Q281JEE Main 2019MCQ

In the given circuit the internal resistance of the18 V cell is negligible. If $R_1$ = 400 Ω, $R_3$ = 100 Ωand $R_4$ = 500 Ω and the reading of an ideal voltmeteracross $R_4$ is 5V, then the value $R_2$ will be :

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📖 Explanation

$V_4$ = 5V $i_1$ = $V_4/R_4$ = 0.01 A $v_3$ = $i_1R_3$ = 1 V $V_3+V_4$ = 6V = $V_2$ $V_1+V_3+V_4$ = 18 V $V_1$ = 12 V i = $V_1/R_1$ = 0.03 Amp $i_2$ = 0.02 Amp , $V_2$ = 6V $R_2$ = $V_2/i_2$ = $6/{0.02}$ = 300 Ω

Q282JEE Main 2019MCQ

Q283JEE Main 2019MCQ

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 Ω/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:

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📖 Explanation

Total resistance of potentiometer wire is 4ΩTotal potential drop across wire is $4/6\times 3V=2V$ For 50cm voltmeter reading $\frac{50}{400} \times 2= \frac{1}{4}=0.25 V$

Q284JEE Main 2019MCQ

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11V is connected across it is :

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📖 Explanation

P = $I^2R$ 4.4 = 4 × $10^{-6}$ R R = = 1.1 × $10^6$ R P' = ${11}^2/R$ = ${11}^2/{1.1}$ × $10^{-6}$ = 11 × $10^{-5}$ W

Q285JEE Main 2019MCQ

A carbon resistance has a following colourcode. What is the value of the resistance ?

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📖 Explanation

R = 53 × $10^4$ ± 5% = 530 kΩ ± 5%

Q286JEE Main 2019MCQ

The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 - 0.5A?

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📖 Explanation

$50(0.002)=R(0.5-0.002)$ 0.1 $\approx R(0.5)$ $R \approx 0.2 \Omega$

Q287JEE Main 2019MCQ

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 lineone may conclude that:

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Q288JEE Main 2019MCQ

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: $R_{1}=15 \Omega, R_{2}=10 \Omega,$ $R_{3}=20 \Omega, R_{4}=5 \Omega,$ $R_{5}=25 \Omega, R_{6}=30 \Omega, E=15 V$

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📖 Explanation

$R_{ {eq}}=15+\;\frac{25}{3}+30=\;\frac{45+25+90}{3}=\;\frac{160}{3}$ $I=\;{E}/{R_{ {eq}}}=\;\frac{15 \times 3}{160}=\;\frac{9}{32} {amp}$

Q289JEE Main 2019MCQ

A moving coil galvanometer has resistance 50Ω and it indicates full deflection at 4 mA current. A voltmeter is made using this galvanometer and a 5 kΩ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to:

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📖 Explanation

$I_{\max }=4 M A$ $R_{G}=50 \Omega$ In voltmeter $V=I_{\max } (R_{s}+S_{K} )$ V = 20

Q290JEE Main 2019MCQ

A moving coil galvanometer, having a resistance G, produces full scale deflection when a current $I_g$ flows through it. This galvanometer can be converted into (i) an anometer of range 0 to $I_{0}\ (I_{0} < I_{g}\ )$ by connecting a shunt resistance $R_A$ to it and connecting a shunt resistance $R_A$ to is and (ii) into a voltmeter of range 0 to $V\ (V=G I_{0}\ )$ by connecting a series of resistance $R_V$ to it. Then,

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📖 Explanation

$S=\ \frac{G}{\ \frac{i}{i g}-1} \Rightarrow R_{A}=\ {G}/{\ \frac{I_{0}}{I_{g}}-1}$ $R_{V}=G [ \frac{V}{V_{g}}-1\ ] \Rightarrow R_{V}=G\ [ \frac{V}{i_{g} G}-1\ ]$ $R_{V}=G\ [ \frac{i_{0}}{i_{g}}\ ]-1$ $\Rightarrow R_{V}=G [\ \frac{i_{0}}{i_{g}}-1\ ]$ $=G\ [\ \frac{G}{R_{A}}\ ]$ $\Rightarrow R_{A} R_{V}=G^{2}$ $\frac{R_{A}}{R_{V}}=$ $\ {G / \frac{i_{0}}{i_{g}}-1}/{G / [ \frac{i_{0}}{i_{g}}-1\ ]}$ $=\ {1}/{\ [\ \frac{i_{0}}{i_{g}}-1\ ]^{2}}$ $\frac{R_{A}}{R_{V}} =\ \frac{i_{g}^{2}}{\ (i_{0}-i_{g}^{2}}$

Q291JEE Main 2019MCQ

In the given circuit,an ideal voltmeter connected across the 10Ω resistance reads 2V. The internal resistance r, of each cell is

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📖 Explanation

${V}_{ {AB}}=2$ , $10 {i}_{1}=2 \Rightarrow {i}_{1}= \frac{1}{5}$ $15i_2=2\Rightarrow {i}_{2}= \frac{2}{15}$ ${i}_{1}+ {i}_{2}= \frac{1}{3}$ $3-2 {r} ( {i}_{1}+ {i}_{2} )-10 {i}_{1}-2 ( {i}_{1}+ {i}_{2} )=0$ $1- \frac{2}{3}= \frac{2 r}{3}\Rightarrow \frac{1}{3}= \frac{2 r}{3}$ $r= \frac{1}{2}=0.5 \Omega$

Q292JEE Main 2019MCQ

The resistive network shown below is connected to a D. C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is.

