal length $$f$$. When it is measured in two different liquids having refractive indices $${4 \over 3}$$ and $${5 \over 3},$$ it has the focal lengths $${f_1}$$ and $${f_2}$$ respectively. The correct relation between the focal leng... | [{"identifier": "A", "content": "$${f_1} = {f_2} < f$$ "}, {"identifier": "B", "content": "$${f_1} > f$$ and $${f_2}$$ becomes negative "}, {"identifier": "C", "content": "$${f_2} > f$$ and $${f_1}$$ becomes negative "}, {"identifier": "D", "content": "$${f_1}\\,$$ and$${f_2}\\,$$ both become negative "}] | ["B"] | null | By Lens maker's formula for convex lens
<br><br>$${1 \over f} = \left( {{\mu \over {{\mu _L}}} - 1} \right)\left( {{2 \over R}} \right)$$
<br><br>for, $$\mu {L_1} = {4 \over 3},{f_1} = 4R$$
<br><br>for $$\mu {L_2} = {5 \over 3},{f_2} = - 5R$$
<br><br>$$ \Rightarrow {f_2} = \left( - \right)ve$$ | mcq | jee-main-2014-offline | 10,871 |
oG9JGEP1uiDq1rmVm22Ym | physics | geometrical-optics | lenses | To find the focal length of a convex mirror, a student records the following data :
<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:... | [{"identifier": "A", "content": "f<sub>1</sub> = 12.7 cm f<sub>2</sub> = 7.8 cm"}, {"identifier": "B", "content": "f<sub>1</sub> = 7.8 cm f<sub>2</sub> = 12.7 cm"}, {"identifier": "C", "content": "f<sub>1</sub> = 7.8 cm f<sub>2</sub> = 25.4 cm"}, {"identifier": "D",... | ["B"] | null | For lens :
<br><br>u<sub>1</sub> = $$-$$ (32.2 $$-$$ 22.2) cm
<br><br>= $$-$$ 10 cm
<br><br>v<sub>1</sub> = (71.2 $$-$$ 32.2) cm
<br><br>= 39 cm
<br><br>$$ \therefore $$ $${1 \over {{f_1}}}$$ = $${1 \over {{v_1}}} - {1 \over {{u_1}}}$$
<br><br>= $${1 \over {39}}$$ + $${1 \over {10}}$$
<br><br>= $${{49... | mcq | jee-main-2016-online-9th-april-morning-slot | 10,873 |
mmI4TdE5otk9XYLM7MAu9 | physics | geometrical-optics | lenses | In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is : | [{"identifier": "A", "content": "27.5 cm"}, {"identifier": "B", "content": "20.0 cm"}, {"identifier": "C", "content": "25.0 cm"}, {"identifier": "D", "content": "30.5 cm"}] | ["A"] | null | <p>The given optical situation is depicted in the following ray diagram:</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l0b0kbts/b7915c52-39cd-425f-ab1e-3c7b5a11f01d/726bae00-9af4-11ec-b7c4-9d40ef99bd3a/file-1l0b0kbtt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l0b0... | mcq | jee-main-2017-online-9th-april-morning-slot | 10,874 |
CgPelWEWGd8PskdE | physics | geometrical-optics | lenses | A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15cm from a converging
lens of magnitude of focal length 20cm. A beam of parallel light falls on the diverging lens. The final
image formed is: | [{"identifier": "A", "content": "real and at a distance of 6 cm from the convergent lens."}, {"identifier": "B", "content": "real and at a distance of 40 cm from convergent lens."}, {"identifier": "C", "content": "virtual and at a distance of 40 cm from convergent lens."}, {"identifier": "D", "content": "real and at a ... | ["B"] | null | As parallel beam incident on diverging lens so the image will be formed
at the focus of diverging lens.
<br><br>$$ \therefore $$ v = –25 cm
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266075/exam_images/tad0rleixsoein2adytk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" l... | mcq | jee-main-2017-offline | 10,875 |
TSx0M3mCVOs7kcSC01lw0 | physics | geometrical-optics | lenses | A planoconvex lens becomes an optical system of $$28$$ $$cm$$ focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A.
