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FhLUL01eNWQmvPMyN9GKz | physics | current-electricity | meter-bridge | In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the variation $${{dR} \over {d\ell }}$$ of its resistance R with length $$\ell $$ is $${{dR} \over {d\ell }}$$ $$ \propto $$ $${1 \over {\sqrt \ell }}$$. Two equal resistances are connected as shown in the figure. The
galvanometer ha... | [{"identifier": "A", "content": "0.3 m"}, {"identifier": "B", "content": "0.25 m"}, {"identifier": "C", "content": "0.35 m"}, {"identifier": "D", "content": "0.2 m"}] | ["B"] | null | For the given wire :
<br><br>dR = C $${{d\ell } \over {\sqrt \ell }}$$,
<br><br>where C = constant.
<br><br>Let resistance of part
<br><br>AP is R<sub>1</sub> and PB is R<sub>2</sub>
<br><br>$$ \therefore $$ $${{R'} \over {R'}} = {{{R_1}} \over {{R_2}}}$$ or R<sub>1</sub> = R<sub... | mcq | jee-main-2019-online-12th-january-morning-slot | 9,914 |
xqRjomA6iVDHm6KzZX18hoxe66ijvzmpx8g | physics | current-electricity | meter-bridge | In a meter bridge experiment, the circuit
diagram and the corresponding observation
table are shown in figure
<img src="data:image/png;base64,UklGRtwRAABXRUJQVlA4INARAABQxQCdASqsArsBP4G+2WQ2MCymoxM50sAwCWlu/CX43A2HZ19fq9+vXef/zsiLbt7O8Qy/s7x/8ejWpJnov/zwif/qLajkpLE1hqYdlRyUliaw1MOyo5KSxD5+7BCZp/fmD5uMcXeK5+7BCZphAo5+7B... | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["A"] | null | $$as\,x = {{R\left( {100 - l} \right)} \over l}$$<br><br>
$$for\,(A)\,\,\,\,x = {{1000 \times \left( {100 - 60} \right)} \over {40}} \approx 667$$<br><br>
$$for\,(B)\,\,\,\,\,x = {{100 \times \left( {100 - 13} \right)} \over {13}} \approx 669$$<br><br>
$$for\,(C)\,\,\,\,\,x = {{10 \times \left( {100 - 1.5} \right)} \ov... | mcq | jee-main-2019-online-10th-april-morning-slot | 9,915 |
p4HS9GP5fPQR3tZrH77k9k2k5lduar7 | physics | current-electricity | meter-bridge | In a meter bridge experiment S is a standard
resistance. R is a resistance wire. It is found
that balancing length is $$l$$ = 25 cm. If R is
replaced by a wire of half length and half
diameter that of R of same material, then the
balancing distance $$l'$$ (in cm) will now
be________.
<img src="data:image/png;base64,Ukl... | [] | null | 40 | $${S \over R} = {{75} \over {25}}$$ = 3
<br><br>R = $${{\rho L} \over A}$$ = $${{4\rho L} \over {\pi {d^2}}}$$
<br><br>R' = $${{4\rho \left( {{L \over 2}} \right)} \over {\pi {{\left( {{d \over 2}} \right)}^2}}}$$ = 2R
<br><br>Then $${S \over {R'}} = {{100 - l} \over l}$$
<br><br>$$ \Rightarrow $$ $${{100 - l} \over l}... | integer | jee-main-2020-online-9th-january-evening-slot | 9,916 |
1l56a4xot | physics | current-electricity | meter-bridge | <p>A meter bridge setup is shown in the figure. It is used to determine an unknown resistance R using a given resistor of 15 $$\Omega$$. The galvanometer (G) shows null deflection when tapping key is at 43 cm mark from end A. If the end correction for end A is 2 cm, then the determined value of R will be ____________ $... | [] | null | 19 | <p>$${{43 + 2} \over {15}} = {{57} \over R}$$</p>
<p>$$R = {{57 \times 15} \over {45}} = 19\,\Omega $$</p> | integer | jee-main-2022-online-28th-june-morning-shift | 9,917 |
1l6go717a | physics | current-electricity | meter-bridge | <p>Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $$l_{1}$$ is $$40 \mathrm{~cm}$$. Now an unknown resistance $$x$$ is connected in series with $$\mathrm{P}$$ and new balancing length is found to be $$80 \mathrm{~cm}$$ measured from the same end. Then the value of $$x$$... | [] | null | 20 | <p>$${P \over {40}} = {Q \over {60}}$$ ...... (1)</p>
<p>$${{P + x} \over {80}} = {Q \over {20}}$$ ..... (2)</p>
<p>$${P \over {P + x}} \times {{80} \over {40}} = {{20} \over {60}}$$</p>
<p>$${4 \over {4 + x}} \times 2 = {1 \over 3}$$</p>
<p>$$24 = 4 + x$$</p>
<p>$$x = 20$$</p> | integer | jee-main-2022-online-26th-july-morning-shift | 9,918 |
1l6jistwh | physics | current-electricity | meter-bridge | <p>In a meter bridge experiment, for measuring unknown resistance 'S', the null point is obtained at a distance $$30 \mathrm{~cm}$$ from the left side as shown at point D. If R is $$5.6$$ $$\mathrm{k} \Omega$$, then the value of unknown resistance 'S' will be __________ $$\Omega$$.</p>
<p><img src="data:image/png;base6... | [] | null | 2400 | <p>$${R \over S} = {{70} \over {30}}$$</p>
<p>$$S = {3 \over 7} \times 5.6 \times {10^3} = 2.4 \times {10^3}\,\Omega $$</p>
<p>$$ = 2400\,\Omega $$</p> | integer | jee-main-2022-online-27th-july-morning-shift | 9,919 |
1l6ko4g6f | physics | current-electricity | meter-bridge | <p>In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is $$40 \mathrm{~cm}$$. The balancing length, if the radius of the wire $$\mathrm{AB}$$ is doubled, will be ______________ $$\mathrm{cm}$$.</p>
<p><img src="data:image/png;base64,UklGRmgJAABXR... | [] | null | 40 | <p>Even if the radius of wire is doubled, the balancing point would not change as $${x \over {l - x}} = {{{R_1}} \over {{R_2}}}$$, which is not including a term of area.</p> | integer | jee-main-2022-online-27th-july-evening-shift | 9,920 |
1ldsbznzk | physics | current-electricity | meter-bridge | <p>When two resistance $$\mathrm{R_1}$$ and $$\mathrm{R_2}$$ connected in series and introduced into the left gap of a meter bridge and a resistance of 10 $$\Omega$$ is introduced into the right gap, a null point is found at 60 cm from left side. When $$\mathrm{R_1}$$ and $$\mathrm{R_2}$$ are connected in parallel and ... | [] | null | 30 | <p>As per given information</p>
<p>$${{{R_1} + {R_2}} \over {10}} = {{0.6} \over {0.4}}$$ ...... (1)</p>
<p>& $${{{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \over 3} = {{0.4} \over {0.6}}$$ ..... (2)</p>
<p>$$ \Rightarrow \left. \matrix{
{R_1} + {R_2} = 15 \hfill \cr
\& \,{R_1}{R_2} = 30 \hfill \cr} \right] \Rightarro... | integer | jee-main-2023-online-29th-january-evening-shift | 9,921 |
lsamtosf | physics | current-electricity | meter-bridge | In a metre-bridge when a resistance in the left gap is $2 \Omega$ and unknown resistance in the right gap, the balance length is found to be $40 \mathrm{~cm}$. On shunting the unknown resistance with $2 \Omega$, the balance length changes by : | [{"identifier": "A", "content": "$62.5 $ "}, {"identifier": "B", "content": "$22.5 \\mathrm{~cm}$"}, {"identifier": "C", "content": "$20 \\mathrm{~cm}$"}, {"identifier": "D", "content": "$65 \\mathrm{~cm}$"}] | ["B"] | null | <p>To solve this problem, let's first understand that a meter bridge setup is based on the principle of a Wheatstone bridge, in which two unknown resistances are in such a configuration that if the bridge is balanced, the ratio of the resistances on one side is equal to the ratio of resistances on the other side. I... | mcq | jee-main-2024-online-1st-february-evening-shift | 9,923 |
jaoe38c1lsd7iy3r | physics | current-electricity | meter-bridge | <p>The resistance per centimeter of a meter bridge wire is $$r$$, with $$X \Omega$$ resistance in left gap. Balancing length from left end is at $$40 \mathrm{~cm}$$ with $$25 \Omega$$ resistance in right gap. Now the wire is replaced by another wire of $$2 r$$ resistance per centimeter. The new balancing length for sam... | [{"identifier": "A", "content": "10 cm"}, {"identifier": "B", "content": "80 cm"}, {"identifier": "C", "content": "40 cm"}, {"identifier": "D", "content": "20 cm"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsiiqszq/b75469fd-7157-4fe4-9ed7-2c53636080e8/1603de60-c96b-11ee-b416-eff853096672/file-6y3zli1lsiiqszr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsiiqszq/b75469fd-7157-4fe4-9ed7-2c53636080e8/1603de60-c96b-11ee... | mcq | jee-main-2024-online-31st-january-evening-shift | 9,924 |
V2UyxyJY4IwPWXSq | physics | current-electricity | ohm's-law | An energy source will supply a constant current into the load if its internal resistance is | [{"identifier": "A", "content": "very large as compared to the load resistance "}, {"identifier": "B", "content": "equal to the resistance of the load "}, {"identifier": "C", "content": "non-zero but less than the resistance of the load"}, {"identifier": "D", "content": "zero "}] | ["D"] | null | $$I = {E \over {R + r}},\,$$ Internal resistance $$\left( r \right)$$ is
<br><br>zero, $$I = {E \over R} = $$ constant. | mcq | aieee-2005 | 9,925 |
gwzftriO0loiJL7b | physics | current-electricity | ohm's-law | When $$5V$$ potential difference is applied across a wire of length $$0.1$$ $$m,$$ the drift speed of electrons is $$2.5 \times {10^{ - 4}}\,\,m{s^{ - 1}}.$$ If the electron density in the wire is $$8 \times {10^{28}}\,\,{m^{ - 3}},$$ the resistivity of the material is close to : | [{"identifier": "A", "content": "$$1.6 \\times {10^{ - 6}}\\Omega m$$ "}, {"identifier": "B", "content": "$$1.6 \\times {10^{ - 5}}\\Omega m$$ "}, {"identifier": "C", "content": "$$1.6 \\times {10^{ - 8}}\\Omega m$$ "}, {"identifier": "D", "content": "$$1.6 \\times {10^{ - 7}}\\Omega m$$ "}] | ["B"] | null | $$V = IR = \left( {neA{v_d}} \right)\rho {\ell \over A}$$
<br><br>$$\therefore$$ $$\rho = {V \over {{V_d}\ln e}}$$
<br><br>Here $$V=$$ potential difference
<br><br>$$l = $$ length of wire
<br><br>$$n=$$ no. of electrons per unit volume of conductor.