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📖 Explanation

$P= \frac{\epsilon ^{2}}{R}\Rightarrow 4= \frac{256}{8 R}$ R = 8Ω

Q293JEE Main 2019MCQ

A metal wire of resistance 3Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be:

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📖 Explanation

$R=\ \frac{\∂ L}{A}=\ \frac{\∂ L^{2}}{V} \;\;\;\;\; R \propto L^{2}$ After doubling $R=12 \Omega$ For ${A} . \ {B}$ wire $R \propto L$ Parallel $\ \frac{1}{ {Reff}}= \frac{1}{2}+ \frac{1}{10}= \frac{3}{5}$ $\;\;\;\;\;\;\;\;\;\;\;\; {Reff}=\ \frac{5}{3} \Omega$

Q294JEE Main 2019MCQ

In a conductor, if the number of conduction electrons per unit volume is $8.5 \times 10^{28} m ^{-3}$ and mean free time is 25 fs (femto second), it's approximate resistivity is: $(m_e = 9.1\times 10^{-31 })$

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📖 Explanation

n=no.of conduction electron per unit vol = $8.5 \times 10^{28} m ^{-3}$ $\tau =25 f s$ Using the formula $\int =\ \frac{m}{n e^{2} \tau }=\ \frac{9.1 \\times 10^{-31}}{8.5 \\times 10^{28} \times (1.6 \times 10^{-19} )^{2} \\times 25 \times 10^{-15}} \approx 10^{-8} \Omega m$

Q295JEE Main 2019MCQ

To verify Ohm'slaw, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained. If $\text{V}_0$ is almost zero, identify the correct statement.

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📖 Explanation

$V=E-I r$ When $I=0, V=E=1.5 V$ When $I=1000 {mA}, V=0$ $\therefore 1.5-1 . r=0$ $r=1.5 \Omega$

Q296JEE Main 2019MCQ

For the circuit shown, with ${R}_{1}=1.0 \Omega, {R}_{2}=2.0 \Omega, {E}_{1}=2 {V}$ and ${E}_{2}= {E}_{3}=4 {V},$ the potential difference between the points 'a' and 'b' is approximately (in V) :

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📖 Explanation

$V_{a}-V_{b}= \frac{ \frac{2}{2}+ \frac{4}{2}+ \frac{4}{2}}{ \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}}$ $= \frac{5 \times 2}{3}=3.3 {V}$

Q297JEE Main 2019MCQ

In the given circuit the cells have zero internalresistance. The currents (in Amperes) passingthrough resistance $R_1$ , and $R_2$ respectively, are:

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📖 Explanation

$i_1$ = $\frac{10}{20}$ = 0.5 A $i_2$ = 0

Q298JEE Main 2019MCQ

A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is :

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📖 Explanation

P = $i^2R$ ∴ for $i_{\max}$ , R must be minimumfrom color coding R = 50 × $10^2$ Ω ∴ $i_{\max}$ = 20mA

Q299JEE Main 2019MCQ

In a meter bridge experiment the circuit diagram and the corresponding observation table are shown in figure . Which of the readings is inconsistent ?

🚀 Solve in Practice Mode

📖 Explanation

We know that in meter bridge $\frac{R}{ l}= \frac{x}{100- l}$ a) $R=1000,1=60 \frac{1000}{60}= \frac{x}{40}$ $\Rightarrow x= \frac{2}{3}\times 1000 \approx 666$ b) $R=100,1=13 \frac{100}{13}= \frac{x}{87}$ $\Rightarrow x= \frac{87}{13} \times 100 \approx 666$ c) $\frac{10}{1.5}= \frac{x}{89.5}$ $\Rightarrow x= \frac{89.5}{1.5} \times 10 \approx 666$ d) $\frac{1}{1}= \frac{x}{99}\Rightarrow x=99$ Option 'd' is odd one So it is incontinent

Q300JEE Main 2019MCQ

The actual value of resistance R, shown in the figure is 30Ω. This is measured in an experiment as shown using the standard formula R = $V/I$ , where V and I are the readingsof the voltmeter and ammeter, respectively. Ifthe measured value of R is 5% less, then theinternal resistance of the voltmeter is :

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📖 Explanation

0.95 R = $\frac{RR_υ}{R+R_υ}$ 0.95 × 30 = 0.05 $R_υ$ $R_υ$ = 19 × 30 = 570 Ω