<br/><br/>If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig-B, it acts like an optical system of focal le... | [{"identifier": "A", "content": "$$1.50$$ "}, {"identifier": "B", "content": "$$1.55$$ "}, {"identifier": "C", "content": "$$1.75$$ "}, {"identifier": "D", "content": "$$1.51$$ "}] | ["B"] | null | When plane durface is silvered then focus length of lens,
<br><br>f<sub>1</sub> = $${R \over {2\left( {\mu - 1} \right)}}$$
<br><br>and when curver surface is silvered then focus length of lens,
<br><br>f<sub>2</sub> = $${R \over {2\mu }}$$
<br><br>$$\therefore\,\,\,\,$$ $${{{f_1}} \over {{f_2}}} = {R \over {2\lef... | mcq | jee-main-2018-online-15th-april-morning-slot | 10,877 |
HANDsjkXe5buQmS0JoLKe | physics | geometrical-optics | lenses | A plano-convex lens (focal length f<sub>2</sub>, refractive index $$\mu $$<sub>2</sub>, radius of curvature R) fits exactly into a plano-concave lens (focal length f<sub>1</sub>, refractive index $$\mu $$<sub>1</sub>, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the... | [{"identifier": "A", "content": "f<sub>1</sub><sub></sub> + f<sub>2</sub>"}, {"identifier": "B", "content": "f<sub>1</sub> $$-$$ f<sub>2</sub>"}, {"identifier": "C", "content": "$${R \\over {{\\mu _2} - {\\mu _1}}}$$"}, {"identifier": "D", "content": "$${{2{f_1}{f_2}} \\over {{f_1} + {f_2}}}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264151/exam_images/mojklubuscboti9ko6j0.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Physics - Geometrical Optics Question 148 English Explanation">
<br><b... | mcq | jee-main-2019-online-12th-january-evening-slot | 10,878 |
9pEyUPYfJSQIhdzETR3rsa0w2w9jwzlyosq | physics | geometrical-optics | lenses | The graph shows how the magnification m produced by a thin lens varies with image distance v. What is the focal length of the lens used?
<img src="data:image/png;base64,UklGRsgJAABXRUJQVlA4ILwJAADwRQCdASpNAfoAPm02mEikIqKhIRF6EIANiWlu4XExG/OZ8H/zv8ffBz/EflB2BnnP2p9JKsovt/xH8p/sf+i/Kj3x/vv8f/Z3z59r/9R/Nv1m+Av0//l/5Z/S/ML... | [{"identifier": "A", "content": "$${{{b^2}} \\over {ac}}$$"}, {"identifier": "B", "content": "$${{{b^2}c} \\over a}$$"}, {"identifier": "C", "content": "$${a \\over c}$$"}, {"identifier": "D", "content": "$${b \\over c}$$"}] | ["D"] | null | As the graph between magnification (m) and
image distance (v) varies linearly, then<br>
m = k<sub>1</sub>v + k<sub>2</sub><br><br>
$$ \Rightarrow {v \over u} = {k_1}v + {k_2}$$<br><br>
$$ \Rightarrow {1 \over u} = {k_1} + {{{k_2}} \over v}$$<br><br>
$$ \Rightarrow {{{k_2}} \over v} - {1 \over u} = {k_1}$$<br><br>
Clear... | mcq | jee-main-2019-online-10th-april-evening-slot | 10,879 |
OVZJZSbwNqaomEGZfH18hoxe66ijvzmsn4h | physics | geometrical-optics | lenses | One plano-convex and one plano-concave lens
of same radius of curvature 'R' but of different
materials are joined side by side as shown in
the figure. If the refractive index of the material
of 1 is $$\mu $$<sub>1 </sub>and that of 2 is $$\mu $$<sub>2</sub>, then the focal
length of the combination is :
<img src="data:... | [{"identifier": "A", "content": "$${2R \\over { {{\\mu _1} - {\\mu _2}}}}$$"}, {"identifier": "B", "content": "$${R \\over {2 - \\left( {{\\mu _1} - {\\mu _2}} \\right)}}$$"}, {"identifier": "C", "content": "$${R \\over { {{\\mu _1} - {\\mu _2}}}}$$"}, {"identifier": "D", "content": "$${R \\over {2 \\left( {{\\mu _1}... | ["C"] | null | For 1<sup>st</sup> lens $${1 \over {{f_1}}} = \left( {{{{\mu _1} - 1} \over 1}} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right) = {{{\mu _1} - 1} \over R}$$<br><br>