<br><br>$$e=$$ no. of electrons
<br><br>Placing the value of above p... | mcq | jee-main-2015-offline | 9,926 |
ymIJC2KG7BBWBhm7nW1kmhorhu5 | physics | current-electricity | ohm's-law | A conducting wire of length 'l', area of cross-section A and electric resistivity $$\rho$$ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current.<br/><br/>If the length of the wire of the same material is doubled and the area of cross-sectio... | [{"identifier": "A", "content": "$$4{{VA} \\over {\\rho l}}$$"}, {"identifier": "B", "content": "$${3 \\over 4}{{VA} \\over {\\rho l}}$$"}, {"identifier": "C", "content": "$${1 \\over 4}{{VA} \\over {\\rho l}}$$"}, {"identifier": "D", "content": "$${1 \\over 4}{{\\rho l} \\over {VA}}$$"}] | ["C"] | null | We know that<br><br>$$R = \rho {l \over A}$$<br><br>Now, new length : $$l' = 2l$$<br><br>new area of cross section : $$A' = A/2$$<br><br>$$ \therefore $$ New resistance : $$R' = \rho .{{2l} \over {A/2}}$$<br><br>$$ \Rightarrow R' = 4{{\rho l} \over A}$$<br><br>$$ \Rightarrow R' = 4R$$<br><br>$$ \therefore $$ Resultant ... | mcq | jee-main-2021-online-16th-march-morning-shift | 9,927 |
PS7TbIqCesBO6stuhw1kmkre8bx | physics | current-electricity | ohm's-law | In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is : | [{"identifier": "A", "content": "3.9"}, {"identifier": "B", "content": "8.4"}, {"identifier": "C", "content": "7.5"}, {"identifier": "D", "content": "3.0"}] | ["A"] | null | $$V = I \times \rho {l \over A}$$<br><br>$$ \Rightarrow \rho = {{VA} \over {Il}} = {\pi \over 4}{{V{d^2}} \over {Il}}$$<br><br>$${{\Delta \rho } \over \rho } = {{2\Delta d} \over d} + {{\Delta V} \over V} + {{\Delta I} \over I} + {{\Delta l} \over l}$$<br><br>$$ = 2\left( {{{0.01} \over 5}} \right) + {{0.1} \over 5} ... | mcq | jee-main-2021-online-18th-march-morning-shift | 9,928 |
1l5482rs2 | physics | current-electricity | ohm's-law | <p>The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x $$\times$$ 10<sup>$$-$$3</sup> $$\Omega$$ cm. The value of x is ____________. [Take $$\pi$$ = 3... | [] | null | 144 | <p>Resistance $$ = \tan 45^\circ = 1\,\Omega $$</p>
<p>$$ \Rightarrow 1 = {{pI} \over A}$$</p>
<p>$$ \Rightarrow p = {{\pi {{(1.2\,cm)}^2}} \over {31.4\,cm}} = 1.44 \times {10^{ - 1}}$$ $$\Omega$$ cm</p>
<p>$$ \Rightarrow x = 144$$</p> | integer | jee-main-2022-online-29th-june-morning-shift | 9,929 |
mRe8l33BFkyzGwBW | physics | current-electricity | potentiometer | The length of a wire of a potentiometer is $$100$$ $$cm$$, and the $$e.$$ $$m.$$ $$f.$$ of its standard cell is $$E$$ volt. It is employed to measure the $$e.m.f.$$ of a battery whose internal resistance in $$0.5\Omega .$$ If the balance point is obtained at $$1=30$$ $$cm$$ from the positive end, the $$e.m.f.$$ of the ... | [{"identifier": "A", "content": "$${{30E} \\over {100.5}}$$ "}, {"identifier": "B", "content": "$${{30E} \\over {\\left( {100 - 0.5} \\right)}}$$ "}, {"identifier": "C", "content": "$${{30\\left( {E - 0.5i} \\right)} \\over {100}}$$ "}, {"identifier": "D", "content": "$${{30E} \\over {100}} - 0.5i$$, where i is the c... | ["D"] | null | Potential gradient along wire, K = $${E \over {100}}$$ volt/cm
<br><br>For battery V = E' – ir, where E' is emf of battery.
<br><br>or K × 30 = E' – ir, where current i is drawn from battery
<br><br>or $${{E \times 30} \over {100}}$$ = E' + 0.5i
<br><br>or E' = $${{30E} \over {100}} - 0.5i$$ | mcq | aieee-2003 | 9,931 |
V6wcr5zbUWy1dmJe | physics | current-electricity | potentiometer | In a potentiometer experiment the balancing with a cell is at length $$240$$ $$cm.$$ On shunting the cell with a resistance of $$2\Omega ,$$ the balancing length becomes $$120$$ $$cm$$. The internal resistance of the cell is | [{"identifier": "A", "content": "$$0.5\\Omega $$ "}, {"identifier": "B", "content": "$$1\\Omega $$"}, {"identifier": "C", "content": "$$2\\Omega $$"}, {"identifier": "D", "content": "$$4\\Omega $$"}] | ["C"] | null | The internal resistance of the cell,
<br><br>$$r = \left( {{{{\ell _1} - {\ell _2}} \over {{\ell _2}}}} \right) \times R$$
<br><br>$$ = {{240 - 120} \over {120}} \times 2 = 2\Omega $$ | mcq | aieee-2005 | 9,932 |
JanVWjpVwdzLFB3tfDL29 | physics | current-electricity | potentiometer | A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key K<sub>3</sub> is open. The balance point is at a length $$\ell $$<sub>1</sub> cm from P when two way key K<sub>3</sub> is plugged in between 2 and 1, while the balance point is at... | [{"identifier": "A", "content": "$${{{l_1}} \\over {{l_1} + {l_2}}}$$ "}, {"identifier": "B", "content": "$${{{l_2}} \\over {{l_2} - {l_1}}}$$"}, {"identifier": "C", "content": "$${{{l_1}} \\over {{l_1} - {l_2}}}$$"}, {"identifier": "D", "content": "$${{{l_1}} \\over {{l_2} - {l_1}}}$$"}] | ["D"] | null | When key K<sub>3</sub> is plugged in between 1 and 2,
<br><br>V<sub>1</sub> = iR<sub>1</sub> = x$$l$$<sub>1</sub> . . . . . (1)
<br><br>When key K<sub>3</sub> is plugged in between 3 and 1,
<br><br>V<sub>2</sub> = i(R<sub>1</sub> + R<sub>2</sub>) = x$$l$$<sub>2</sub> . . . . (2)
<br><br>On dividing (1) and (2),
<br><... | mcq | jee-main-2017-online-8th-april-morning-slot | 9,933 |
R9UdGikCD5CKM2yU | physics | current-electricity | potentiometer | In a potentiometer experiment, it is found that no current passes through the galvanometer when the
terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a
resistance of 5 $$\Omega$$, a balance is found when the cell is connected across 40 cm of the wire. Find the interna... | [{"identifier": "A", "content": "2.5 $$\\Omega$$"}, {"identifier": "B", "content": "1 $$\\Omega$$"}, {"identifier": "C", "content": "1.5 $$\\Omega$$"}, {"identifier": "D", "content": "2 $$\\Omega$$"}] | ["C"] | null | Internal resistance of potentiometer,
<br><br>r = $$\left( {{E \over V} - 1} \right) \times R$$
<br><br>Initially when no current passes through the galvanometer then
<br><br>emf, E = K (52)
<br><br>here K = potential gradient
<br><br>After cell is shunted by a resistance 5 $$\Omega $$, then,
<br><br>Terminal voltage... | mcq | jee-main-2018-offline | 9,934 |
wLMGxFVcNkie0EQSJwhn0 | physics | current-electricity | potentiometer | A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf $$\varepsilon $$ and internal resistance r. A cell C having emf $$\varepsilon $$/2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is –
<br/><br/><img src="dat... | [{"identifier": "A", "content": "$${{11} \\over {12}}L$$"}, {"identifier": "B", "content": "$${{13} \\over {24}}L$$"}, {"identifier": "C", "content": "$${{5} \\over {12}}L$$"}, {"identifier": "D", "content": "$${{11} \\over {24}}L$$"}] | ["B"] | null | $$i = {\varepsilon \over {13r}}$$
<br><br>$$i\left( {{x \over L}12r} \right) = {\varepsilon \over 2}$$
<br><br>$${\varepsilon \over {13r}}\left[ {{x \over L}.12r} \right] = {\varepsilon \over 2}$$
$$ \Rightarrow \,\,\,\,x = {{13L} \over {24}}$$ | mcq | jee-main-2019-online-10th-january-morning-slot | 9,935 |
r3rKRvu4wCsV4VHjXDf1k | physics | current-electricity | potentiometer | An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and esistance 5 $$\Omega $$. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is : | [{"identifier": "A", "content": "480 $$\\Omega $$"}, {"identifier": "B", "content": "495 $$\\Omega $$"}, {"identifier": "C", "content": "490 $$\\Omega $$"}, {"identifier": "D", "content": "395 $$\\Omega $$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265790/exam_images/w6xlxbddqdjpsipdd2ho.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Physics - Current Electricity Question 243 English Explanation">
<br>L... | mcq | jee-main-2019-online-12th-january-morning-slot | 9,936 |
WxPScHOVKWTRyiIT1rMrK | physics | current-electricity | potentiometer | In the circuit shown, a four-wire potentiometer
is made of a 400 cm long wire, which extends
between A and B. The resistance per unit length
of the potentiometer wire is r = 0.01 $$\Omega $$/cm. If
an ideal voltmeter is connected as shown with
jockey J at 50 cm from end A, the expected
reading of the voltmeter will be ... | [{"identifier": "A", "content": "0.25 V"}, {"identifier": "B", "content": "0.20 V"}, {"identifier": "C", "content": "0.50V"}, {"identifier": "D", "content": "0.75 V"}] | ["A"] | null | Resistance of wire AB = 400 × 0.01 = 4 $$\Omega $$<br><br>
i = $${3 \over 6} = 0.5A$$<br><br>
Now voltmeter reading = i (Resistance of 50 cm
length)<br>
= (0.5 A) (0.01 × 50) = 0.25 volt | mcq | jee-main-2019-online-8th-april-evening-slot | 9,937 |
qhVXBy9DGpT9ehdVIm7k9k2k5fd7ekv | physics | current-electricity | potentiometer | The balancing length for a cell is 560 cm in a potentiometer experiment. When an external
resistance of 10 $$\Omega $$ is connected in parallel to the cell, the balancing length changes by 60 cm. If
the internal resistance of the ceil is $${N \over {10}}$$
$$\Omega $$ , where N is an integer then value of N is ........... | [] | null | 12 | Let the emf of cell is $$\varepsilon $$ internal resistance is 'r' and potential gradient is x.
