Q161JEE Main 2022MCQ

Two cells of same emf but different internal resistances $r_1$ and $r_2$ are connected in series with a resistance $R$ . The value of resistance ${\text{R}}$ , for which the potential difference across second cell is zero, is : [25-Jun-2022-Shift-2]

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📖 Explanation

$I=\;\frac{2 \epsilon }{R+r_1+r_2}$ As per the question, $\;\frac{2 \epsilon }{R+r_1+r_2} \times r_2-\epsilon =0$ $\Rightarrow R=r_2-r_1$

Q162JEE Main 2022MCQ

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2 Ω$ . The value of internal resistance of each cell is [29-Jun-2022-Shift-2]

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📖 Explanation

From diagram $i_p=\;\frac{E}{2+\;\frac{r}{2}}$ and $i_s=\;\frac{2 E}{2+2 r}$ given $i_p=i_s$ $\;\frac{1}{2+\;\frac{r}{2}}=\;\frac{1}{1+r}$ $1+r=2+\;\frac{r}{2}$ $r=2 Ω$

Q163JEE Main 2022NAT

In a meter bridge experiment, for measuring unknown resistance ' $S$ ', the null point is obtained at a distance $30 {\text{cm}}$ from the left side as shown at point ${\text{D}}$ . If ${\text{R}}$ is $5.6 {\text{k}} Ω$ , then the value of unknown resistance 'S' will be _______ $Ω$ . [27-Jul-2022-Shift-1]

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📖 Explanation

$\;{ {\text{S}}}/{30}=\;\frac{5.6 \times 10^3}{70}$ ${\text{S}}=\;\frac{3}{7} \times 5.6 \times 10^3=2400$

Q164JEE Main 2022MCQ

(A) The drift velocity of electrons decreases with the increase in the temperature of conductor. (B) The drift velocity is inversely proportional to the area of cross-section of given conductor. (C) The drift velocity does not depend on the applied potential difference to the conductor. (D) The drift velocity of electron is inversely proportional to the length of the conductor. (E) The drift velocity increases with the increase in the temperature of conductor. Choose the correct answer from the options given below : [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Drift velocity $=(\;\frac{e \tau }{m}) E$ ${\text{v}}_{ {\text{d}}}=(\;{ {\text{e}} \tau }/{ {\text{m}}})(\;{\Delta {\text{V}}}/{ℓ})$ $\Delta {\text{V}}=$ Potential difference applied across the wireAs temperature increases, relaxation time decreases, hence ${\text{V}}_{ {\text{d}}}$ decreases.As per formula, ${\text{V}}_{ {\text{d}}} \propto \;\frac{1}{ℓ}$ ${\text{v}}_{ {\text{d}}}=\;{ {\text{I}}}/{\;\text{ neA }\;}$ , as it is not mentioned that current is at steady state neither it is mentioned that ${\text{n}}$ is constant for given conductor. So it can't be said that ${\text{v}}_{ {\text{d}}}$ is inversely proportional to ${\text{A}}$ . ${\text{I}}=n e A v_d=\;\frac{V}{R}=\;{ {\text{V}}}/{\rho ℓ} {\text{A}}$ ${\text{v}}_{ {\text{d}}}=\;{ {\text{V}}}/{\rho ℓ \;\text{ ne }\;} \;\; ( {\text{E}}=\;{ {\text{V}}}/{ℓ})$ ${\text{v}}_{ {\text{d}}}=\;{ {\text{eE}} \tau }/{ {\text{m}}}$ $\tau$ decrease with temperature increase.First and fourth statements are correct.

Q165JEE Main 2022MCQ

An electric cable of copper has just one wire of radius $9 {\text{mm}}$ . Its resistance is $14 Ω$ . If this single copper wire of the cable is replaced by seven identical well insulated copper wires each of radius $3 {\text{mm}}$ connected in parallel, then the new resistance of the combination will be : [30-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Initially, copper wire radius $(r_1) =9 {\text{mm}}$ Resistance $(R)=14 Ω$ We know, $R=\;\frac{\rho L}{A}=\;\frac{\rho L}{\pi r_1^2}=14$ Now this copper wire is replaced by 7 parallel copper wire of resistance $R_1$ . $\therefore$ Equivalent resistance of 7 parallel copper wire, $\;\frac{1}{R_{e q}}=\;\frac{1}{R_1}+\;\frac{1}{R_1}+\;\frac{1}{R_1}+\;\frac{1}{R_1}+\;\frac{1}{R_1}+\;\frac{1}{R_1}+\;\frac{1}{R_1}$ $\Rightarrow \;\frac{1}{R_{e q}}=\;\frac{7}{R_1}$ $\Rightarrow R_{e q}=\;\frac{R_1}{7}$ $=\;\frac{1}{7} \times \;\frac{\rho L}{\pi r_2^2}$ $=\;\frac{1}{7} \times \;\frac{\rho L}{\pi (\;\frac{r_1}{3}) ^2} [\text{as} \; r_2=\;\frac{r_1}{3} ; r_1=9 {\text{m}} \;\text{ and }\; r_2=3 ]$ $=\;\frac{1}{7} \times \;\frac{\rho L}{\pi r_1^2} \times 9$ $=\;\frac{1}{7} \times 14 \times 9$ $=18 Ω$