For 2<sup>nd</sup> lens $${1 \over {{f_2}}} = \left( {{{{\mu _2} - 1} \over 1}} \right)\left( {{1 \over { - R}}} \right) - 0 = {{{\mu _2} - 1}... | mcq | jee-main-2019-online-10th-april-morning-slot | 10,880 |
IEKDHT0eM8TDV5qSHppje | physics | geometrical-optics | lenses | A convex lens of focal length 20 cm produces
images of the same magnification 2 when an
object is kept at two distances x<sub>1</sub> and x<sub>2</sub>
(x<sub>1</sub> > x<sub>2</sub>) from the lens. The ratio of x<sub>1</sub> and x<sub>2</sub>
is :- | [{"identifier": "A", "content": "3 : 1"}, {"identifier": "B", "content": "2 : 1"}, {"identifier": "C", "content": "5 : 3"}, {"identifier": "D", "content": "4 : 3"}] | ["A"] | null | Magnification is 2<br><br>
If image is real, $${x_1} = {{3f} \over 2}$$<br><br>
If image is virtual, $${x_2} = {f \over 2}$$<br><br>
$${{{x_1}} \over {{x_2}}} = 3:1$$ | mcq | jee-main-2019-online-9th-april-evening-slot | 10,881 |
W4ZcpX60BGJC3e308a5S9 | physics | geometrical-optics | lenses | A thin convex lens L (refractive index = 1.5)
is placed on a plane mirror M. When a pin is
placed at A, such that OA = 18 cm, its real
inverted image is formed at A itself, as shown
in figure. When a liquid of refractive index μ1
is put between the lens and the mirror, The pin
has to be moved to A', such that OA' = 27 ... | [{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "3 / 2"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "4 / 3"}] | ["D"] | null | For image to form at object itself, says must retrace their path back to object. Hence
must incident on mirror normally.<br><br>
Case 1: Object will be at focus of lens <br>
$${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right) = {1 \over { - 18}}$$
$$ \Rightarrow $$ R = 18 cm<br><br>... | mcq | jee-main-2019-online-9th-april-evening-slot | 10,882 |
vMmFWJZKzLPyuBsfdGl8U | physics | geometrical-optics | lenses | A convex lens (of focal length 20 cm) and a
concave mirror, having their principal axes
along the same lines, are kept 80 cm apart from
each other. The concave mirror is to the right
of the convex lens. When an object is kept at
a distance of 30 cm to the left of the convex
lens, its image remains at the same position
... | [{"identifier": "A", "content": "25 cm"}, {"identifier": "B", "content": "10 cm"}, {"identifier": "C", "content": "20 cm"}, {"identifier": "D", "content": "30 cm"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265550/exam_images/jypm5yq892xxv7umqtpr.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Physics - Geometrical Optics Question 144 English Explanation"><br><br>
Ima... | mcq | jee-main-2019-online-8th-april-evening-slot | 10,883 |
Zly3XvQjqaA02vIrdIVd6 | physics | geometrical-optics | lenses | An upright object is placed at a distance of
40 cm in front of a convergent lens of focal
length 20 cm. A convergent mirror of focal
length 10 cm is placed at a distance of 60 cm
on the other side of the lens. The position and