<br><br>When only cell connected :
<br><br>$$\varepsilon $$ = 560x .....(1)
<br><br>After connecting the resistor
<br><br>$${{\varepsilon \times 10} \over {10 + r}}$$ = 500x ....(2)
<br><br>from (1) and (2)
<br><br>56 = 50 ... | integer | jee-main-2020-online-7th-january-evening-slot | 9,938 |
aamzmExiqp1NE2DDqk7k9k2k5gv5a7t | physics | current-electricity | potentiometer | The length of a potentiometer wire is 1200 cm
and it carries a current of 60 mA. For a cell of
emf 5V and internal resistance of 20$$\Omega $$, the null
point on it is found to be a 1000 cm. The
resistance of whole wire is : | [{"identifier": "A", "content": "80$$\\Omega $$"}, {"identifier": "B", "content": "60$$\\Omega $$"}, {"identifier": "C", "content": "120$$\\Omega $$"}, {"identifier": "D", "content": "100$$\\Omega $$"}] | ["D"] | null | Let Resistance per unit length of potentiometer wire = $$\lambda $$
<br><br>5 = $$\lambda $$ $$ \times $$ 1000 $$ \times $$ 60 $$ \times $$ 10<sup>-3</sup>
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${5 \over {60}}$$
<br><br>Resistance of potentiometer wire = 1200 $$ \times $$ $${5 \over {60}}$$ = 100 $$\Omega $$ | mcq | jee-main-2020-online-8th-january-morning-slot | 9,939 |
sqAZuyzwMG245E9Lbbjgy2xukexxi6z4 | physics | current-electricity | potentiometer | A potentiometer wire PQ of 1 m length is
connected to a standard cell E<sub>1</sub>. Another cell E<sub>2</sub>
of emf 1.02 V is connected with a resistance ‘r’
and switch S (as shown in figure). With switch
S open, the null position is obtained at a
distance of 49 cm from Q. The potential
gradient in the potentiometer... | [{"identifier": "A", "content": "0.04 V/cm"}, {"identifier": "B", "content": "0.01 V/cm"}, {"identifier": "C", "content": "0.02 V/cm"}, {"identifier": "D", "content": "0.03 V/cm"}] | ["C"] | null | Balancing length is measured from P
<br><br>PQ = 1m
<br><br>QJ = 49 cm
<br><br>$$ \therefore $$ PJ = 51 cm
<br><br>Potential drop = Potential gradient($$\phi $$) $$ \times $$ length
<br><br>$$ \Rightarrow $$ 1.02 = $$\phi $$ $$ \times $$ 51
<br><br>$$ \Rightarrow $$ $$\phi $$ = 0.02 V/cm | mcq | jee-main-2020-online-2nd-september-evening-slot | 9,940 |
HxofUToorNUo4AROU01klrz57i6 | physics | current-electricity | potentiometer | In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E<sub>1</sub> and E<sub>2</sub> as well. For key K<sub>1</sub> (closed), the jockey is adjusted to touch the wire at point J<sub>1</sub> so that there is no deflection in the galvanometer. Now the first battery (E<s... | [] | null | 1 | Length of AB = 10 m<br><br>For battery E<sub>1</sub>, balancing length is l<sub>1</sub><br><br>l<sub>1</sub> = 380 cm [from end A]<br><br>For battery E<sub>2</sub>, balancing length is l<sub>2</sub><br><br>l<sub>2</sub> = 760 cm [from end A]<br><br>Now, we know that $${{{E_1}} \over {{E_2}}} = {{{l_1}} \over {{l_2}}}$$... | integer | jee-main-2021-online-25th-february-morning-slot | 9,941 |
1krw705e8 | physics | current-electricity | potentiometer | In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, $${{{\varepsilon _1}} \over {{\varepsilon _2}}}$$ is :<b... | [{"identifier": "A", "content": "$${5 \\over 3}$$"}, {"identifier": "B", "content": "$${8 \\over 5}$$"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["A"] | null | $${E_1} = k{l_1}$$ .... (i)<br><br>$${E_1} + {E_2} = k{l_2}$$ .... (ii)<br><br>$${{{E_1}} \over {{E_1} + {E_2}}} = {{{l_1}} \over {{l_2}}} = {{250} \over {400}} = {5 \over 8}$$<br><br>$$8{E_1} = 5{E_1} + 5{E_2}$$<br><br>$$3{E_1} = 5{E_2}$$<br><br>$${{{E_1}} \over {{E_2}}} = {5 \over 3}$$ | mcq | jee-main-2021-online-25th-july-evening-shift | 9,943 |
1krw9qlan | physics | current-electricity | potentiometer | The given potentiometer has its wire of resistance 10$$\Omega$$. When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2$$\Omega$$ resistor is :<br/><br/><img src="data:image/png;base64,UklGRpIJAABXRUJQVlA4IIYJAAAwQwCdASpAAckAPm02lkikIyIhIZQK6IANiWlu4W5xG/N18cf0f8Y/A7+y/jH1uXi72... | [{"identifier": "A", "content": "10 V"}, {"identifier": "B", "content": "5 V"}, {"identifier": "C", "content": "$${{40} \\over 9}$$ V"}, {"identifier": "D", "content": "$${{40} \\over 11}$$ V"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267029/exam_images/sqhgsgspfmebpb8sxr3h.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265317/exam_images/pvtpysfruoid7eel5y63.webp"><source media="(max-wid... | mcq | jee-main-2021-online-25th-july-evening-shift | 9,944 |
1l57qrdjk | physics | current-electricity | potentiometer | <p>A cell, shunted by a 8 $$\Omega$$ resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 $$\Omega$$ resistance. The value of internal resistance of the cell will be ____________ $$\Omega$$.</p> | [] | null | 8 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5jqaucn/96f18807-6859-40ab-bd14-b64bed306520/72161370-02bc-11ed-95db-c3fa9a0f41ba/file-1l5jqauco.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5jqaucn/96f18807-6859-40ab-bd14-b64bed306520/72161370-02bc-11ed-95db-c3fa9a0f41ba... | integer | jee-main-2022-online-27th-june-morning-shift | 9,945 |
1l5bcfhc0 | physics | current-electricity | potentiometer | <p>A potentiometer wire of length 10 m and resistance 20 $$\Omega$$ is connected in series with a 25 V battery and an external resistance 30 $$\Omega$$. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is $${x \over {10}}$$. The value of x is __________.</p> | [] | null | 25 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5l6kgp9/4955abcb-b0c1-4312-b9c2-87b467ab097c/d85973d0-0388-11ed-bfd1-873560f9d960/file-1l5l6kgpa.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5l6kgp9/4955abcb-b0c1-4312-b9c2-87b467ab097c/d85973d0-0388-11ed-bfd1-873560f9d960... | integer | jee-main-2022-online-24th-june-evening-shift | 9,946 |
1l5c4rucc | physics | current-electricity | potentiometer | <p>In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3 : 2, the difference in the balancing length of the potentiometer wire in above two cases will be ___________ cm.</... | [] | null | 25 | <p>At balancing point, we know that emf is proportional to the balancing length. i.e.,</p>
<p>emf $$\propto$$ balancing length</p>
<p>Now, let the emf's be 3$$\varepsilon $$ and 2$$\varepsilon $$.</p>
<p>$$\Rightarrow$$ 3$$\varepsilon $$ = k(75) ..... (1)</p>
<p>and 2$$\varepsilon $$ = k(l) ....... (2)</p>
<p>$$\Righta... | integer | jee-main-2022-online-24th-june-morning-shift | 9,947 |
1l5w3lr3t | physics | current-electricity | potentiometer | <p>The circuit diagram of potentiometer used to measure the internal resistance of a cell (E) is shown in figure. The key 'K' is kept closed so as to send constant current through potentiometer wire. When key 'K<sub>1</sub>' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K... | [] | null | 2 | <p><b>Shortcut Method :</b></p>
<p>Internal Resistance of Unknown Battery</p>
<p>$r=\left(\frac{\ell_1-\ell_2}{\ell_2}\right) \mathrm{R}$</p>
<p>Where l<sub>1</sub> means balanced length when key K<sub>1</sub> is open</p>
<p>Where l<sub>2</sub> means balanced length when key K<sub>1</sub> is closed</p>
<p>Here l<sub>1<... | integer | jee-main-2022-online-30th-june-morning-shift | 9,948 |
1l6f5oe1v | physics | current-electricity | potentiometer | <p>In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ___________ cm.</p> | [] | null | 18 | <p>$$E \propto I$$</p>
<p>$${{1.2} \over {1.8}} = {{36} \over {I'}}$$</p>
<p>$$I' = {3 \over 2} \times 36 = 54$$ cm</p>
<p>$$\Delta I = I' - I = 54 - 36 = 18$$ cm</p> | integer | jee-main-2022-online-25th-july-evening-shift | 9,949 |
1l6i426wu | physics | current-electricity | potentiometer | <p>A potentiometer wire of length $$300 \mathrm{~cm}$$ is connected in series with a resistance 780 $$\Omega$$ and a standard cell of emf $$4 \mathrm{V}$$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $$20\, \mathrm{mV}$$ is found to be $$60 \mathrm{~cm}$$. The resis... | [] | null | 20 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6wir8x4/ff3dc388-37a3-4dc9-b20b-2dfaeb9bbc0c/59131680-1d91-11ed-b1e3-c3a54149d5d6/file-1l6wir8x5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6wir8x4/ff3dc388-37a3-4dc9-b20b-2dfaeb9bbc0c/59131680-1d91-11ed-b1e3-c3a54149d5d6... | integer | jee-main-2022-online-26th-july-evening-shift | 9,950 |
1l6mbfpwx | physics | current-electricity | potentiometer | <p>As shown in the figure, a potentiometer wire of resistance $$20 \,\Omega$$ and length $$300 \mathrm{~cm}$$ is connected with resistance box (R.B.) and a standard cell of emf $$4 \mathrm{~V}$$. For a resistance '$$R$$' of resistance box introduced into the circuit, the null point for a cell of $$20 \,\mathrm{mV}$$ is... | [] | null | 780 | <p>$$l = 3m$$, $${R_w} = 20\,\Omega $$</p>
<p>$${\varepsilon _0} = 4V$$</p>
<p>$${{4 \times 20} \over {20 + R}} \times {{60} \over {300}} = 20 \times {10^{ - 3}}$$</p>
<p>$${4 \over {20 + R}} = 5 \times {10^{ - 3}}$$</p>
<p>$$20 + R = 800$$</p>
<p>$$R = 780\,\Omega $$</p> | integer | jee-main-2022-online-28th-july-morning-shift | 9,951 |
1ldogpz1r | physics | current-electricity | potentiometer | <p>In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $$1.5 \mathrm{~V}$$ is found to be $$60 \mathrm{~cm}$$. If this cell is replaced by another cell of emf E, the length-of null point increases by $$40 \mathrm{~cm}$$. The value of $$E$$ is $$\frac{x}{10} V$$. The va... | [] | null | 25 | E<sub>1</sub> = 1.5 V, l<sub>1</sub> = 60 cm, l<sub>2</sub> = 40 cm + 60 cm = 100 cm
<br/><br/>$E \propto l$
<br/><br/>$$
\begin{aligned}
& \frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}} \\\\
& \frac{1.5}{E}=\frac{60}{100} \\\\
& E=\frac{150}{60}=\frac{5}{2}=\frac{25}{10} \\\\
& \text { so } x=25
\end{aligned}
$$ | integer | jee-main-2023-online-1st-february-morning-shift | 9,952 |
1lds9clec | physics | current-electricity | potentiometer | <p>With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is</p>
<p>(A) directly proportional to the length of the potentiometer wire</p>
<p>(B) directly proportional to the potential gradient of the wire</p>
<p>(C) inversely proportional to the potential... | [{"identifier": "A", "content": "A only"}, {"identifier": "B", "content": "B and D only"}, {"identifier": "C", "content": "A and C only"}, {"identifier": "D", "content": "inversely C only"}] | ["C"] | null | <p>A potentiometer is an electrical instrument used to measure the electromotive force (EMF) of a cell.</p>
<p>The sensitivity of a potentiometer is defined as the change in potential difference per unit length of the wire.</p>
<p>It is directly proportional to the length of the potentiometer wire (Option A), because a... | mcq | jee-main-2023-online-29th-january-evening-shift | 9,953 |
1ldsbgn5v | physics | current-electricity | potentiometer | <p>A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5$$\Omega$$. When a resistance of 15$$\Omega$$ is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________$$\Omega$$.</p> | [] | null | 5 | <p>Let the emf is E and internal resistance is r of this secondary cell so</p>
<p>$${{RE} \over {r + R}} \propto l$$</p>
<p>so $${{{R_1}E} \over {r + {R_1}}} \propto {l_1}$$</p>
<p>& $${{{R_2}E} \over {r + {R_2}}} \propto {l_2}$$</p>
<p>$$ \Rightarrow {{{R_1}(r + {R_2})} \over {{R_2}(r + {R_1})}} = {{{l_1}} \over {{l_2... | integer | jee-main-2023-online-29th-january-evening-shift | 9,954 |
TXWeNjjwXkpJKaG4 | physics | current-electricity | resistance-and-resistivity | The length of a given cylindrical wire is increased by $$100\% $$. Due to the consequent decrease in diameter the change in the resistance of the wire will be | [{"identifier": "A", "content": "$$200\\% $$"}, {"identifier": "B", "content": "$$100\\% $$ "}, {"identifier": "C", "content": "$$50\\% $$"}, {"identifier": "D", "content": "$$300\\% $$"}] | ["D"] | null | $${R_f} = {n^2}{R_1}$$
<br><br>Here $$n=2$$ (length becomes twice)
<br><br>$$\therefore$$ $${R_f} = 4{R_i}$$
<br><br>New resistance $$=400$$ of $${R_i}$$