Q166JEE Main 2022MCQ

A $72 Ω$ galvanometer is shunted by a resistance of $8 Ω$ . The percentage of the total current which passes through the galvanometer is : [27-Jun-2022-Shift-1]

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📖 Explanation

From the given setup $y \times R_G=(x-y) (R_S)$ $\Rightarrow y \times 72=(x-y) \times 8$ $\Rightarrow 9 y=x-y$ $\Rightarrow y=\;\frac{x}{10} \;\text{ or }\; 10 \% \;\text{ of }\; {\text{x}}$

Q167JEE Main 2022NAT

As shown in the figure, a potentiometer wire of resistance $20 Ω$ and length $300 {\text{cm}}$ is connected with resistance box (R.B.) and a standard cell of emf 4 V. For a resistance ' $R$ ' of resistance box introduced into the circuit, the null point for a cell of $20 {\text{mV}}$ is found to be $60 {\text{cm}}$ . The value of ' $R$ ' _______ is $Ω$ . [28-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

${\text{E}}=\;{ {\text{AC}}}/{ {\text{AB}}}( {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{B}}})$ $\therefore 20 \times 10^{-3}=\;\frac{60}{300} \times \;{4 \times 20}/{ {\text{R}}+20}$ $\therefore {\text{R}}=780$

Q168JEE Main 2022NAT

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_1$ is $40 {\text{cm}}$ . Now an unknown resistance $x$ is connected in series with ${\text{P}}$ and new balancing length is found to be $80 {\text{cm}}$ measured from the same end. Then the value of $x$ will be _________ $Ω$ . [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\frac{P}{40}=\;\frac{Q}{60} . . . . . .(1)$ $\;\frac{P+x}{80}=\;\frac{Q}{20} . . . .(2)$ $\;\frac{P}{P+x} \times \;\frac{80}{40}=\;\frac{20}{60}$ $\;\frac{4}{4+x} \times 2=\;\frac{1}{3}$ $24=4+x$ $x=20$

Q169JEE Main 2022MCQ

Given below are two statements :Statement I : A uniform wire of resistance $80 Ω$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be $5 Ω$ .Statement II: Two resistances $2 {\text{R}}$ and $3 {\text{R}}$ are connected in parallel in a electric circuit. The value of thermal energy developed in $3 R$ and $2 R$ will be in the ratio $3: 2$ .In the light of the above statements, choose the most appropriate answer from the option given below [28-Jul-2022-Shift-2]

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📖 Explanation

Statement I: $R_{1 \;\text{ part }\;}=\;\frac{80}{4}=20 Ω$ $\Rightarrow R_{e f f}=\;\frac{20}{4}=5 Ω$ Statement II : Ratio $=\;\frac{\;\frac{(\Delta V)^2}{3 R}}{\frac{\Delta \Delta V)^2}{2 R}}$ $=\;\frac{2}{3}$

Q170JEE Main 2022MCQ

A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for $\;\frac{1}{3}$ deflection in the galvanometer. Which of the below is true for measuring value of ${\text{G}}$ ? [25-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

The circuit for the given situation is:Since $G$ and $S$ are in parallel, $\Rightarrow \;\frac{i}{3} \times {\text{G}}=\;\frac{2 i}{3} \times {\text{S}}$ $\Rightarrow {\text{G}}=2 {\text{S}}$ $\Rightarrow G$ equals twice the value of shunt resistance.

Q171JEE Main 2022MCQ

The current I in the given circuit will be : [26-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

The grouping of resistance is a wheatstone bridgeSo, $R_{\;\text{net }\;}=4 Ω$ So, $i=\;{V}/{R_{\;\text{net }\;}}=10 {\text{A}}$

Q172JEE Main 2022NAT

A $1 {\text{m}}$ long copper wire carries a current of $1 {\text{A}}$ . If the cross section of the wire is $2.0 {\text{mm}}^2$ and the resistivity of copper is $1.7 \times 10^{-8} Ω {\text{m}}$ , the force experienced by moving electron in the wire is _______ $\times 10^{-23} {\text{N}}$ . (charge on electron $=1.6 \times 10^{-19} {\text{C}}$ ) [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$l=1 {\text{m}}$ ${\text{i}}=1 {\text{A}}$ Area $=2 \times 10^{-6}$ $\rho =1.7 \times 10^{-8}$ ${\text{R}}=\;{\rho ℓ}/{ {\text{A}}}=\;\frac{1.7 \times 10^{-8} \times 1}{2 \times 10^{-5}}=\;\frac{1.7}{2} \times 10^{-2}$ ${\text{v}}=\;\frac{1.7}{2} \times 10^{-2}$ ${\text{F}}=1.6 \times 10^{-19} \times \;\frac{1.7}{2} \times 10^{-2}$ $=1.36 \times 10^{-21}$ $=136 \times 10^{-23}$