size of the final image will be : | [{"identifier": "A", "content": "20 cm from the convergent mirror, same\nsize as the object"}, {"identifier": "B", "content": "40 cm from the convergent mirror, same\nsize as the object"}, {"identifier": "C", "content": "40 cm from the convergent lens, same\nsize as the object"}, {"identifier": "D", "content": "20 cm f... | ["C"] | null | <p>In given system of lens and mirror, position of object O in front of lens is at a distance 2f. i.e. u = 2f = 40 cm</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l35inazs/a048151d-56b2-4eba-8b91-8feeb29d77cf/1bc563b0-d353-11ec-ada9-55bd22062331/file-1l35inazv.png?format=png" data-orsrc="http... | mcq | jee-main-2019-online-8th-april-morning-slot | 10,884 |
wEzErLuc1SUAsMXI9fIcl | physics | geometrical-optics | lenses | Formation of real image using a biconvex lens is shown below :
<br/><br/><img src="data:image/png;base64,UklGRr4NAABXRUJQVlA4ILINAAAQjQCdASoAAygBP4G+1mU2L6wnIhMJssAwCWlu/FuX9DYHZ1+frhjB+B/6Dc3dm8uj8DjGp1qymef/9vB/Ezv+dllVYB9u/52WVVgH27/nY5pa+qEYJK47LKqdfVuMNZRmAW+vc8MtxhrKMwC317m8D8AhNoV+OGsozGmiKAd9e5w1lGYBZfIkW+Yph+U... | [{"identifier": "A", "content": "Image disappears "}, {"identifier": "B", "content": "Magnified image"}, {"identifier": "C", "content": "Erect real image"}, {"identifier": "D", "content": "No change "}] | ["A"] | null | From $${1 \over F} = \left( {{\mu _{rel}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
<br><br>Focal length of lens will change hence image disappears from the screen. | mcq | jee-main-2019-online-12th-january-evening-slot | 10,885 |
aMJtQ6yfnGvOeYjfJLeyy | physics | geometrical-optics | lenses | What is the position and nature of image formed by lens combination shown in figure ? (f<sub>1</sub>, f<sub>2</sub> are focal lengths)
<br/><br/><img src="data:image/png;base64,UklGRgYSAABXRUJQVlA4IPoRAADQqQCdASoAA04BP4HA12S2MCwnIZPqQsAwCWlu+EQM0UAQis4JjcLneaP8F25/7L8sOz6DzZ4f3nfv+4eIXlH2H3a+YRfp4Q/H/svJ5X/38HsWINOFHJO... | [{"identifier": "A", "content": "$${{20} \\over 3}$$ cm from point B at right, real "}, {"identifier": "B", "content": "70 cm from point B at right ; real"}, {"identifier": "C", "content": "40 cm from point B at right; real "}, {"identifier": "D", "content": "70 cm from point B at left ; virtual"}] | ["B"] | null | For first lens
<br><br>$${1 \over V} - {1 \over { - 20}} = {1 \over 5}$$
<br><br>V $$=$$ $${{20} \over 3}$$
<br><br>For second lens
<br><br>V = $${{20} \over 3}$$ $$-$$ 2 $$=$$ $${{14} \over 3}$$
<br><br>$${1 \over V} - {1 \over {{{14} \over 3}}} = {1 \over { - 5}}$$
<br><br>V $$=$$ 70cm | mcq | jee-main-2019-online-12th-january-morning-slot | 10,886 |
7p8e61TpaHO7YAaYwjSBZ | physics | geometrical-optics | lenses | A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is : | [{"identifier": "A", "content": "1.1 cm away from the lens"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "0.55 cm towards the lens"}, {"identifier": "D", "content": "0.55 cm away from the lens"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267378/exam_images/dqzebqidyn1wn19qpyr5.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Physics - Geometrical Optics Question 157 English Explanation 1">
<br><... | mcq | jee-main-2019-online-9th-january-morning-slot | 10,890 |
5SpfAsKpCKEQJp72q77k9k2k5f7jlte | physics | geometrical-optics | lenses | A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid
of refractive index 1.42. If its focal length in liquid is f<sub>1</sub>
, then the ratio $${{{f_1}} \over f}$$ is closest to the
integer : | [{"identifier": "A", "content": "17"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "5"}] | ["C"] | null | Using formula
<br>$${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
<br><br>$${1 \over {{f}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$ ...(1)