<br><br>$$\therefore$$ Increase $$ = 300\% $$ | mcq | aieee-2003 | 9,956 |
WB8X5uHFdgLImheC | physics | current-electricity | resistance-and-resistivity | Thermistors are usually made of | [{"identifier": "A", "content": "metal oxides with high temperature coefficient of resistivity "}, {"identifier": "B", "content": "metals with high temperature coefficient of resistivity "}, {"identifier": "C", "content": "metals with low temperature coefficient of resistivity "}, {"identifier": "D", "content": "semico... | ["A"] | null | Thermistors are usually made of metal-oxides with high temperature coefficient of resistivity. | mcq | aieee-2004 | 9,957 |
qkfzXkk4NtqmJUoP | physics | current-electricity | resistance-and-resistivity | Consider a block of conducting material of resistivity $$'\rho '$$ shown in the figure. Current $$'I'$$ enters at $$'A'$$ and leaves from $$'D'$$. We apply superposition principle to find voltage $$'\Delta V'$$ developed between $$'B'$$ and $$'C'$$. The calculation is done in the following steps:
<br/>(i) Take current... | [{"identifier": "A", "content": "$${{\\rho I} \\over {8\\pi {r^2}}}$$ "}, {"identifier": "B", "content": "$${{\\rho I} \\over {{r^2}}}$$ "}, {"identifier": "C", "content": "$${{\\rho I} \\over {2\\pi {r^2}}}$$ "}, {"identifier": "D", "content": "$${{\\rho I} \\over {4\\pi {r^2}}}$$ "}] | ["C"] | null | As shown above $$E = {{\rho I} \over {2\pi {r^2}}}$$ | mcq | aieee-2008 | 9,959 |
sgUjbwzNCmrOI2y6lNHln | physics | current-electricity | resistance-and-resistivity | A uniform wire of length 1 and radius r has a resistance of 100 $$\Omega $$. It is recast into a wire of radius $${r \over 2}.$$ The resistance of new wire will be : | [{"identifier": "A", "content": "1600 $$\\Omega $$ "}, {"identifier": "B", "content": "400 $$\\Omega $$"}, {"identifier": "C", "content": "200 $$\\Omega $$"}, {"identifier": "D", "content": "100 $$\\Omega $$"}] | ["A"] | null | Resistance of a wire of length l and radius r is given
by
<br><br>R = $${{\rho l} \over A}$$ = $${{\rho l} \over A} \times {A \over A} = {{\rho V} \over {{A^2}}} = {{\rho V} \over {{\pi ^2}{r^4}}}$$
<br><br>$$ \Rightarrow $$ R $$ \propto $$ $${1 \over {{r^4}}}$$
<br><br>$$ \therefore $$ $${{{R_1}} \over {{R_2}}} = {\le... | mcq | jee-main-2017-online-9th-april-morning-slot | 9,961 |
MNHy91FwKSD3f1uwlCAO6 | physics | current-electricity | resistance-and-resistivity | In a conductor, if the number of conduction
electrons per unit volume is 8.5 × 10<sup>28</sup> m<sup>–3</sup> and
mean free time is 25ƒs (femto second), it's
approximate resistivity is :-<br/>
(m<sub>e</sub> = 9.1 × 10<sup>–31</sup> kg) | [{"identifier": "A", "content": "10<sup>\u20138</sup> $$\\Omega $$m"}, {"identifier": "B", "content": "10<sup>\u20137</sup> $$\\Omega $$m"}, {"identifier": "C", "content": "10<sup>\u20135</sup> $$\\Omega $$m"}, {"identifier": "D", "content": "10<sup>\u20136</sup> $$\\Omega $$m"}] | ["A"] | null | $$\rho = {{2m} \over {n{e^2}\tau }}$$<br><br>
= 3.34 × 10<sup>–8</sup> $$\Omega $$ m | mcq | jee-main-2019-online-9th-april-evening-slot | 9,963 |
ylwu9ATRfrMmV19byR18hoxe66ijvwwkzsm | physics | current-electricity | resistance-and-resistivity | A current of 5 A passes through a copper
conductor (resistivity = 1.7 × 10<sup>–8</sup> $$\Omega $$m) of radius
of cross-section 5 mm. Find the mobility of the
charges if their drift velocity is 1.1 × 10<sup>–3</sup> m/s. | [{"identifier": "A", "content": "1.3 m<sup>2</sup>/Vs"}, {"identifier": "B", "content": "1.0 m<sup>2</sup>/Vs"}, {"identifier": "C", "content": "1.8 m<sup>2</sup>/Vs"}, {"identifier": "D", "content": "1.5 m<sup>2</sup>/Vs"}] | ["B"] | null | $$\mu = {{{V_d}} \over E}\,\,\,\,\,\,E = \rho J$$<br><br>
$$ = {{1.1 \times {{10}^{ - 3}}} \over {1.7 \times {{10}^{ - 8}} \times {5 \over {\pi \times 25 \times {0^{ - 6}}}}}}$$<br><br>
$$ = {{1.1 \times {{10}^{ - 3}} \times \pi \times 25 \times {0^{ - 6}}} \over {1.7 \times {{10}^{ - 8}} \times 5}} \approx 1.01\,{m... | mcq | jee-main-2019-online-10th-april-morning-slot | 9,964 |
Lkglb7x9eP3YPu86NA18hoxe66ijvzsy5bj | physics | current-electricity | resistance-and-resistivity | In an experiment, the resistance of a material
is plotted as a function of temperature (in some
range). As shown in the figure, it is a straight
line. One may conclude that :
<img src="data:image/png;base64,UklGRrIGAABXRUJQVlA4IKYGAABQRwCdASrsAQsBPm02mkikIyKhJPKYaIANiWlu4XShG/OZ8EfyrtG/tnR4eLspv+A1W343+Qf1f0v/QDeX/5T+O... | [{"identifier": "A", "content": "$$R(T) = {R_0}{e^{ - {T^2}/T_0^2}}$$"}, {"identifier": "B", "content": "$$R(T) = {{{R_0}} \\over {{T^2}}}$$"}, {"identifier": "C", "content": "$$R(T) = {R_0}{e^{ {T^2}/T_0^2}}$$"}, {"identifier": "D", "content": "$$R(T) = {R_0}{e^{ - T_0^2/{T^2}}}$$"}] | ["D"] | null | $${{{1 \over {{T^2}}}} \over {{1 \over {T_0^2}}}} + {{\ln \,R\left( T \right)} \over {\ln \,R\left( {{T_o}} \right)}} = 1$$<br><br>
$$ \Rightarrow $$ ln R(T) = ln R(T<sub>o</sub>) $$\left( {1 - {{T_o^2} \over {{T^2}}}} \right)$$<br><br>
$$R(T) = {R_o}{e^{ - \left( {{{T_o^2} \over {{T^2}}}} \right)}}$$ | mcq | jee-main-2019-online-10th-april-morning-slot | 9,965 |
98vIQwSQdX5DxVcjkS3rsa0w2w9jwzhx53v | physics | current-electricity | resistance-and-resistivity | Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of
resistivity $$\rho $$. The resistance between the two spheres will be : | [{"identifier": "A", "content": "$${\\rho \\over {2\\pi }}\\left( {{1 \\over a} + {1 \\over b}} \\right)$$"}, {"identifier": "B", "content": "$${\\rho \\over {4\\pi }}\\left( {{1 \\over a} + {1 \\over b}} \\right)$$"}, {"identifier": "C", "content": "$${\\rho \\over {2\\pi }}\\left( {{1 \\over a} - {1 \\over b}} \\r... | ["D"] | null | $$R = \int\limits_a^b {{{\rho \,dx} \over {4\pi {x^2}}}} $$<br><br>
$$ = {\rho \over {4\pi }}\left( {{1 \over a} - {1 \over b}} \right)$$ | mcq | jee-main-2019-online-10th-april-evening-slot | 9,966 |
A2kPmvArT3fBp9ggf3jgy2xukev21dnd | physics | current-electricity | resistance-and-resistivity | Consider four conducting materials copper,
tungsten, mercury and aluminium with
resistivity $$\rho $$<sub>C</sub>, $$\rho $$<sub>T</sub>, $$\rho $$<sub>M</sub> and $$\rho $$<sub>A</sub> respectively. Then : | [{"identifier": "A", "content": "$$\\rho $$<sub>C</sub> > $$\\rho $$<sub>A</sub> > $$\\rho $$<sub>T</sub>"}, {"identifier": "B", "content": "$$\\rho $$<sub>M</sub> > $$\\rho $$<sub>A</sub> > $$\\rho $$<sub>C</sub>"}, {"identifier": "C", "content": "$$\\rho $$<sub>A</sub> > $$\\rho $$<sub>T</sub> >... | ["B"] | null | ρ<sub>M</sub> = 98 × 10<sup>–8</sup>
<br>ρ<sub>A</sub> = 2.80 × 10<sup>–8</sup>
<br>ρ<sub>C</sub> = 1.72 × 10<sup>–8</sup>
<br>ρ<sub>T</sub> = 5.65 × 10<sup>–8</sup>
<br><br>$$ \therefore $$ $$\rho $$<sub>M</sub> > $$\rho $$<sub>T</sub> > $$\rho $$<sub>A</sub> > $$\rho $$<sub>C</sub> | mcq | jee-main-2020-online-2nd-september-morning-slot | 9,967 |
k7H73ooaxw5SRjDAoH1klulf7ut | physics | current-electricity | resistance-and-resistivity | A wire of 1$$\Omega$$ has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is : | [{"identifier": "A", "content": "76%"}, {"identifier": "B", "content": "12.5%"}, {"identifier": "C", "content": "25%"}, {"identifier": "D", "content": "56%"}] | ["D"] | null | R<sub>0</sub> = 1$$\Omega$$<br><br>R<sub>1</sub> = ?<br><br>l<sub>0</sub> = 1m<br><br>l<sub>1</sub> = 1.25 m<br><br>A<sub>0</sub> = A<br><br>As volume of wire remains constant so<br><br>A<sub>0</sub>l<sub>0</sub> = A<sub>1</sub>l<sub>1</sub> $$ \Rightarrow $$ A<sub>1</sub> = $${{{l_0}{A_0}} \over {{l_1}}}$$<br><br>Now<... | mcq | jee-main-2021-online-26th-february-evening-slot | 9,968 |
1kryt5g00 | physics | current-electricity | resistance-and-resistivity | In the given figure, a battery of emf E is connected across a conductor PQ of length 'l' and different area of cross-sections having radii r<sub>1</sub> and r<sub>2</sub> (r<sub>2</sub> < r<sub>1</sub>). <br/><br/><img src="data:image/png;base64,UklGRtoIAABXRUJQVlA4IM4IAACQPACdASoBAccAPm02mEgkIyKhIlUKmIANiWlu+F7MgPV... | [{"identifier": "A", "content": "Drift velocity of electron increases."}, {"identifier": "B", "content": "Electric field decreases."}, {"identifier": "C", "content": "Electron current decreases."}, {"identifier": "D", "content": "All of these"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267265/exam_images/drjaakuyfr5xcwddg9qq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Physics - Current Electricity Question 166 English Explanation"><br>Curren... | mcq | jee-main-2021-online-27th-july-morning-shift | 9,969 |
1l58bvl9r | physics | current-electricity | resistance-and-resistivity | <p>An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :</p> | [{"identifier": "A", "content": "0.4%"}, {"identifier": "B", "content": "0.2%"}, {"identifier": "C", "content": "0.8%"}, {"identifier": "D", "content": "0.6%"}] | ["C"] | null | <p>$$R = {{\rho l} \over A}$$</p>
<p>Also volume will remain constant</p>
<p>i.e., Al = constant $$ \Rightarrow A \propto {1 \over l}$$</p>
<p>$$\therefore$$ $$R \propto {l^2}$$</p>
<p>$${{\Delta R} \over R} = 2{{\Delta l} \over l} = 0.8$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift | 9,970 |
1l6p50ibg | physics | current-electricity | resistance-and-resistivity | <p>Two metallic wires of identical dimensions are connected in series. If $$\sigma_{1}$$ and $$\sigma_{2}$$ are the conductivities of the these wires respectively, the effective conductivity of the combination is :</p> | [{"identifier": "A", "content": "$$\n\\frac{\\sigma_{1} \\sigma_{2}}{\\sigma_{1}+\\sigma_{2}}\n$$"}, {"identifier": "B", "content": "$$\n\\frac{2 \\sigma_{1} \\sigma_{2}}{\\sigma_{1}+\\sigma_{2}}\n$$"}, {"identifier": "C", "content": "$$\n\\frac{\\sigma_{1}+\\sigma_{2}}{2 \\sigma_{1} \\sigma_{2}}\n$$"}, {"identifier": ... | ["B"] | null | <p>$$R = {R_1} + {R_2}$$</p>
<p>$$ \Rightarrow {{{l_1} + {l_2}} \over {\sigma A}} = {{{l_1}} \over {{\sigma _1}A}} + {{{l_2}} \over {{\sigma _2}A}}$$</p>
<p>$$ \Rightarrow {2 \over \sigma } = {1 \over {{\sigma _1}}} + {1 \over {{\sigma _2}}}$$</p>
<p>$$ \Rightarrow \sigma = {{2{\sigma _1}{\sigma _2}} \over {{\sigma _1... | mcq | jee-main-2022-online-29th-july-morning-shift | 9,973 |
1l6rgp9n8 | physics | current-electricity | resistance-and-resistivity | <p>A $$1 \mathrm{~m}$$ long wire is broken into two unequal parts $$\mathrm{X}$$ and $$\mathrm{Y}$$. The $$\mathrm{X}$$ part of the wire is streched into another wire W. Length of $$W$$ is twice the length of $$X$$ and the resistance of $$\mathrm{W}$$ is twice that of $$\mathrm{Y}$$. Find the ratio of length of $$\math... | [{"identifier": "A", "content": "1 : 4"}, {"identifier": "B", "content": "1 : 2"}, {"identifier": "C", "content": "4 : 1"}, {"identifier": "D", "content": "2 : 1"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7m6bkq7/a6d47508-08e1-44c8-ad64-3b608b9c37ae/1895a3f0-2bad-11ed-bad3-59534b1d6f8c/file-1l7m6bkq8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7m6bkq7/a6d47508-08e1-44c8-ad64-3b608b9c37ae/1895a3f0-2bad-11ed-bad3-59534b1d6f8c/fi... | mcq | jee-main-2022-online-29th-july-evening-shift | 9,975 |
1ldtyrs3j | physics | current-electricity | resistance-and-resistivity | <p>The resistance of a wire is 5 $$\Omega$$. It's new resistance in ohm if stretched to 5 times of it's original length will be :</p> | [{"identifier": "A", "content": "25"}, {"identifier": "B", "content": "625"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "125"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ledln4ww/f3c91b62-ea60-4cdd-8190-c3805dd18cc4/78d2ba00-b18b-11ed-a682-13f364283dca/file-1ledln4wx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ledln4ww/f3c91b62-ea60-4cdd-8190-c3805dd18cc4/78d2ba00-b18b-11ed-a682-13f364283dca... | mcq | jee-main-2023-online-25th-january-evening-shift | 9,976 |
1ldws3r8d | physics | current-electricity | resistance-and-resistivity | <p>If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is __________%.</p> | [] | null | 44 | Let $\ell_{0}$ be its initial length and $A_{0}$ be initial area.
