Practice free Physics NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Two resistances $R_1 and R_ 2 $have effective resistance $R_s$ when connected in series combination and $R_p$ when connected in parallel combination if $R_s R_p= 16 and R_1/ R_ 2 = 4$ the values of $R_1 and R_ 2$ are
$R_p = { R_1 R_2 \over R_1 + R_2 }$ $Rs = R_1 +R_2 $ $ Use the \, {R_1 \over R_2 } = 4 $
Three identical resistors connected in series with a battery, together dissipate 10W of power. What will be the power dissipated, if the same resistors are connected in parallel across the same battery?
$ R_s = R_1 + R_2 + R_3 = R +R+R = 3 R $ $R_p = { 1 \over R_1} + { 1 \over R_2 } + { 1 \over R_3 } = { 1\over R }+ {1 \over R} + {1 \over R } = { 3 \over R } $ $Or R_p = { R \over 3 } $ $ P = {V^2 \over R} \alpha { 1 \over R} $ $ {P_s \over P_p} = { R_p \over R_s} Or P_p = { R_s \over R_p } \times P_s = { 3R \over R/3} \times 10 = 90 W$
A potentiometer wire of length 1 m and resistance $10 \Omega$ is connected in series with a cell of e.m.f 2V with internal resistance $ 1 \Omega $ and a resistance box of a resistance R if potential difference between ends of the wire is 1V the value of R is.
$ I = { E \over x + R+r} = { 2 \over 10+R+1} = { 2 \over 11+R} $ $ \Rightarrow V = I \times X $ $ Or I = { 2 \over 11 + R } \times 10 = { 20 \over 11+R} $ $ Or 11 + R = 20 $ $ R = 20 -11 = 9 \Omega $
For a cellof e.m.f 2V, a balance is obtained for 50 cmof the potentiometer wire If the cell is shunted by a $2 \Omega$ resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is.
$ r = { I_1 -I_2 \over l_2 } \times R = 0.5 \Omega $
n identical cells each of e.m.f E and internal resistance r are connected in series An external resistance R is connected in series to this combination the current through R is
Total e.m.f = nE Total resistance $ R +nr \Rightarrow i = {nE \over R+nr } $
4 cell each of emf 2v and internal resistance of $1 \Omega $ are connected in parallel to a load resistor of $ 2 \Omega$ Then the current through the load resistor is....
$ I = { E \over R+{1 \over 4}} = { 2 \over 2 + {1 \over 4 }} = { 2 \over 2.25 } = 0.888A$
A Potentiometer wire, 10mlong, has a resistance of $40 \Omega $ It is connected in series with a resistance box and a 2V storage cell If the potential gradient along the wire is 0.1mv/cm, the resistance unplugged in the box is.
Potential gardient along wire = Potential difference along wire length of wire $ = 0.1 \times 10^ {-3} = {I \times 40 \over 1000} V/cm $ $ \Rightarrow current in wire , I = { 1 /400 A}$ $ {2 \over 40+R} = { 1 \over 400} or R = 800 -40 = 760 \Omega $
The resistivity of a potentiometer wire is $ 40 \times 10 ^ {-8} ohm m$ and its area of cross-section is $ 8 \times 10 ^ { -6 } m^2 $. If 0.2 amp current is flowing through the wire, the potential gradient will be.
Potential gradient = $ { V \over L } = { iR \over L } = { ipl \over AL } = { ip \over A} $
Potentiometer wire of length 1mis connected in series with $490 \Omega $ resistance and 2V battery If 0.2 mv/ cm is the potential gradient, then resistance of the potentiometer wire is.
Potential gradient x = $ { e \over ( R +Rh +r ) } {R \over L }$
A cell supplies a current I, through aresistance$ R_1$ and a current$ I_2$ through a resistance $R_ 2$ the internal resistance of a cell is....
$ E = I_1(R_1+r ) = I_2 (R_2+r)$
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.