<br><br>and $${1 \over {{f_1}}} = \left( {{{1.5} \o... | mcq | jee-main-2020-online-7th-january-evening-slot | 10,891 |
zC4Xqdd4aPJaZLYGNc7k9k2k5gy43sb | physics | geometrical-optics | lenses | A point object in air is in front of the curved
surface of a plano-convex lens. The radius of
curvature of the curved surface is 30 cm and
the refractive index of the lens material is 1.5,
then the focal length of the lens (in cm)
is ______. | [] | null | 60 | Lens-maker formula
<br><br>$${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
<br><br>for plano-convex lens
<br><br>$${{R_1}}$$ = $$\infty $$ and R<sub>2</sub> = -R
<br><br>$$ \therefore $$ f = $${R \over {\mu - 1}}$$ = $${{30} \over {1.5 - 1}}$$ = 60 cm | integer | jee-main-2020-online-8th-january-morning-slot | 10,892 |
RZBFFArevFDXMHs4lbjgy2xukfayia4h | physics | geometrical-optics | lenses | The distance between an object and a screen is 100 cm. A lens can produce real image of the
object on the screen for two different positions between the screen and the object. The distance
between these two positions is 40 cm. If the power of the lens is close to $$\left( {{N \over {100}}} \right)D$$ where N is an
inte... | [] | null | 476 | Using displacement method<br><br>$$f = {{{D^2} - {d^2}} \over {4D}}$$<br><br>Here, D = 100 cm<br><br>and d = 40 cm<br><br>$$f = {{{{100}^2} - {{40}^2}} \over {4(100)}} = 21\,cm$$<br><br>$$P = {1 \over f} = {{100} \over {21}}D$$<br><br>$${N \over {100}} = {{100} \over {21}}$$<br><br>$$N = 476$$ | integer | jee-main-2020-online-4th-september-evening-slot | 10,893 |
6M9U3zpBj2SyZvh6RHjgy2xukg09z0ui | physics | geometrical-optics | lenses | A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of
curvature of a surface of a plano-convex lens made of the same material with power 1.5 P is : | [{"identifier": "A", "content": "$${R \\over 3}$$"}, {"identifier": "B", "content": "$${{3R} \\over 2}$$"}, {"identifier": "C", "content": "$${R \\over 2}$$"}, {"identifier": "D", "content": "2R"}] | ["A"] | null | Assume refractive index = $$\mu $$<sub>$$l$$</sub>
<br><br>P = $$\left( {{\mu _l} - 1} \right)\left( {{2 \over R}} \right)$$ .....(1)
<br><br>$${3 \over 2}P = \left( {{\mu _l} - 1} \right)\left( {{1 \over {{R_1}}}} \right)$$ ......(2)
<br><br>from (1)/(2)
<br><br>$${P \over {{3 \over 2}P}} = {{\left( {{2 \over R}} \rig... | mcq | jee-main-2020-online-6th-september-evening-slot | 10,896 |
1f68Q6TWCISDhItzx71klryu4jm | physics | geometrical-optics | lenses | The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is __________ cm. | [] | null | 15 | $${1 \over v} - {1 \over u} = {1 \over f}$$ .... (1)
<br><br>m = $${v \over u}$$ ..... (2)
<br><br>from (1) and (2) we get
<br><br>m = $${f \over {f + u}}$$
<br><br>given conditions
<br><br>m<sub>1</sub> = -m<sub>2</sub>
<br><br>$${f \over {f - 10}} = {{ - f} \over {f - 20}}$$
<br><br>$$ \Rightarrow $$ f – 20 = -f + 10... | integer | jee-main-2021-online-25th-february-morning-slot | 10,897 |
sk6OHjgasMT1o2GEQW1kmiogq4k | physics | geometrical-optics | lenses | The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R<sub>1</sub> and R<sub>2</sub> respectively. | [{"identifier": "A", "content": "Zero"}, {"identifier": "B", "content": "Infinite"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${{{R_1}{R_2}} \\over {{R_1} - {R_2}}}$$"}] | ["B"] | null | Given, initially refractive index (n<sub>1</sub>) = 1.4
<br><br>Then it placed in medium of same refractive index.