<br/><br/>
Considering volume to be conserved
<br/><br/>
$$
\begin{aligned}
& \text { Vol. }=\ell_{0} A_{0}=\left(1.2 \ell_{0}\right) \mathrm{A} \\\\
& A_{\text {final }}=\frac{A_{0}}{1.2} \\\\
& R_{\text {in }}=\frac{\rho \ell_{0}}{A_{0}} \\\\
& R_{\tex... | integer | jee-main-2023-online-24th-january-evening-shift | 9,977 |
1ldyeqsgs | physics | current-electricity | resistance-and-resistivity | <p>A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is $$n\times10^{-3}\Omega$$. If the resistivity of the material is $$\mathrm{2.4\times10^{-8}\Omega m}$$. The value of $$n$$ is ___________.</p> | [] | null | 2 | Resistance of the hollow cylindrical conductor is given by,
<br/><br/>$R=\rho \frac{l}{\pi\left(r_2^2-r_1^2\right)}$
<br/><br/>where $r_2=$ outer radius
<br/><br/>$$
\begin{gathered}
r_1=\text { inner radius } \\\\
\rho=\text { resistivity, } l=\text { length } \\\\
\therefore \rho=\frac{2.4 \times 10^{-8} \times 3.14... | integer | jee-main-2023-online-24th-january-morning-shift | 9,978 |
1lgriitoy | physics | current-electricity | resistance-and-resistivity | <p>A wire of resistance $$160 ~\Omega$$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be</p> | [{"identifier": "A", "content": "$$640 ~\\Omega$$"}, {"identifier": "B", "content": "$$40 ~\\Omega$$"}, {"identifier": "C", "content": "$$16 ~\\Omega$$"}, {"identifier": "D", "content": "$$10 ~\\Omega$$"}] | ["D"] | null | Let the original length of the wire be L and its cross-sectional area be A. Then, its resistance R is given by:<br/><br/>
$$R = \frac{\rho L}{A}$$<br/><br/>
where $$\rho$$ is the resistivity of the material of the wire.
<br/><br/>
When the wire is melted and drawn into a wire of one-fourth of its length, its new length... | mcq | jee-main-2023-online-12th-april-morning-shift | 9,979 |
1lgvtgqlg | physics | current-electricity | resistance-and-resistivity | <p>A rectangular parallelopiped is measured as $$1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}$$. If its specific resistance is $$3 \times 10^{-7} ~\Omega \mathrm{m}$$, then the resistance between its two opposite rectangular faces will be ___________ $$\times 10^{-7} ~\Omega$$.</p> | [] | null | 3 | <p>The resistance of a material can be calculated using the formula:</p>
<p>$ R = \rho \frac{L}{A} $</p>
<p>where </p>
<ul>
<li>$R$ is the resistance, </li>
<li>$\rho$ (rho) is the resistivity or specific resistance of the material, </li>
<li>$L$ is the length (or distance over which the resistance is being measured), ... | integer | jee-main-2023-online-10th-april-evening-shift | 9,980 |
jaoe38c1lscpqh3c | physics | current-electricity | resistance-and-resistivity | <p>Wheatstone bridge principle is used to measure the specific resistance $$\left(S_1\right)$$ of given wire, having length $$L$$, radius $$r$$. If $$X$$ is the resistance of wire, then specific resistance is ; $$S_1=X\left(\frac{\pi r^2}{L}\right)$$. If the length of the wire gets doubled then the value of specific re... | [{"identifier": "A", "content": "$$\\frac{S_1}{4}$$\n"}, {"identifier": "B", "content": "$$2 \\mathrm{~S}_1$$\n"}, {"identifier": "C", "content": "$$\\frac{\\mathrm{S}_1}{2}$$\n"}, {"identifier": "D", "content": "$$S_1$$"}] | ["D"] | null | <p>The specific resistance (or resistivity) of a material is a fundamental property that describes how much the material resists the flow of electric current. The resistivity is typically denoted by the symbol $$\rho$$ (rho), and it can be calculated by using the resistance $$X$$ of a uniform specimen of the material, ... | mcq | jee-main-2024-online-27th-january-evening-shift | 9,982 |
jaoe38c1lse6jbrc | physics | current-electricity | resistance-and-resistivity | <p>Two conductors have the same resistances at $$0^{\circ} \mathrm{C}$$ but their temperature coefficients of resistance are $$\alpha_1$$ and $$\alpha_2$$. The respective temperature coefficients for their series and parallel combinations are :</p> | [{"identifier": "A", "content": "$$\\alpha_1+\\alpha_2, \\frac{\\alpha_1 \\alpha_2}{\\alpha_1+\\alpha_2}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\alpha_1+\\alpha_2}{2}, \\frac{\\alpha_1+\\alpha_2}{2}$$\n"}, {"identifier": "C", "content": "$$\\alpha_1+\\alpha_2, \\frac{\\alpha_1+\\alpha_2}{2}$$\n"}, {"identifie... | ["B"] | null | <p>Series :</p>
<p>$$\begin{aligned}
& \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\
& 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=\mathrm{R}\left(1+\alpha_1 \Delta \theta\right)+\mathrm{R}\left(1+\alpha_2 \Delta \theta\right) \\
& 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=... | mcq | jee-main-2024-online-31st-january-morning-shift | 9,983 |
luxwco6g | physics | current-electricity | resistance-and-resistivity | <p>At room temperature $$(27^{\circ} \mathrm{C})$$, the resistance of a heating element is $$50 \Omega$$. The temperature coefficient of the material is $$2.4 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$$. The temperature of the element, when its resistance is $$62 \Omega$$, is __________$${ }^{\circ} \mathrm{C}$$.</p> | [] | null | 1027 | <p>We can start solving this problem by first understanding that the resistance of a material changes with temperature, and this change can be quantified using the temperature coefficient of resistance $ \alpha $. The relationship between the resistance of a material at any temperature $ T $ and its resistance at a ref... | integer | jee-main-2024-online-9th-april-evening-shift | 9,984 |
lv5gt1ps | physics | current-electricity | resistance-and-resistivity | <p>Resistance of a wire at $$0^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C}$$ and $$t^{\circ} \mathrm{C}$$ is found to be $$10 \Omega, 10.2 \Omega$$ and $$10.95 \Omega$$ respectively. The temperature $$t$$ in Kelvin scale is _________.</p> | [] | null | 748 | <p>To determine the temperature $$t$$ in the Kelvin scale, we need to use the relationship between the resistance of a wire and temperature. The general formula for the resistance $R$ of a wire as a function of temperature is:</p>
<p>$$ R_t = R_0 (1 + \alpha t) $$</p>
<p>where:</p>
<ul>
<li>$$R_t$$ is the resistanc... | integer | jee-main-2024-online-8th-april-morning-shift | 9,986 |
lvc58ec5 | physics | current-electricity | resistance-and-resistivity | <p>A wire of resistance $$R$$ and radius $$r$$ is stretched till its radius became $$r / 2$$. If new resistance of the stretched wire is $$x ~R$$, then value of $$x$$ is ________.</p> | [] | null | 16 | <p>The resistance $$R$$ of a wire is given by the formula:</p>
$$
R = \rho \frac{l}{A},
$$
<p>where:</p>
<ul>
<li>$$\rho$$ is the resistivity of the material,</li>
<li>$$l$$ is the length of the wire,</li>
<li>$$A$$ is the cross-sectional area of the wire.</li>
</ul>
<p>If we have a cylindrical wire, the cross-section... | integer | jee-main-2024-online-6th-april-morning-shift | 9,987 |
z3fhLKP1hY20Sqg8 | physics | current-electricity | wheatstone-bridge | The current $${\rm I}$$ drawn from the $$5$$ volt source will be
<img src="data:image/png;base64,UklGRu4MAABXRUJQVlA4IOIMAAAQdACdASrhAYEBP4HA1mW2MCwnIZUJ2sAwCWlu4W+zHmNwvV5c/yncX4B11Ho+6N7E/4fxCMk+uvAB3NcwiUeySjzPfv/cv/XwQRYJIUwADvYwibmmMpIIyTYWNpAz1Rc0Oy0LZVZaFsm6hwBnZu5Rof9OsRs2O5eAzK9cPnd+cvdhj785fXBdg1BQfCU9DVh19P... | [{"identifier": "A", "content": "$$0.33$$ $$A$$ "}, {"identifier": "B", "content": "$$0.5$$ $$A$$ "}, {"identifier": "C", "content": "$$0.67$$ $$A$$ "}, {"identifier": "D", "content": "$$0.17$$ $$A$$ "}] | ["B"] | null | The network of resistors is a balanced wheatstone bridge. The equivalent circuit is
<br><br><img class="question-image" src="https://imagex.cdn.examgoal.net/u0grPX1JblmtziFul/zO02ieOHwrqVvktIOF5I7UvzdhDu0/dOIrz8ZY9JjmNZNTdEO9xm/image.svg" loading="lazy" alt="AIEEE 2006 Physics - Current Electricity Question 291 Englis... | mcq | aieee-2006 | 9,989 |
gAjeS9CRTM5Z7YrMuYx87 | physics | current-electricity | wheatstone-bridge | The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as R<sub>1</sub> has the colour code (Orange, Red, Brown). The resistors R<sub>2</sub> and R<sub>4</sub> are 80$$\Omega $$ and 40$$\Omega $$, respectively. Assuming that the colour code for the carbon resistors gives their accurate... | [{"identifier": "A", "content": "Brown, Blue, Brown "}, {"identifier": "B", "content": "Grey, Black, Brown "}, {"identifier": "C", "content": "Red, Green, Brown"}, {"identifier": "D", "content": "Brown, Blue, Black"}] | ["A"] | null | R<sub>1</sub> = 32 $$ \times $$ 10 = 320
<br><br>for wheatstone bridge
<br><br>$$ \Rightarrow $$ $${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$$
<br><br>$${{320} \over {{R_3}}} = {{80} \over {40}}$$
<br><br>$${R_3} = 160$$
<br><br>$$ \therefore $$ Correct answer is Brown Blue ... | mcq | jee-main-2019-online-10th-january-evening-slot | 9,990 |
BcGiaIUTxqMbTW1Soy7k9k2k5gy3haa | physics | current-electricity | wheatstone-bridge | Four resistances of 15$$\Omega $$, 12$$\Omega $$, 4$$\Omega $$ and 10$$\Omega $$
respectively in cyclic order to form
Wheatstone's network. The resistance that
is to be connected in parallel with the
resistance of 10$$\Omega $$ to balance the network is
_____$$\Omega $$. | [] | null | 10 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267599/exam_images/xf5ezrbsu9exsycwrfqp.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Morning Slot Physics - Current Electricity Question 214 English Explanation">
<br><br>... | integer | jee-main-2020-online-8th-january-morning-slot | 9,992 |