Q173JEE Main 2022NAT

Current measured by the ammeter $( {\text{A}})$ in the reported circuit when no current flows through $10 Ω$ resistance, will be_____A. [28-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Using the condition of a balanced wheat stone bridge, $\Rightarrow \;{ {\text{R}}}/{3}=\;\frac{4}{6} \Rightarrow {\text{R}}=2 Ω$ So the effective resistance of the circuit is ${\text{R}}_{ {\text{eq}}}=\;\frac{6 \times 9}{6+9}=\;\frac{18}{5} Ω$ ${\text{i}}=\;{36}/{ {\text{R}}_{ {\text{eq}}}}=10 {\text{A}}$

Q174JEE Main 2022MCQ

Which of the following physical quantities have the same dimensions? [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Electric displacement $(\vec{D})=\epsilon _{0} \vec{E}$ $\Rightarrow [\overline{D}]=[\epsilon _{0}][\overline{E}]$ $= M^{-1} L^{-3} T^{4} A^{2}][M^{1} L^{1} A^{-1} T^{-3}$ ${[\overline{D}]=\$ [L^{-2} T^{1} A^{1}]$} $[Surface charge density]$ =;\frac{[Q]}{[A]}{[\sigma ]=$[A T L^{-2}]$}\Rightarrow \vec{D} ;\text{ and };[\sigma ] ;\text{ have same dimensions };$

Q175JEE Main 2022NAT

The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be _____ $\%$ . [25-Jun-2022-Shift-2]

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📖 Explanation

Volume is constant so on length doubledArea is halved so $R=\rho \;\frac{l}{A}$ and $R^{'}=\rho \;\frac{2 l}{\frac{A}{2}}=4 \rho \;\frac{l}{A}=4 R$ So percentage increase will be $R \%=\;\frac{4 R-R}{R} \times 100=300 \%$

Q176JEE Main 2022MCQ

An aluminium wire is stretched to make its length, $0.4 \%$ larger. The percentage change in resistance is : [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$R=\;\frac{\rho l}{A}$ Also volume will remain constanti.e., ${\text{Al}}=$ constant $\Rightarrow A \propto \;\frac{1}{l}$ $\therefore R \propto l^2$ $\;\frac{\Delta R}{R}=2 \;\frac{\Delta l}{l}=0.8$

Q177JEE Main 2022MCQ

A current of $15 {mA}$ flows in the circuit as shown in figure. The value of potential difference between the points $A$ and $B$ will be: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Effective $R=[5+\;\frac{5 \times 10}{5+10}+10] k Ω$ $=\;\frac{275}{15} k Ω$ $\Rightarrow \Delta V_{A B}=15 m A \times \;\frac{275}{15} k Ω$ $=275 {V}$

Q178JEE Main 2022NAT

In the given figure, the value of ${V}_{0}$ will be V. [25-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Using Kirchhoff's junction rule. $\;\frac{2-V_{0}}{1}+\;\frac{4-V_{0}}{1}+\;\frac{6-V_{0}}{1}=0$ $12-3 V_{0}=0$ $V_{0}=4 V$

Q179JEE Main 2022NAT

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 ' ${\text{K}}_1$ ' is kept open the null point is found to be at $120 {\text{cm}}$ on the potentiometer wire. When the key ' ${\text{K}}_1$ ' is closed the null point is shifted at $80 {\text{cm}}$ at the potentiometer wire. The internal resistance of the given cell is_____ $Ω$ [30-Jun-2022-Shift-1]

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Q180JEE Main 2022MCQ

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 : [29-Jul-2022-Shift-1]

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📖 Explanation

Let length of wire be ' $ℓ$ 'Area of wire as 'A'For equivalent wire length $=2 ℓ \&$ area will be ${\text{A}}$ Thermal resistance ${\text{R}}_{ {\text{eq}}}= {\text{R}}_1+ {\text{R}}_2$ $\;{2 ℓ}/{\sigma _{ {\text{eq}}} {\text{A}}}=\;{ℓ}/{\sigma _1 {\text{A}}}+\;{ℓ}/{\sigma _1 {\text{A}}}$ $\;{2 ℓ}/{\sigma _{ {\text{eq}}}}=\;\frac{ℓ}{\sigma _1}+\;\frac{ℓ}{\sigma _2} \Rightarrow \sigma _{ {\text{eq}}}=\;\frac{2 \sigma _1 \sigma _2}{\sigma _1+\sigma _2}$