<br><br>$$ \therefore $$ n<sub>2</sub> = 1.4
<br><br>We know, Focal length
<br><br>$${1 \over f} = \left( {{{{n_1}} \over {{n_2}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
<br><... | mcq | jee-main-2021-online-16th-march-evening-shift | 10,898 |
UXW2043csIElKMbMBx1kmj21n72 | physics | geometrical-optics | lenses | The thickness at the centre of a plane convex lens is 3 mm and the diameter is 6 cm. If the speed of light in the material of the lens is 2 $$\times$$ 10<sup>8</sup> ms<sup>$$-$$1</sup>. The focal length of the lens is ____________. | [{"identifier": "A", "content": "0.30 cm"}, {"identifier": "B", "content": "30 cm"}, {"identifier": "C", "content": "15 cm"}, {"identifier": "D", "content": "1.5 cm"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266212/exam_images/osor7k0wtqqcp5iin9uu.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Morning Shift Physics - Geometrical Optics Question 105 English Explanation"><br><br>$$... | mcq | jee-main-2021-online-17th-march-morning-shift | 10,899 |
1krpnyi1z | physics | geometrical-optics | lenses | Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :<br/><br/><img src="data:image/png;base64,UklGRkYJAABXRUJQVlA4IDoJAABwQQCdASq5AZ8APm02mEikIyKhIROaUIANiWlu4XExG/Of8G/0T8d/Az+39AF4o9... | [{"identifier": "A", "content": "55.44 cm"}, {"identifier": "B", "content": "18.23 cm"}, {"identifier": "C", "content": "9.52 cm"}, {"identifier": "D", "content": "37.58 cm"}] | ["D"] | null | As we know that, the equation of refraction at spherical surface is :<br/><br/>$${{{\mu _{II}}} \over v} - {{{\mu _I}} \over u} = {{{\mu _{II}} - {\mu _I}} \over R}$$<br/><br/>where, <br/><br/>$$\mu$$<sub>II</sub> = refractive index of region II = 1.4<br/><br/>$$\mu$$<sub>I</sub> = refractive index of region I = 1.25<b... | mcq | jee-main-2021-online-20th-july-morning-shift | 10,900 |
1ktbvub35 | physics | geometrical-optics | lenses | An object is placed at a distance of 12 cm from a convex lens. A convex mirror of focal length 15 cm is placed on other side of lens at 8 cm as shown in the figure. Image of object coincides with the object.<br/><br/><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v... | [] | null | 50 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264474/exam_images/vwgxzpcbt833t4jos3cc.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264585/exam_images/eqnll2t8oofxpitmvuo7.webp"><source media="(max-wid... | integer | jee-main-2021-online-26th-august-evening-shift | 10,901 |
1kte6cx6h | physics | geometrical-optics | lenses | Find the distance of the image from object O, formed by the combination of lenses in the figure :<br/><br/><img src="data:image/png;base64,UklGRq4YAABXRUJQVlA4IKIYAAAwjwCdASoKAioBPm02l0ikIyKhIdQqoIANiWlu/HyYsF9ZMo7nw//PPx7943eR9s/sv7X/2nyI/NP2L8mvSw/a/4v+yHsE5Q/xf9S/bf3T/lH1K+6fxr+p/7r+4/uT98P0T/F/xv9m/Qv3tf0nqEfr/8u/z... | [{"identifier": "A", "content": "75 cm"}, {"identifier": "B", "content": "10 cm"}, {"identifier": "C", "content": "20 cm"}, {"identifier": "D", "content": "infinity"}] | ["A"] | null | $${1 \over {{V_1}}} + {1 \over {30}} = {1 \over {10}}$$<br><br>$${1 \over {{V_1}}} = {2 \over {30}} \Rightarrow {V_1} = 15$$ cm<br><br>$${1 \over {{V_2}}} - {1 \over {10}} = - {1 \over {10}}$$<br><br>$${1 \over {{V_2}}} = 0$$<br><br>V<sub>2</sub> = $$\infty$$<br><br>V<sub>3</sub> = 30 cm<br><br>OV<sub>3</sub> = 75 cm | mcq | jee-main-2021-online-27th-august-morning-shift | 10,902 |
1ktfj43dp | physics | geometrical-optics | lenses | Curved surfaces of a plano-convex lens of refractive index $$\mu$$<sub>1</sub> and a plano-concave lens of refractive index $$\mu$$<sub>2</sub> have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.<br/><br/><img src="data:image/png;base64,Uk... | [{"identifier": "A", "content": "$${1 \\over {{\\mu _2} - {\\mu _1}}}$$"}, {"identifier": "B", "content": "$${\\mu _1} - {\\mu _2}$$"}, {"identifier": "C", "content": "$${1 \\over {{\\mu _1} - {\\mu _2}}}$$"}, {"identifier": "D", "content": "$${\\mu _2} - {\\mu _1}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263800/exam_images/a2fkusiflxmdouzjzrit.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Physics - Geometrical Optics Question 88 English Explanation"><br>$${1 \... | mcq | jee-main-2021-online-27th-august-evening-shift | 10,903 |
