MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 11
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
The resistance in the two arms of the meter-bridge are 5Ω and RΩ, respectively.When the resistance R is shunted with an equal resistance,the new balance point is at 1.6l1.The resistance 'R', is
0%
a) 20 Ω
0%
b) 25Ω
0%
c) 10Ω
0%
d) 15Ω
Explanation
According to Wheatstone bridge balance condition When another resistance of R is shunted by another R Multiply (i) by 1.6 Comparing (ii) and (iii) 160 -1.6l1 = 200 -3.2l1 1.6l1 = 40; l1=25 substituting l1 is (i) R = 15Ω Answer:(d)
Q.2
Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is0.5 立. The power loss in the wire is Answer : (d)
0%
a) 19.2J
0%
b) 12.2 kW
0%
c) 19.2W
0%
d) 19.2 kW
Explanation
Total resistance = 150 × 0.5 =75 Current in the wire = 8/0.5= 16 A Power loss = I2R = 162× 75 = 19200 w = 19.2kW Answer:(d)
Q.3
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4m long.When the resistance R, connected across the given cell, has values of(i)infinity(ii) 9.5Ω, the 'balancing lengths', on the potentiometer wire are found to be 3m and 2.85 m, respectively. The value of internal resistance of the cell is
0%
a) 0.5Ω
0%
b) 0.75Ω
0%
c) 0.25Ω
0%
d) 0.95Ω
Explanation
Value of resistance is infinity means key is open length l1 = 3 Value of resistance is 9.5 Ω means key is closed length l2 =2.85 Answer:(a)
Q.4
A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. Then :
0%
a) VA ≠ VB ≠ VC
0%
b) VA = VB = VC
0%
c) VA ≠ VB = VC
0%
d) VA = VB ≠ VC
Explanation
B and C are parallel thus VB = VC Effective resistance of parallel combination of B and C = 1R Thus VA = VB = VC Answer:(b)
Q.5
A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1 mV per cm on the wire is :
0%
a) 48 Ω
0%
b) 32 Ω
0%
c) 40 Ω
0%
d) 44 Ω
Explanation
Required potential across wire = 400×1mV=0.4V And current in wire =0.4/8= 0.05 A same current flow through resistance Current drop in resistance connected = 2-0.4=1.6 V Resistance =1.6/0.05=32 Ω Answer:(b)
Q.6
s a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is :
0%
a) electric filed
0%
b) current density
0%
c) current
0%
d) drift velocity
Explanation
Answer:(c)
Q.7
A potentiometer wire of length L and a resistance are connected in series with a battery of e.m.f. E0 and a resistance rAn unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by :
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Current in wire of length L and resistance ‘r’ and connected with resistance r1 because of E0 Now potential is balanced by length ‘l’, E= I ρl …(ii) Now ρ=r/L Substituting value of I from (i) and ρ in (ii) Answer:(c)
Q.8
Two metal wires of identical dimensions are connected in series. If σ1 and σ2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is :-
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let R1 and R2 are the resistance having conductivity σ1 and σ2 Since resistance connected in series, resistance of the combination R = R1+R2 If σ is the conductivity of combination, Length of resistance is now 2l as both resistance connected in series, but area A will be same thus From (i) and (ii) we get Answer:(b)
Q.9
A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be :-
0%
a) 1 A
0%
b) 0.5 A
0%
c) 0.25 A
0%
d) 2 A
Explanation
Resistance of coil of ammeter and shunt are parallel thus effective resistance Rp=480/25 Now Rp and 40.8 ohm are connected in series Current = V/R = 30/60 = 0.5 A Answer:(b)
Q.10
The charge flowing through a resistance R varies with time t as Q = at – bt2, where a and b are positive constants. The total heat produced in R is:
0%
a) (a³ R)/6b
0%
b) (a³ R)/3b
0%
c) (a³ R)/2b
0%
d) (a³ R)/b
Explanation
If I =0 then 0 =a-2bt t=a/2b Heat produced = I2Rt But current is time dependent thus we will integrate (a3 R)/6b Answer:(a)
Q.11
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is :- [ AIPMT 2015]
0%
a) 5 : 1
0%
b) 5 : 4
0%
c) 3 : 4
0%
d) 3 : 2
Explanation
10E1+10E2=50E1-50E2 E1+E2=5E1-5E2 4E1=6E2 E1/E2=3/2 Answer:(d)
Q.12
The potential difference (VA-VB) between the points A and B in the given figure is …[NEET II 2016]
0%
a) +6V
0%
b) +9V
0%
c) -3V
0%
d) +3V
Explanation
Applying Kirchhoff’s law VA-4 – 3-2=VB VB - VA = 9V Answer:(b)
Q.13
A filament bulb (500 W, 100 V) is be used in a 230V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500W. The value of R is … [NEET II 2016]
0%
a) 26 Ω
0%
b) 13 Ω
0%
c) 230 Ω
0%
d) 46 Ω
Explanation
Power = VI 500 = 100I I=5A Voltage across Resistance 130 = 5R R=26Ω Answer:(a)
Q.14
A potentiometer is an accurate and versatile deviceto make electrical measurements of E.M.F, becausethe method involves:
0%
a) Cells
0%
b) Potential gradients
0%
c) A condition of no current flow through thegalvanometer
0%
d) A combination
Explanation
When no current flows through the galvanometer, current from battery is zero, thus do not have effect of internal resistance of battery E= V- Ir as I = 0 E = V Answer:(c)
Q.15
The resistance of a wire is ‘R’ ohm. If it is meltedand stretched to ‘n’ times its original length, its newresistance will be
0%
a) nR
0%
b) R2
0%
c) n2R
0%
d) 2Rn
Explanation
Volume of the wire before melting = Al Volume of the wire of length = l1A1 Since volume is same Al = A1l1 Al = nA1l Thus A1 = A/n New resistance Answer:(c)
Q.16
This question has statement I and Statement II. Of the four choices given after the Statements,choose the one that best describes the two statements. Statement − I : Higher the range, greater is the resistance of ammeter. Statement − II : To increase the range of ammeter, additional shunt needs to be used across it.[ IIT Mains 2013]
0%
a) Statement - I is true, Statements - II is true, Statement - II is the correct explanation of Statements - I
0%
b) Statement - I is true, Statement - II is true, Statement - II is not the correct explanation ofStatement - I.