dSbAQiKANGGcftImN01kltjhz5q | physics | current-electricity | wheatstone-bridge | Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is :<br/><br/><img src="data:image/png;base64,UklGRqYMAABXRUJQVlA4IJoMAAAQQwCdASrWAMUAPm0ylkikIqIhJFFaUIANiWlu4W8hG/Np8Gfz3tV/r35QeI75j+1/knuVv9v5KPqn93/jf9n/3vx0/T/7n/KP2g8/eAR6Z/tv8j/bf+v+pX9KP673+Ndf0d... | [{"identifier": "A", "content": "$${{3R} \\over 2}$$"}, {"identifier": "B", "content": "$${{R} \\over 2}$$"}, {"identifier": "C", "content": "2R"}, {"identifier": "D", "content": "R"}] | ["D"] | null | <p>Given all resistances have same resistance R.</p>
<p>Now, we can redraw the circuit as below</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3hevh7q/bee31c2a-f73f-4182-84a3-d00f539f63a2/c252f860-d9dd-11ec-8ea5-5be6b5e3368b/file-1l3hevh7r.png?format=png" data-orsrc="https://app-content.cdn.ex... | mcq | jee-main-2021-online-26th-february-morning-slot | 9,994 |
1l6gmmm0m | physics | current-electricity | wheatstone-bridge | <p>The current I in the given circuit will be :</p>
<p><img src="data:image/png;base64,UklGRiIJAABXRUJQVlA4IBYJAABwhQCdASoAA0ABP4HA2WS2MDmnIbJJ8zAwCWlu4W5TbmNwvH5y/2VqbaId0P8P34/Vkfse9djvzSwxVxlbG+XQdytfrGacJG48zD49PGHmfv3GpiY3y6DuVsb5dB3K2NJ5lyF7k3P3Jq14f5Rs9YCZ7YVq1J44dKgSJvl0HcrY3y2K59hsi0no6Zn+IdLVodI8zlqhfQXjrAAK+... | [{"identifier": "A", "content": "10 A"}, {"identifier": "B", "content": "20 A"}, {"identifier": "C", "content": "4 A"}, {"identifier": "D", "content": "40 A"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6v69sjm/27a47107-2a15-470c-8996-0f72b9195635/bd787a20-1cd3-11ed-843d-81ad9f680592/file-1l6v69sjn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6v69sjm/27a47107-2a15-470c-8996-0f72b9195635/bd787a20-1cd3-11ed-843d-81ad9f680592... | mcq | jee-main-2022-online-26th-july-morning-shift | 9,996 |
1l6i19h3h | physics | current-electricity | wheatstone-bridge | <p>A battery of $$6 \mathrm{~V}$$ is connected to the circuit as shown below. The current I drawn from the battery is :</p>
<p><img src="data:image/png;base64,UklGRnwRAABXRUJQVlA4IHARAADwFgGdASoAA7QCP4HA2WU2MLmnIfM5uzAwCWlu++7NRJ/pDXgd+PnX6k/4/W+8y7RCO7b17pcVfBI1/8P/npmT3//6yPv8TAktC2VUxy4VXg2l1HOpaQzPdo6q78zeBJCkJLQrm... | [{"identifier": "A", "content": "1A"}, {"identifier": "B", "content": "2A"}, {"identifier": "C", "content": "$$\\frac{6}{11}$$ A"}, {"identifier": "D", "content": "$$\\frac{4}{3}$$ A"}] | ["A"] | null | <p>Balance wheatstone</p>
<p>$$ \Rightarrow {R_{eff}} = {{3 \times 6} \over {3 + 6}} \times 2 + 2$$</p>
<p>$$ = 6\,\Omega $$</p>
<p>$$ \Rightarrow I = {V \over R} = 1\,A$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift | 9,997 |
1ldofd09r | physics | current-electricity | wheatstone-bridge | <p>The equivalent resistance between $$A$$ and $$B$$ of the network shown in figure;</p>
<p><img src="data:image/png;base64,UklGRnQIAABXRUJQVlA4IGgIAACQegCdASoAAxMBP4HA2mS2MK0nIrWZUsAwCWlu+F9KJTF6ctGSnX6P3O7NE3h2/4m6Cr5Wp/+uQyjxmdkHAqHuDgTXTPrSH9qaIpHsg4FQ94xA4FQ94xA31/A4HA4HA4HA2bk/2k0dpmj4m1gp6B4zOyDgVD3jEDgUwNTut1ut... | [{"identifier": "A", "content": "21 R"}, {"identifier": "B", "content": "$$\\frac{8}{3}$$ R"}, {"identifier": "C", "content": "11$$\\frac{2}{3}$$ R "}, {"identifier": "D", "content": "14 R"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le7d3y76/1a793b27-692c-48c5-91c3-08497cd4aed7/66556330-ae1d-11ed-8ebb-e5c62db85e5e/file-1le7d3y77.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le7d3y76/1a793b27-692c-48c5-91c3-08497cd4aed7/66556330-ae1d-11ed-8ebb-e5c62db85e5e/fi... | mcq | jee-main-2023-online-1st-february-morning-shift | 9,998 |
ldquz9x2 | physics | current-electricity | wheatstone-bridge | The equivalent resistance between $A$ and $B$ is _________.<br/><br/>
<img src="data:image/png;base64,UklGRh4aAABXRUJQVlA4IBIaAADwZAGdASoAA/gCP4HA2mQ2MS0movPpcsAwCWlu4WxEDmNwvH57/4nXJ2sTFH8f2ilAna/im4eCv/oy//+v1zJ///qe+/BP9ui/REXCkzovzyK+Ze7mPoI2YvrJH7dF+iIuFJnRfnuVqLtQ5E8F9rUXaYJ3FqolEVboo4PXKHXFyw7IuFJnRfoiLhQ4e7UORP... | [{"identifier": "A", "content": "$\\frac{1}{2} \\Omega$"}, {"identifier": "B", "content": "$\\frac{2}{3} \\Omega$"}, {"identifier": "C", "content": "$\\frac{3}{2} \\Omega$"}, {"identifier": "D", "content": "$\\frac{1}{3} \\Omega$ "}] | ["B"] | null | <p>Equivalent circuit can be drawn as</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leoi54ri/6dbd00fd-9061-4abc-9be2-5e05f3252362/67e627e0-b78a-11ed-8877-d5d81ac5ff88/file-1leoi54rj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leoi54ri/6dbd00fd-9061-4abc-9be2-5e05f32... | mcq | jee-main-2023-online-30th-january-evening-shift | 9,999 |
ldqw5clm | physics | current-electricity | wheatstone-bridge | <p>If the potential difference between $\mathrm{B}$ and $\mathrm{D}$ is zero, the value of $x$ is $\frac{1}{n} \Omega$. The value of $n$ is __________.</p>
<p><img src="data:image/png;base64,UklGRnQXAABXRUJQVlA4IGgXAADQUQGdASrOAgADP4HA2WS2MCynI3KZcsAwCWlu/CoYS1/P6B4T8PpX/g+qa4hXxYp/oe03nd7W/mz91dDTEF1+9BTzz6MH/3sJcyf//... | [] | null | 2 | <p>The circuit is a Wheatstone bridge, so</p>
<p>$${{{{6 \times 3} \over {6 + 3}}} \over {{{x \times 1} \over {x + 1}}}} = {{1 + 2} \over x}$$</p>
<p>$$ \Rightarrow {{2(x + 1)} \over x} = {3 \over x}$$</p>
<p>$$ \Rightarrow x = {1 \over 2}$$</p>
<p>So $$n = 2$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 10,000 |
jaoe38c1lsc3vcj8 | physics | current-electricity | wheatstone-bridge | <p>A wire of length $$10 \mathrm{~cm}$$ and radius $$\sqrt{7} \times 10^{-4} \mathrm{~m}$$ connected across the right gap of a meter bridge. When a resistance of $$4.5 \Omega$$ is connected on the left gap by using a resistance box, the balance length is found to be at $$60 \mathrm{~cm}$$ from the left end. If the resi... | [{"identifier": "A", "content": "63"}, {"identifier": "B", "content": "70"}, {"identifier": "C", "content": "66"}, {"identifier": "D", "content": "35"}] | ["C"] | null | <p>For null point,</p>
<p>$$\begin{aligned}
& \frac{4.5}{60}=\frac{R}{40} \\
& \text { Also, } R=\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^2} \\
& 4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60 \\
& \rho=66 \times 10^{-7} \Omega \times \mathrm{m}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift | 10,001 |
lvc58eq8 | physics | current-electricity | wheatstone-bridge | <p>The value of unknown resistance $$(x)$$ for which the potential difference between $$B$$ and $$D$$ will be zero in the arrangement shown, is :</p>
<p><img src="data:image/png;base64,UklGRpwVAABXRUJQVlA4IJAVAAAQLgGdASrtAgADP4HA2GW2MCynIbDpasAwCWlu/DT34KidQehCvlTdB2+jv+L4Ne3vWB4Q/kt0P2u6vf3N4Q6sfns0w/Xtj3iCGiFuG59QPg1... | [{"identifier": "A", "content": "3 $$\\Omega$$"}, {"identifier": "B", "content": "42 $$\\Omega$$"}, {"identifier": "C", "content": "6 $$\\Omega$$"}, {"identifier": "D", "content": "9 $$\\Omega$$"}] | ["C"] | null | <p>Balanced wheatstone bridge.</p>
<p>$$\Rightarrow \quad 12 \times 0.5=(x+6) \times \frac{1}{2} \Rightarrow x=6 \Omega$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift | 10,002 |
dyu6Mmd1u34EYLZC | physics | dual-nature-of-radiation | davisson-and-germer-experiment | Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively... | [{"identifier": "A", "content": "$$d\\,\\sin \\,i = n{\\lambda _{dB}}$$ "}, {"identifier": "B", "content": "$$2d\\,\\cos \\,i = n{\\lambda _{dB}}$$ "}, {"identifier": "C", "content": "$$2d\\,\\sin \\,i = n{\\lambda _{dB}}$$ "}, {"identifier": "D", "content": "$$d\\,\\cos \\,i = n{\\lambda _{dB}}$$ "}] | ["B"] | null | $$2d\,\cos \,i = n{\lambda _{dB}}$$ | mcq | aieee-2008 | 10,004 |
j0GobDyBXbEQ1R4TLr1qpqahkk8fiiqbl | physics | dual-nature-of-radiation | davisson-and-germer-experiment | This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.
<br/><br/> <b>Statement 1 :</b> Davisson - Germer experiment established the wave nature of electrons. <br/><br/><b>Statement 2 :</b> If electrons have wave nature, they ... | [{"identifier": "A", "content": "Statement 1 is true, Statement 2 is false"}, {"identifier": "B", "content": "Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1. "}, {"identifier": "C", "content": "Statement 1 is true, Statement 2 is true, Statement 2 is not the correct exp... | ["B"] | null | Davisson-Germer experiment showed that electron beams can undergo diffraction when passed through atomic crystals. This shows the wave nature of electrons as waves can exhibit interference and diffraction. | mcq | aieee-2012 | 10,005 |
lWQrCVkYdBhBHkm5 | physics | dual-nature-of-radiation | davisson-and-germer-experiment | Match <b>List - $${\rm I}$$</b> (Fundamental Experiment) with <b>List - $${\rm II}$$</b> (its conclusion) and select the correct option from the choices given below the list:<br/><br/><img src="data:image/png;base64,UklGRjoaAABXRUJQVlA4IC4aAADQEwGdASoAA7QBP4G40mM2LrYoorM8QsAwCWlu++w4L++9ooEPnUNX9bP8v/cfW98y/h/+lkHWf/rv... | [{"identifier": "A", "content": "$$A - ii;\\,\\,B - i,\\,\\,C - iii$$ "}, {"identifier": "B", "content": "$$A - iv;\\,\\,B - iii,\\,\\,C - ii$$ "}, {"identifier": "C", "content": "$$A - i;\\,\\,B - iv,\\,\\,C - iii$$ "}, {"identifier": "D", "content": "$$A - ii;\\,\\,B - iv,\\,\\,C - iii$$ "}] | ["A"] | null | Frank-Hertz experiment - Discrete energy levels of atom
<br><br>Photoelectric effects - Particle nature of light
<br><br>Davison - Germer experiment - wave nature of electron. | mcq | jee-main-2015-offline | 10,006 |
AX1UQfubUh1KCUlfk7jgy2xukfi5o53x | physics | dual-nature-of-radiation | davisson-and-germer-experiment | A beam of electrons of energy E scatters from
a target having atomic spacing of 1 $$\mathop A\limits^o $$. The first
maximum intensity occurs at $$\theta $$ = 60<sup>o</sup>. Then E (in
eV) is ______.