1kth17ivn | physics | geometrical-optics | lenses | An object is placed at the focus of concave lens having focal length f. What is the magnification and distance of the image from the optical centre of the lens? | [{"identifier": "A", "content": "1, $$\\infty$$"}, {"identifier": "B", "content": "Very high, $$\\infty$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$, $${f \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$, $${f \\over 4}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263869/exam_images/j5kj3zbqnpkfn4wd4ipc.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Physics - Geometrical Optics Question 87 English Explanation"><br>U = $$... | mcq | jee-main-2021-online-31st-august-morning-shift | 10,904 |
1l5489lgg | physics | geometrical-optics | lenses | <p>A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is _____________ mm.</p> | [] | null | 225 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5grbhxq/331088c1-e677-4427-83c5-a1030fb4cbdc/137c8ae0-011a-11ed-9784-f3e7c455b437/file-1l5grbhxr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5grbhxq/331088c1-e677-4427-83c5-a1030fb4cbdc/137c8ae0-011a-11ed-9784-f3e7c455b437... | integer | jee-main-2022-online-29th-june-morning-shift | 10,905 |
1l56v9yke | physics | geometrical-optics | lenses | <p>A convex lens has power P. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figures). Choose the incorrect option for the reported pieces.</p>
<p> <img src="data:image/png;base64,UklGRm4QAABXRUJQVlA4I... | [{"identifier": "A", "content": "Power of $${L_1} = {P \\over 2}$$"}, {"identifier": "B", "content": "Power of $${L_2} = {P \\over 2}$$"}, {"identifier": "C", "content": "Power of $${L_3} = {P \\over 2}$$"}, {"identifier": "D", "content": "Power of L<sub>1</sub> = P"}] | ["A"] | null | <p>We know $$P = {1 \over f} = (\mu - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$</p>
<p>$${L_1}:{1 \over {{f_1}}} = (\mu - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) = {P_1} = (\mu - 1)\left( {{2 \over R}} \right) = P$$</p>
<p>$${L_2}:{1 \over {{f_2}}} = (\mu - 1)\left( {{1 \over {{R_1... | mcq | jee-main-2022-online-27th-june-evening-shift | 10,906 |
1l5c4foeo | physics | geometrical-optics | lenses | <p>Two identical thin biconvex lens of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is __________ cm.</p> | [] | null | 10 | <p>$${1 \over {{f_l}}} = \left( {{{{\mu _e}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$</p>
<p>here $$|{R_1}| = |{R_2}| = R$$</p>
<p>$$ \Rightarrow {1 \over {{f_{{l_1}}}}} = (1.5 - 1)\left( {{2 \over R}} \right) = {1 \over {15}}$$</p>
<p>$$ \Rightarrow {1 \over R} = {1 \over ... | integer | jee-main-2022-online-24th-june-morning-shift | 10,907 |
1l6f4ohfa | physics | geometrical-optics | lenses | <p>For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance s... | [{"identifier": "A", "content": "0.8 m"}, {"identifier": "B", "content": "3.2 m"}, {"identifier": "C", "content": "1.2 m"}, {"identifier": "D", "content": "5.6 m"}] | ["B"] | null | <p>The shift produced by the glass plate is</p>
<p>$$d = t\left( {1 - {1 \over \mu }} \right) = 1 \times \left( {1 - {1 \over {1.5}}} \right) = {1 \over 3}$$ cm</p>
<p>So final image must be produced at $$\left( {12 - {1 \over 3}} \right)$$ cm $$ = {{35} \over 3}$$ cm from lens so that glass plate must shift it to prod... | mcq | jee-main-2022-online-25th-july-evening-shift | 10,908 |