0%
c) Statement - I is true, Statement - II is false.
0%
d) Statement - I is false, Statement - II is true.
Explanation
I - False, II – True Answer:(d)
Q.17
than one correct answer Q341) Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by anew heater having two wires of the same material, each of length L anddiameter 2d. Theway these wires are connected is given in the options. How much time in minutes will ittake to raise the temperature of the same amount of water by 40 K?[ IIT Advance 2014]
0%
a) 4 if wires are in parallel
0%
b) 2 if wires are in series
0%
c) 1 if wires are in series
0%
d) 0.5 if wires are in parallel
Explanation
Amount of heat produced in given case H=0.5C40 Resistance of new wires Option a If connected in parallel resultant resistance = R/8 0.5×C×40=8×5C×t t= 0.5 minutes Option a is wrong but option d is correct Option b: Resistance in series = R/2 0.5C40=2×5C×t T= 2 minutes option b correct Option C is wrong Answer:(b, d)
Q.18
than one correct answer Q342) Two ideal batteries of emf V1 and V2 and three resistances R1, R2 and R3 are connected as shown in the figure. The current in resistance R2 would be zero if ..[IIT Advance 2014]
0%
c) V1 = 2V2 and 2R1 = 2R2 = R3
0%
d) 2V1= V2 and 2R1 = R2 = R3
0%
a) V₁ = V₂ and R₁ = R₂ = R₃
0%
b) V₁ = V₂ and R₁ = 2R2 = R₃
Explanation
iR1 + i1R2 = V1 If i1 = 0 then i= V1/R1…(i) i1R2 – (i – i1)R3 = -V2 If i1 = 0 then i= V2/R3…(ii) From (i) and (ii) Option “a” correct Option “b” correct Option c Option d correct Answer:(a, b, d)
Q.19
During an experiment with a metre bridge, the galvanometer shows a null point when thejockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure.The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is…[ IIT Advance 2014]
0%
a) 60 ± 0.15 Ω
0%
b) 135 ±0.56 Ω
0%
c) 60 ± 0.25 Ω
0%
d) 135 ± 0.23 Ω
Explanation
R= 60Ω Now error in measurement is least count Δl= 1mm = 0.1cm We will assure R2 have no error R = 60±0.25Ω Answer:(c)
Q.20
An infinite line charge of uniform electric charge density 位 lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity ε and electrical conductivity σ. The electrical conduction in the material follows Ohm鈥檚 law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
It may be considered as charging of capacitor and graph given in option 'a' depects the same However detail calculations are as follows Let initial charge density be λ0 Consider inner shell of thickness dr and radius r and unit height Electric filed produced by infinitely long linear charge distribution |dv| = Edr Now resistance of cylindrical element of unit length and radius r and thickness dr which will be our length. Here current flowing radially out ward., also height of cylinder is unit. Thus Area = 2πr Now I = dq/dt, Charge for unit length= 位, as charge density is reducing On integrating Now J = I/A Option a follows the conditions Answer:(a)
Q.21
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true?