<br/>(Planck constant h = 6.64 × 10<sup>–34</sup> Js, <br/>1 eV =
1.6 × 10<sup>–19</sup> J, electron <br/>mass m = 9.1 ... | [] | null | 50.47 | Given d = 1 $$\mathop A\limits^o $$
<br><br>For first maxima, $$\theta $$ = 60<sup>o</sup>
<br><br>$$ \therefore $$ $$\theta $$<sub>1</sub> = 90 - $${\theta \over 2}$$
<br><br>= $$90 - {{60} \over 2}$$ = 60<sup>o</sup>
<br><br>and $$2d\sin \theta = \lambda = {h \over {\sqrt {2mE} }}$$<br><br>$$ \Rightarrow $$ $$2 \t... | integer | jee-main-2020-online-5th-september-morning-slot | 10,007 |
y9UgqES3uXbL76U97R1kmlvx8ti | physics | dual-nature-of-radiation | davisson-and-germer-experiment | The speed of electrons in a scanning electron microscope is 1 $$\times$$ 10<sup>7</sup> ms<sup>-1</sup>. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of : | [{"identifier": "A", "content": "$${1 \\over {1837}}$$"}, {"identifier": "B", "content": "1837"}, {"identifier": "C", "content": "$${\\sqrt {1837} }$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt {1837} }}$$"}] | ["B"] | null | Resolving power (RP) $$ \propto $$ $${1 \over \lambda }$$<br><br>We know, de-Broglie wavelength<br><br>$$\lambda = {h \over {mv}}$$<br><br>$$ \therefore $$ RP $$ \propto $$ $$ {mv \over {h}}$$<br><br>$$ \therefore $$ $${{R{P_e}} \over {R{P_p}}} = {{{m_e}} \over {{m_p}}} = 1837$$ | mcq | jee-main-2021-online-18th-march-evening-shift | 10,008 |
1l5akxdim | physics | dual-nature-of-radiation | davisson-and-germer-experiment | <p>Given below are two statements :</p>
<p>Statement I : Davisson-Germer experiment establishes the wave nature of electrons.</p>
<p>Statement II : If electrons have wave nature, they can interfere and show diffraction.</p>
<p>In the light of the above statements choose the correct answer from the option given below :<... | [{"identifier": "A", "content": "Both Statement I and Statement II are true."}, {"identifier": "B", "content": "Both Statement I and Statement II are false."}, {"identifier": "C", "content": "Statement I is true but Statement II is false."}, {"identifier": "D", "content": "Statement I is false but Statement II is true.... | ["A"] | null | <p>Davisson-Germer experiment is done and establishes the wave nature of electrons. Interference and diffraction establishes wave nature.</p> | mcq | jee-main-2022-online-25th-june-morning-shift | 10,009 |
1ldpjswzk | physics | dual-nature-of-radiation | davisson-and-germer-experiment | <p>Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R</p>
<p>Assertion A : The beam of electrons show wave nature and exhibit interference and diffraction.</p>
<p>Reason R : Davisson Germer Experimentally verified the wave nature of electrons.</p>
<p>In the light of th... | [{"identifier": "A", "content": "A is not correct but R is correct."}, {"identifier": "B", "content": "Both A and R are correct and R is the correct explanation of A"}, {"identifier": "C", "content": "Both A and R are correct but R is Not the correct explanation of A"}, {"identifier": "D", "content": "A is correct but ... | ["B"] | null | <p>The assertion A and the reason R are both correct statements, and the reason R is the correct explanation of the assertion A.</p>
<p><b>Explanation :</b></p>
<p>The assertion A states that the beam of electrons exhibit wave nature and show interference and diffraction. This statement is correct because electrons e... | mcq | jee-main-2023-online-31st-january-morning-shift | 10,010 |
FsFDTRuhDRQdfGrH | physics | dual-nature-of-radiation | matter-waves | Formation of covalent bonds in compounds exhibits | [{"identifier": "A", "content": "wave nature of electron "}, {"identifier": "B", "content": "particle nature of electron "}, {"identifier": "C", "content": "both wave and particle nature of electron "}, {"identifier": "D", "content": "none of these "}] | ["A"] | null | Formation of covalent bond is best explained by molecular orbital theory. | mcq | aieee-2002 | 10,011 |
alES5b54AmPqiO3X | physics | dual-nature-of-radiation | matter-waves | If the kinetic energy of a free electron doubles, it's deBroglie wavelength changes by the factor | [{"identifier": "A", "content": "$$2$$ "}, {"identifier": "B", "content": "$${1 \\over 2}$$ "}, {"identifier": "C", "content": "$${\\sqrt 2 }$$ "}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 2 }}$$ "}] | ["D"] | null | de-Broglie wavelength,
<br><br>$$\lambda = {h \over p} = {h \over {\sqrt {2.m,\left( {K.E} \right)} }}$$
<br><br>$$\therefore$$ $$\lambda \propto {1 \over {\sqrt {K.E} }}$$
<br><br>If $$K.E$$ is doubled, wavelength becomes $${\lambda \over {\sqrt 2 }}$$ | mcq | aieee-2005 | 10,012 |
neXZCyPZN0StxkzJ | physics | dual-nature-of-radiation | matter-waves | In an experiment, electrons are made to pass through a narrow slit of width $$'d'$$ comparable to their de Broglie wavelength. They are detected on a screen at a distance $$'D'$$ from the slit (see figure).
<img src="data:image/png;base64,UklGRnwNAABXRUJQVlA4IHANAAAQiACdASq1AkcBP4G41WQ2LjemoXLbUvAwCWlu+DnrATD3V9GSnX5W/... | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264590/exam_images/xjawgr7tgetmgewdih8o.webp\" loading=\"lazy\" alt=\"AIEEE 2008 Physics - Dual Nature of Radiation Question 172 English Option 1\"> "}, {"identifier": "B", "content": "<img class=... | ["D"] | null | The electron beam will be diffracted and the maxima is obtained at $$y=0.$$ Also the distance between the first minima on both side will be greater than $$d.$$ | mcq | aieee-2008 | 10,013 |
mgydk8X4u7rD2Q4z | physics | dual-nature-of-radiation | matter-waves | A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The
collision is head on, and elastic. The ratio of the de-Broglie wavelengths $${\lambda _A}$$ to $${\lambda _B}$$ after the collision is: | [{"identifier": "A", "content": "$${{{\\lambda _A}} \\over {{\\lambda _B}}} = {1 \\over 3}$$ "}, {"identifier": "B", "content": "$${{{\\lambda _A}} \\over {{\\lambda _B}}} = 2$$"}, {"identifier": "C", "content": "$${{{\\lambda _A}} \\over {{\\lambda _B}}} = {2 \\over 3}$$"}, {"identifier": "D", "content": "$${{{\\lambd... | ["B"] | null | From question, m<sub>A</sub> = M; m<sub>B</sub> = $${m \over 2}$$
<br><br>u<sub>A</sub> = V and u<sub>B</sub> = 0
<br><br>Let after collision velocity of A = V<sub>1</sub> and
<br><br>velocity of B = V<sub>2</sub>
<br><br>Applying law of conservation of momentum,
<br><br>mu = mv<sub>1</sub> + $$\left( {{m \over 2}} \ri... | mcq | jee-main-2017-offline | 10,014 |
G7bISf9Jmp18tn2w | physics | dual-nature-of-radiation | matter-waves | An electron beam is accelerated by a potential difference V to hit a metallic target to produce X–rays. It
produces continuous as well as characteristic X-rays. If $$\lambda $$<sub>min</sub> is the smallest possible wavelength of X-ray in the spectrum, the variation of log$$\lambda $$<sub>min</sub> with log V is correc... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ymqtm4/f6d3ea65-8ac0-4be3-be22-1d8071b4682e/ba721ac0-3286-11ed-8893-19b23ee4c66d/file-1l7ymqtm5.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l7ymqtm4/f6d3ea65-8ac0-4be3-be22-1d8071b4682e/ba7... | ["B"] | null | In X-ray tube, $${\lambda _{\min }} = {{hc} \over {eV}}$$
<br><br>$$\log \left( {{\lambda _{\min }}} \right) = \log \left( {{{hc} \over e}} \right) - \log V$$
<br><br>Clearly, log ($$\lambda $$<sub>min</sub>) versus log V graph
slope is negative hence option (b) is correct. | mcq | jee-main-2017-offline | 10,015 |
DYdWfFiwVfYSaj1TM9Fxg | physics | dual-nature-of-radiation | matter-waves | Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are $${\lambda _1}$$ and $${\lambda _2},$$ their de Broglie wavelength in the frame of reference attached to their center of masses : | [{"identifier": "A", "content": "$${\\lambda _{CM}} = {\\lambda _1} = {\\lambda _2}$$ "}, {"identifier": "B", "content": "$${\\lambda _{CM}} = {{2{\\lambda _1}{\\lambda _2}} \\over {\\sqrt {\\lambda _1^2 + \\lambda _2^2} }}$$ "}, {"identifier": "C", "content": "$${1 \\over {{\\lambda _{CM}}}} = {1 \\over {{\\lambda _1}... | ["B"] | null | As we know,
<br><br>momentum (p) = $${h \over \lambda }$$
<br><br>Let one perticle is moving in x direction and other is in y dirrection.
<br><br>$$\therefore\,\,\,\,$$ momentum of each electrons $${h \over {{\lambda _1}}}\widehat i$$ and $${h \over {{\lambda _2}}}\widehat j$$
<br><br>$$\therefore\,\,\,\,$$ Veloci... | mcq | jee-main-2018-online-15th-april-morning-slot | 10,016 |
I2FGpkExYmcIqpafLOzzT | physics | dual-nature-of-radiation | matter-waves | If the de Broglie wavelengths associated with a proton and an $$\alpha $$-particle are equal, then the ratio of velocities of the proton and the $$\alpha $$-particle will be : | [{"identifier": "A", "content": "4 : 1"}, {"identifier": "B", "content": "2 : 1"}, {"identifier": "C", "content": "1 : 2"}, {"identifier": "D", "content": "1 : 4"}] | ["A"] | null | We know, $${\lambda _p} = {h \over {{p_p}}}$$ = $$ {h \over {{m_p}{v_p}}}$$
<br><br>Similarly, $${\lambda _\alpha } = {h \over {{m_\alpha }{v_\alpha }}}$$
<br><br>Given, $${\lambda _p} = {\lambda _\alpha }$$
<br><br>$$ \Rightarrow $$ $${h \over {{m_p}{v_p}}} = {h \over {{m_\alpha }{v_\alpha }}}$$
<br><br>$$ \therefore ... | mcq | jee-main-2018-online-15th-april-evening-slot | 10,017 |
D0GFpBeIrXG4xMsU1E3hh | physics | dual-nature-of-radiation | matter-waves | The de-Broglie wavelength ($$\lambda $$<sub>B</sub>) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($$\lambda $$<sub>G</sub>) by : | [{"identifier": "A", "content": "$$\\lambda $$<sub>B</sub> = 2$$\\lambda $$<sub>G</sub>"}, {"identifier": "B", "content": "$$\\lambda $$<sub>B</sub> = 3$$\\lambda $$<sub>G</sub>"}, {"identifier": "C", "content": "$$\\lambda $$<sub>B</sub> = $$\\lambda $$<sub>G/2</sub>"}, {"identifier": "D", "content": "$$\\lambda $$<su... | ["B"] | null | <p>We know that, $$\lambda = {h \over {mv}}$$</p>
<p>From third Bohr's postulate, we have</p>
<p>$$mvr = n{h \over {2\pi }}$$</p>
<p>$${h \over {mv}} = {{2\pi r} \over n} \Rightarrow \lambda = {{2\pi r} \over n}$$</p>
<p>Since, $$r = {a_0}{{{n^2}} \over Z}$$, where a<sub>0</sub> is radius of Bohr's orbit having value... | mcq | jee-main-2018-online-16th-april-morning-slot | 10,018 |
DaDijtRYe9XUHGUzdkg6y | physics | dual-nature-of-radiation | matter-waves | Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths <sup>$$\lambda $$</sup>N, <sup>$$\lambda $$</sup>A respectively. The ratio $${{{}^\lambda N} \over {{}^\lambda A}}$$is closest to : | [{"identifier": "A", "content": "10<sup>$$-$$6</sup>"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "10<sup>$$-$$10</sup>"}, {"identifier": "D", "content": "10<sup>$$-$$1</sup>"}] | ["A"] | null | <p>We know that $$E = {{hc} \over \lambda }$$</p>
<p>So, for atom $${E_A} = {{hc} \over {{\lambda _A}}}$$</p>
<p>And for neutron $${E_N} = {{hc} \over {{\lambda _N}}}$$</p>
<p>Then, $${{{E_A}} \over {{E_N}}} = {{hc} \over {{\lambda _A}}} \times {{{\lambda _N}} \over {hc}} \Rightarrow {{{\lambda _N}} \over {{\lambda _A}... | mcq | jee-main-2018-online-16th-april-morning-slot | 10,019 |
veJ5gtjAQ0lcQMShGoJUf | physics | dual-nature-of-radiation | matter-waves | In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10<sup>–12</sup> m, the minimum electron energy required is close to - | [{"identifier": "A", "content": "25 keV"}, {"identifier": "B", "content": "500 keV"}, {"identifier": "C", "content": "100 keV"}, {"identifier": "D", "content": "1 keV"}] | ["A"] | null | $$\lambda $$ = $${h \over p}$$ {$$\lambda $$ = 7.5 $$ \times $$ 10<sup>$$-$$12</sup>}
<br><br>P = $${h \over \lambda }$$
<br><br>KE = $${{{P^2}} \over {2m}} = {{{{\left( {h/\lambda } \right)}^2}} \over {2m}}$$
<br><br>$$ = {{\left\{ {{{6.6 \times {{10}^{ - 34}}} \ov... | mcq | jee-main-2019-online-10th-january-morning-slot | 10,020 |
gkvdIXp9RRo0p7DYCik64 | physics | dual-nature-of-radiation | matter-waves | If the de Broglie wavelength of an electron is equal to the 10<sup>–3</sup> times the wavelength of a photon of frequency 6 $$ \times $$ 10<sup>14</sup> Hz, then the speed of electron is equal to : (Speed of light = 3 $$ \times $$ 10<sup>8</sup> m/s, Planck's constant = 6.63 $$ \times $$
10<sup>–34</sup> J.s, Mass o... | [{"identifier": "A", "content": "1.7 $$ \\times $$ 10<sup>6</sup> m/s"}, {"identifier": "B", "content": "1.45 $$ \\times $$ 10<sup>6</sup> m/s"}, {"identifier": "C", "content": "1.1 $$ \\times $$ 10<sup>6</sup> m/s"}, {"identifier": "D", "content": "1.8 $$ \\times $$ 10<sup>6</sup> m/s"}] | ["B"] | null | $${h \over {mv}} = {10^{ - 3}}\left( {{{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}}} \right)$$
<br><br>v $$ = {{6.63 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^5}}}$$
<br><br>v $$ = 1.45 \times {10^6}$$ m/s | mcq | jee-main-2019-online-11th-january-morning-slot | 10,021 |
OdwULAdXOexjvxOmahFAV | physics | dual-nature-of-radiation | matter-waves | A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths $${{{\lambda _A}} \over {{\lambda _B}}}$$ is close to : | [{"identifier": "A", "content": "4.47"}, {"identifier": "B", "content": "10.00"}, {"identifier": "C", "content": "14.14"}, {"identifier": "D", "content": "0.07"}] | ["C"] | null | K.E. acquired by charge = K = qV
<br><br>$$\lambda $$ = $${h \over P}$$ = $${h \over {\sqrt {2mK} }}$$ = $${h \over {\sqrt {2mqV} }}$$
<br><br>$$ \therefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2... | mcq | jee-main-2019-online-12th-january-morning-slot | 10,022 |
EEzB0y2MGQ0q5jvXbodPz | physics | dual-nature-of-radiation | matter-waves | Two particles move at right angle to each other.
