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Physics NEET Questions — Free MCQ Practice

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Two wires of resistances R1 and R 2 have temperature coeffcient of resistances $ \alpha _1 $ and $ \alpha_2 $ respectively they are joined in series the effective temperature coefficient of resistance is ....

A. $ {\rho_1 + \rho_2 \over 2 } $ B. $ \sqrt{ \rho_1 \rho_2}$ C. $ { \rho_1 R_1 + \rho_2 R_2 \over R_1 +R_2}$ D. $ {\sqrt{R_1 R_2 \alpha_1 \alpha_2} \over \sqrt{ R_1^2 + R_2^2}}$

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Explanation

$ R_it = R_1 ( 1 + \alpha it ) R_2 = R_2 ( 1 + \alpha_2 t )$ $ R_st = R_1 t + R_2 t $ $ = R_1[ 1 + \alpha_1 t ] + R_2[ 1 + \alpha_2 t ] $ $ = R_s [ 1+ ({(R_1 \alpha_1 +R_1 \alpha_2 t )\over R_1+R_2})]$

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The resistance of the series combination of two resistances is S, when they are joined in parallel the total resistance is P If S=n P, then the minimum possible value of n is....

A. 4 B. 3 C. 2 D. 1

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Explanation

$ S = R_1 + R_2 ; P = { R_1R_2 \over R_1 +R_2 } ; S = nP$ $ R_1 + R_2 = { nR_1R_2 \over R_1 + R_2 } $ or $ R_1^2 + R_2^2+2R_1R_2 = nR_1R_2$ or $(R_1 -R_2) ^2 + 4R_1R_2 = nR_1R_2$ or $ (R_1 -R_2)^2 -R_1R_2 (n-4)$ $ If R_1 = R_2 , Then (n-4) = 0 or (n=4) $

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Two sources of equal emf are connected to an external resistance R the internal resistance of the two soureces are $R_1 and R_2 ( R_ 2 > R_1 )$ if the potential difference across the source having internalresistance R 2 is Zero, then

A. $ R = R_1R_2 / (R_2 - R_1)$ B. $ R = R_1R_2 / (R_1 + R_2)$ C. $ R = R_2 -R_1 $ D. $ R = R_2(R_1 + R_2) / (R_2 -R_1)$

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Explanation

$ I = { E+E \over R_1 + R_2 +R } $ $ R_2 = { 2E \times R_2 \over (R_1 + R_2 + R)}$ $ \Rightarrow E - { 2ER_2 \over R_1 + R _2 +R } = 0 or E = { 2ER_2 \over R_1 + R_2 +R} $ $ R = R_2 - R_1 $

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In a wheatstone's bridge, three resistance P, Q and R connected in three are a and the fourth arm is formed by two resistances $S_1 and S_2$ connected in paralled The condition for bridge to be balanced will be.

A. $ {P \over Q} = { R \over S_1+S_2}$ B. $ {P \over Q} = { 2R \over S_1+S_2}$ C. $ {P \over Q} = { R ( S_1+S_2) \over \rho_1 S_2}$ D. $ {P \over Q} = { R ( S_1+S_2) \over 2 S_1S_2}$

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Explanation

$ S = { S_1 S_2 \over S_1 + S_2} $ $ {P \over Q } = { R \over S}$

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The resistance of a wire is $5 \Omega $ at $ 50 ^\circ C $ and $6 \Omega $ at $ 100 ^\circ C $ The resistance of the wire at $ 0 ^ \circ C $ will be

A. $ 3 \Omega $ B. 2 C. $ 1 \Omega $ D. $ 4 \Omega $

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Explanation

$ { P \over 1/3 } = { Q \over 1 - (1/3)} , 1 =1 m$ or 3P = 3/2 Q or P = Q/2 $ { P+6 \over 2/3 } = { Q \over 1/3 } $ $ P+6 = 2 Q $ $ 6 = 2 Q - { Q \over 2 } = { 3Q \over 2 } $ So Q = 4 and P =2

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Resistors P and Q connected in the gaps of the meter bridge. the balancing point is obtained 1/3 m from the zero end If a $6 \Omega $ resistance is connected in series with p the balance point shifts to 2/3m form same end P and Q are.