1l6f5v1ju | physics | geometrical-optics | lenses | <p>A convex lens of focal length 20 cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60 cm from the convex lens. The image formed by combination coincides the object itself. ... | [] | null | 10 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6umex33/b64b6a21-31fd-4076-9f18-72715abb5a3c/1570e8f0-1c86-11ed-b633-f353ad3cb8e4/file-1l6umex34.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6umex33/b64b6a21-31fd-4076-9f18-72715abb5a3c/1570e8f0-1c86-11ed-b633-f353ad3cb8e4... | integer | jee-main-2022-online-25th-july-evening-shift | 10,909 |
1l6gooyuf | physics | geometrical-optics | lenses | <p>The graph between $$\frac{1}{u}$$ and $$\frac{1}{v}$$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is $$1.5$$ and its both the surfaces have same radius of curvature $$R$$. The value of $$R$$ will be ____________ $$\mathrm{cm} .$$ (wher... | [] | null | 10 | <p>$$f = 10$$ cm</p>
<p>$${1 \over f} = (\mu - 1)\left( {{1 \over R} - {1 \over { - R}}} \right)$$</p>
<p>$${1 \over {10}} = {{1.5 - 1} \over 1} \times {2 \over R}$$</p>
<p>$$ {1 \over {10}} = {1 \over R}$$</p>
<p>$$R = 10$$ cm</p> | integer | jee-main-2022-online-26th-july-morning-shift | 10,910 |
1l6nsxeio | physics | geometrical-optics | lenses | <p>The power of a lens (biconvex) is $$1.25 \mathrm{~m}^{-1}$$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are $$20 \mathrm{~cm}$$ and $$40 \mathrm{~cm}$$ respectively. The refractive index of surrounding medium:</p> | [{"identifier": "A", "content": "1.0"}, {"identifier": "B", "content": "$$\\frac{9}{7}$$"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "$$\\frac{4}{3}$$"}] | ["B"] | null | <p>$$\because$$ $${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$</p>
<p>$$ \Rightarrow {{1.25} \over {100}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {20}} + {1 \over {40}}} \right)$$</p>
<p>$$ \Rightarrow {1 \over {80}} = \... | mcq | jee-main-2022-online-28th-july-evening-shift | 10,911 |
1l6nu6klw | physics | geometrical-optics | lenses | <p>In an experiment with a convex lens, The plot of the image distance $$\left(v^{\prime}\right)$$ against the object distance ($$\left.\mu^{\prime}\right)$$ measured from the focus gives a curve $$v^{\prime} \mu^{\prime}=225$$. If all the distances are measured in $$\mathrm{cm}$$. The magnitude of the focal length of ... | [] | null | 15 | <p>Using Newton's formula for lenses,</p>
<p>$$v'\mu ' = {f^2} = 225 \Rightarrow f = 15$$</p> | integer | jee-main-2022-online-28th-july-evening-shift | 10,912 |
1ldr2gyi9 | physics | geometrical-optics | lenses | <p>A person has been using spectacles of power $$-1.0$$ dioptre for distant vision and a separate reading glass of power $$2.0$$ dioptres. What is the least distance of distinct vision for this person :</p> | [{"identifier": "A", "content": "50 cm"}, {"identifier": "B", "content": "40 cm"}, {"identifier": "C", "content": "30 cm"}, {"identifier": "D", "content": "10 cm"}] | ["A"] | null | <p>u = 25 cm</p>
<p>$$f=\frac{1}{2}$$ m = 50 cm</p>
<p>$${1 \over v} - {1 \over u} = {1 \over f}$$</p>
<p>$$ \Rightarrow {1 \over v} + {1 \over {25}} = - {1 \over {50}}$$</p>
<p>$${1 \over v} = - {1 \over {50}}$$</p>
<p>$$ \Rightarrow u = - 50$$ cm</p> | mcq | jee-main-2023-online-30th-january-morning-shift | 10,913 |
1ldtzs86q | physics | geometrical-optics | lenses | <p>An object is placed on the principal axis of convex lens of focal length 10cm as shown. A plane mirror is placed on the other side of lens at a distance of 20 cm. The image produced by the plane mirror is 5cm inside the mirror. The distance of the object from the lens is ___________ cm.</p>
<p><img src="data:image/p... | [] | null | 30 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ledm3868/df6832b9-547b-438d-a902-5c53474cd5e1/38498200-b18d-11ed-a682-13f364283dca/file-1ledm3869.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ledm3868/df6832b9-547b-438d-a902-5c53474cd5e1/38498200-b18d-11ed-a682-13f364283dca... | integer | jee-main-2023-online-25th-january-evening-shift | 10,914 |
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