0%
a) The temperature distribution over the filament is uniform
0%
b) The resistance over small sections of the filament decreases with time
0%
c) The filament emits more light at higher band of frequencies before it breaks up
0%
d) The filament consumes less electrical power towards the end of the life of the bulb
Explanation
a) If we assume resistivity and cross-section is uniform then temperature will be uniform But it is given that wire break up at random locations thus option a wrong b) is wrong as cross section of wire gets reduced resistance increases c) is correct because temperature is not uniform, and frequency ∝ T d) correct as resistance increases with decreasing thickness Answer:(c, d)
Q.22
Resistivity of iron is 1×10⁻⁷ ohm×metre. The resistance of the given wire of a particular thickness and length is one ohm. If the diameter and the length of wire both are doubled, the resistivity will be in in ohm ×metre
0%
a)1×10⁻⁷
0%
b) 2×10⁻⁷
0%
c)4×10⁻⁷
0%
d)8×10⁻⁷
Explanation
Answer: (a)
Q.23
A copper wire ( specific resistance 1.7×10⁻⁸ Ωm) has a mass per unit length of 10-7 kg/cm. What is the resistance of a wire 200 m length? ( Density of copper=8.9×103 kg/m3)
0%
a) 3 KΩ approx
0%
b) 0.3 mΩ approx
0%
c)30mΩ approx
0%
d)0.3Ω approx
Explanation
LEt d be the density of commerλ be the mass per unit length m be the mass of conductor Answer: (a)
Q.24
It is required to construct a coil having resistance of 35Ω from a wire 0.2mm diameter. If specific resistance of the material of wire is 1.10×10⁻⁶ ohm×metre, the length of the wire required is
0%
a) 2.2 m
0%
b) 1.0 m
0%
c) 49.0 m
0%
d) 3.5 m
Explanation
Answer:(b)
Q.25
64 cells, each of emf 2volt and internal resistance 2Ω are connected to supply a maximum current through an external resistance of 8×10⁻³ Ω. Then the cells must be connected in
0%
a)series only
0%
b) parallel only
0%
c)mixed series and parallel arrangement
0%
d)any of the above three combination
Explanation
Answer: (c)
Q.26
In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivity’s of Al and Fe are 2.7 × 10⁻⁸Ωm and 1.0 × 10⁻⁷Ωm, respectively. The electrical resistance between the two faces P and Q of the composite bar is … [ IIT Advance 2015]
0%
a) (2475/64)µΩ
0%
b) (1875/64) µΩ
0%
c) (1875/49)µΩ
0%
d) (2475/132)µΩ
Explanation
Answer:(b)
Q.27
A resistance wire connected in the left gap of a metre bridge balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3 :If the length of the resistance wire is 1.5 m, then the length of 1 Ω of the resistance wire is : [ NEET 2020]
0%
a) 1.0×10⁻² m
0%
b) 1.0×10⁻¹ m
0%
c) 1.5×10⁻¹ m
0%
d) 1.5×10⁻² m
Explanation
Length of wire of resistance R = 1.5 m, S = 10 Ω R = 15 Ω = 1.5 m 1 Ω = 1.0×10⁻¹ m Ans : (b)
Q.28
Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1mm diameter, then the drift velocity (approx) will be (density of copper=9 x 103 kg m-3 and atomic weight of copper=63) [AFMC 2009]
0%
a)0.1 mm/s
0%
b)0.2 mm/s
0%
c)0.3 mm/s
0%
d) 0.2 cm/s
Explanation
To calculate drift velocity we have to find electron number density ‘n’ firstIn 63 gm copper=6.022x1023 atoms ,Hence 9x103 Kg will contain On substituting the values we getIt is given for every atom makes one free electron, hence number of electrons per m3=no of atoms per m3[OR you may use formula for free electrons ]Here MO is molecular weight; W is weight of sample;NAAvogadro's number, p is valancy of element; d is density;Now Current I=neAVd ; Vd=I /neA here Vd is drift velocityCross sectional area of conductor A=πr2Given current I=1.1 A, charge on electron=1.6 x 10⁻¹⁶Substituting value in equation for Vdwe getVd=1x10⁻⁴ m/s=0.1mm/s Answer : (a)
Q.29
A nichrome wire 1m long and 1 mm2 in cross¬sectional area draws 4 ampere at 2 volt. The resistivity of nichrome is [AFMC 2009]
0%
a) 1x10⁻⁷Ωm
0%
b) 2 x10⁻⁷Ωm
0%
c) 4x10⁻⁷Ωm
0%
d) 5x10⁻⁷Ωm
Explanation
Resistance of wire=V/I=2/4=0.5 ΩHere a=1 mm2=10-6 m2 , l=1 m by substituting the values, we getΡ=5x10⁻⁷ Ω mAnswer: (d)
Q.30
If a negligible small current is passed through wire of length 15m and of resistance 5Ω having uniform cross-section of 6×10⁻⁷ m2, the coefficient of resistivity of material is [ CBSE PMT 1996]
0%
a)1×10⁻⁷ Ω-m
0%
b) 2×10⁻⁷ Ω-m
0%
c)3×10⁻⁷ Ω-m
0%
d)4×10⁻⁷ Ω-m
Explanation
Given length of wire (l)=15 mArea (A)=6×10⁻⁷ m2 Resistance (R)=5 Ω We know that resistance of the wire material R=ρl / A Answer: (b)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)