Their de-Broglie wavelengths are $$\lambda _1$$ and $$\lambda _2$$
respectively. The particles suffer perfectly
inelastic collision. The de-Broglie wavelength
$$\lambda _2$$ of the final particle, is given by : | [{"identifier": "A", "content": "$$\\lambda = {{{\\lambda _1} + {\\lambda _2}} \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over {{\\lambda ^2}}} = {1 \\over {\\lambda _1^2}} + {1 \\over {\\lambda _2^2}}$$"}, {"identifier": "C", "content": "$$\\lambda = \\sqrt {{\\lambda _1}{\\lambda _2}} $$"}, {"identifier"... | ["B"] | null | Let the two particles be moving along x-direction
and y-direction.
<br><br>So, the net momentum initially is $$\sqrt {{{{h^2}} \over {\lambda _1^2}} + {{{h^2}} \over {\lambda _2^2}}} $$
<br><br>and final momentum will be $${h \over \lambda }$$.
<br><br>Applying momentum conservation,
<br><br>$${h \over \lambda } = \sqr... | mcq | jee-main-2019-online-8th-april-morning-slot | 10,023 |
xalFQITtlDzcmeosCIMGX | physics | dual-nature-of-radiation | matter-waves | A nucleus A, with a finite de-broglie
wavelength $$\lambda $$<sub>A</sub>, undergoes spontaneous fission
into two nuclei B and C of equal mass. B flies
in the same direction as that of A, while C flies
in the opposite direction with a velocity equal
to half of that of B. The de-Broglie wavelengths
$$\lambda $$<sub>B</s... | [{"identifier": "A", "content": "$$\\lambda $$<sub>A</sub>, 2$$\\lambda $$<sub>A</sub>"}, {"identifier": "B", "content": "2$$\\lambda $$<sub>A</sub>, $$\\lambda $$<sub>A</sub>"}, {"identifier": "C", "content": "$$\\lambda $$<sub>A</sub>, $$\\lambda $$<sub>A</sub>/2"}, {"identifier": "D", "content": "$$\\lambda $$<sub>A... | ["D"] | null | Let mass of B and C is m each. By momentum conservation<br><br>
$$2m{v_0} = mv - {{mv} \over 2}$$<br><br>
v = 4v<sub>0</sub><br>
P<sub>A</sub> = 2mv<sub>0</sub> pB = 4mv<sub>0</sub> pc = 2mv<sub>0</sub><br><br>
De-Broglie wavelength $$\lambda = {h \over p}$$<br><br>
$${\lambda _A} = {h \over {2m{v_0}}}$$; $${\lambda _... | mcq | jee-main-2019-online-8th-april-evening-slot | 10,024 |
asqToXkbxZF251dVWAPp3 | physics | dual-nature-of-radiation | matter-waves | A particle 'P' is formed due to a completely
inelastic collision of particles 'x' and 'y' having
de-Broglie wavelengths '$$\lambda $$<sub>x</sub>' and '$$\lambda $$<sub>y</sub>'
respectively. If x and y were moving in opposite
directions, then the de-Broglie wavelength of
'P' is :- | [{"identifier": "A", "content": "$${\\lambda _x} - {\\lambda _y}$$"}, {"identifier": "B", "content": "$${{{\\lambda _x}{\\lambda _y}} \\over {\\left| {{\\lambda _x} - {\\lambda _y}} \\right|}}$$"}, {"identifier": "C", "content": "$${\\lambda _x} + {\\lambda _y}$$"}, {"identifier": "D", "content": "$${{{\\lambda _x}{\\l... | ["B"] | null | Conservation of momentum<br><br>
$$\overrightarrow {{p_x}} + \overrightarrow {{p_y}} = {\overrightarrow p _{final}}$$<br><br>
m<sub>x</sub>v<sub>x</sub> – m<sub>y</sub>v<sub>y</sub> = (m<sub>x</sub> + m<sub>y</sub>) V<br><br>
$${h \over {{\lambda _x}}} - {h \over {{\lambda _y}}} = {h \over \lambda }$$<br><br>
$$ \Rig... | mcq | jee-main-2019-online-9th-april-evening-slot | 10,025 |
ei2rUmc6Jy9P32f82P7k9k2k5ld8cvl | physics | dual-nature-of-radiation | matter-waves | An electron of mass m and magnitude of charge
|e| initially at rest gets accelerated by a constant
electric field E. The rate of change of de-Broglie
wavelength of this electron at time t ignoring
relativistic effects is : | [{"identifier": "A", "content": "$${{ - h} \\over {\\left| e \\right|Et}}$$"}, {"identifier": "B", "content": "$${{ - h} \\over {\\left| e \\right|E\\sqrt t }}$$"}, {"identifier": "C", "content": "$${{ - h} \\over {\\left| e \\right|E{t^2}}}$$"}, {"identifier": "D", "content": "$${{\\left| e \\right|Et} \\over h}$$"}] | ["C"] | null | F = |e| E
<br><br>$$a = {F \over m}$$ = $${{\left| e \right|E} \over m}$$
<br><br>V = $$at = $$ $${{\left| e \right|E} \over m}t$$
<br><br>$$\lambda $$ = $${h \over {mV}}$$ = $${h \over {\left| e \right|Et}}$$
<br><br>$${{d\lambda } \over {dt}}$$ = $${{ - h} \over {\left| e \right|E{t^2}}}$$ | mcq | jee-main-2020-online-9th-january-evening-slot | 10,026 |
i9RlreoY3rNewPKMnyjgy2xukg0cq8qq | physics | dual-nature-of-radiation | matter-waves | Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength
of nitrogen molecule is close to :
<br/>(Given : nitrogen molecule weight : 4.64 $$ \times $$ 10<sup>–26</sup> kg, <br/>Boltzman
constant: 1.38 $$ \times $$ 10<sup>–23</sup> J/K, <br/>Planck constant : 6.63 $$ \times $$ 1... | [{"identifier": "A", "content": "0.44 $$\\mathop A\\limits^o $$"}, {"identifier": "B", "content": "0.34 $$\\mathop A\\limits^o $$"}, {"identifier": "C", "content": "0.20 $$\\mathop A\\limits^o $$"}, {"identifier": "D", "content": "0.24 $$\\mathop A\\limits^o $$"}] | ["D"] | null | We know, the de-Broglie
wavelength<br><br>
$$\lambda $$ = $${h \over {m{v_{rms}}}}$$
<br><br>also V<sub>rms</sub> = $$\sqrt {{{3kT} \over m}} $$
<br><br>$$ \therefore $$ $$\lambda $$ = $${h \over {\sqrt {3mkT} }}$$
<br><br>= $${{6.63 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 4.6 \times {{10}^{ - 26}} \times 1.38 \t... | mcq | jee-main-2020-online-6th-september-evening-slot | 10,027 |
klPxGf59fWvLLQxIwAjgy2xukfrmlo12 | physics | dual-nature-of-radiation | matter-waves | An electron, a doubly ionized helium ion (He<sup>++</sup>) and a proton are having the same kinetic energy.
The relation between their respective de-Broglie wavelengths $$\lambda $$<sub>e</sub>, $$\lambda $$<sub>He<sup>++</sup></sub> and $$\lambda $$<sub>p</sub> is : | [{"identifier": "A", "content": "$$\\lambda $$<sub>e</sub> > $$\\lambda $$<sub>He<sup>++</sup></sub> > $$\\lambda $$<sub>p</sub>"}, {"identifier": "B", "content": "$$\\lambda $$<sub>e</sub> < $$\\lambda $$<sub>p</sub> < $$\\lambda $$<sub>He<sup>++</sup></sub>"}, {"identifier": "C", "content": "$$\\lambda $$... | ["C"] | null | $$\lambda $$ = $${h \over P}$$ = $${h \over {\sqrt {2m\left( {KE} \right)} }}$$
<br><br>$$ \therefore $$ $$\lambda $$ $$ \propto $$ $${1 \over {\sqrt m }}$$
<br><br>m<sub>He<sup>++</sup></sub> > m<sub>p</sub> > m<sub>e</sub>
<br><br>$$ \therefore $$ $$\lambda $$<sub>e</sub> > $$\lambda $$<sub>p</sub> > $$\l... | mcq | jee-main-2020-online-6th-september-morning-slot | 10,028 |
on0C9OUQaeBK1eukKojgy2xukeu4wpq1 | physics | dual-nature-of-radiation | matter-waves | Particle A of mass m<sub>A</sub> = $${m \over 2}$$ moving along the x-axis with velocity v<sub>0</sub> collides elastically with another particle B at rest having mass m<sub>B</sub> = $${m \over 3}$$. If both particles move along the x-axis after the collision, the change $$\Delta $$$$\lambda $$ in de-Broglie wavlength... | [{"identifier": "A", "content": "$$\\Delta $$$$\\lambda $$ = $${5 \\over 2}{\\lambda _0}$$"}, {"identifier": "B", "content": "$$\\Delta $$$$\\lambda $$ = $${3 \\over 2}{\\lambda _0}$$"}, {"identifier": "C", "content": "$$\\Delta $$$$\\lambda $$ = 2$$\\lambda $$<sub>0</sub>"}, {"identifier": "D", "content": "$$\\Delta $... | ["D"] | null | Applying momentum conservation
<br><br>$${m \over 2} \times {V_0} + {m \over 3} \times 0 = {m \over 2}{V_A} + {m \over 3}{V_B}$$
<br><br>$$ \Rightarrow $$ $${{{V_0}} \over 2} = {{{V_A}} \over 2} + {{{V_B}} \over 3}$$ .....(1)
<br><br>Since, collision is elastic (e = 1)
<br><br>e = 1 = $${{{V_B} - {V_A}} \over {{V_0}}}$... | mcq | jee-main-2020-online-4th-september-morning-slot | 10,030 |
Z5WfFTlP0aFOz9TVJC7k9k2k5i8f500 | physics | dual-nature-of-radiation | matter-waves | A particle moving with kinetic energy E has
de Broglie wavelength $$\lambda $$. If energy $$\Delta $$E is added
to its energy, the wavelength become $$\lambda $$/2. Value
of $$\Delta $$E, is : | [{"identifier": "A", "content": "E"}, {"identifier": "B", "content": "3E"}, {"identifier": "C", "content": "2E"}, {"identifier": "D", "content": "4E"}] | ["B"] | null | $$\lambda = {h \over {\sqrt {2mE} }}$$
<br><br>Also, $${h \over {\sqrt {2m\left( {E + \Delta E} \right)} }}$$ = $${\lambda \over 2}$$
<br><br>$$ \therefore $$ $${{E + \Delta E} \over E} = 4$$
<br><br>$$ \Rightarrow $$ $$\Delta $$E = 3E | mcq | jee-main-2020-online-9th-january-morning-slot | 10,031 |
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