A. 4,2 B. 2,4 C. both (a) and (b) D. neither (a) nor (b)

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Explanation

$ R = R_0 [1 + 2t ] $ $ for series connection R_s = R_1 + R_2 $ $ at 0 ^ \circ C tap R_s = R_o + R_0 + 2 R_0 $ $2R_0 [ 1 + \alpha st ] = R_0 [ 1 + \alpha_1 t ] + R_0[1 + \alpha _2 t ]$ $ \therefore \alpha _3 = { 1 \over \alpha } (\alpha_1 + \alpha_2) ...(1) $ $ for parallel connection { 1 \over R_p} = {1 \over R_1} + {1 \over R_2}$ $ at 0 ^\circ C temp RP = R_0 /2 $ $ \therefore { 1 \over R_0/2(1 + \alpha_p t ) }= { 1 \over R_0[ 1 + \alpha_1 t]} + { 1 \over R_0[1+\alpha_2 t ]} $ $ \therefore \alpha_p = {1 \over \alpha} (\alpha_1 + \alpha _2)$

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2 A current is obtained when a $2 \Omega $ resistor is connectd with battery having $ r \Omega $ as internal resistance 0.5A current is obtained if the above battery is connected to $ 9 \Omega $ resistor. Culculate the internal resistance of the battery.

A.

$ 0.5 \Omega $

B.

$ { 1 \over 3} \Omega $

C.

$ { 1 \over 4 } \Omega $

D.

$ 1 \Omega $

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Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature It at room temperature, 100w, 60w and 40w bulbs have filament resistances $R_100 , R_60 and R_ 40$ respectively the relation between these resistances is

A. $ {1 \over R_{100}} = {1 \over R_{40}} + {1 \over R_{60}}$ B. $ R_{100} = R_{40} + R_{60}$ C. $ R_{100} \gt R_{40} \gt R_{60}$ D. $ {1 \over R_{100}} \gt {1 \over R_{60}} \gt {1 \over R_{40}}$

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Explanation

The power rating of a bulb is given by P = V^2 / R, where P is the power, V is the voltage, and R is the resistance. For bulbs with different wattages at the same voltage, the resistance is inversely proportional to the power rating. Therefore, the bulb with the highest wattage will have the lowest resistance. Thus, R_100 < R_60 < R_40, which implies 1/R_100 > 1/R_60 > 1/R_40.

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Column I
(a) The unit of electrical resistivity is (b) The unit of current density is (c) The unit of electrical conductivity is (d) The unit of electric mobility is Column II (p) $ m^2S^-1V^-1$ (q) $ \Omega ^ {-1} m^{-1}$ (r.) $ Am^{-2}$ (s) $ \Omega m$

A. a - p, b - q, c - r, d - s B. a - s, b - r, c - q, d - p C. a - r, b - q, c - p, d - s D. a - q, b - r, c - s, d - p

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Explanation

The units of the given physical quantities are: (a) Electrical resistivity: Î©m, (b) Current density: A/m^2, (c) Electrical conductivity: Î©^-1m^-1, (d) Electric mobility: m^2s^-1V^-1. Matching these units to the corresponding quantities in Column II, we get: a - s, b - r, c - q, d - p.

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Match the physical quantities given in column I with their dimensional formulae given in column II -I stands for the dimension of current. Column I (a) Electromotive force (emf) (b) Resistance (c) Resistivity (d) Conductivity Column II (p) $ [ M L^2T^{3} A^{-2}]$ (q) $ [ M L^3T^{-3} A^{-2}]$ (r ) $ [ M^{-1} L^{-3}T^{3} A^{2}]$ (s) $ [ M L^2T^{-3} A^{-1}]$

A. a - s, b - p, c - q, d - r B. a - p, b - s, c - r, d - q C. a - p, b - s, c - r, d - q D. a - r, b - p, c - q, d - s

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Explanation

The dimensional formulas for the given physical quantities are: (a) Electromotive force (emf): [ML^2T^-3A^-1], (b) Resistance: [ML^2T^-3A^-2], (c) Resistivity: [ML^3T^-3A^-2], (d) Conductivity: [M^-1L^-3T^3A^2]. Matching these dimensional formulas to the quantities in Column II, we get: a - s, b - p, c - q, d - r.

Physics MCQs for NEET — Practice Questions with Answers

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