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Physics NEET MCQ
Quiz 12
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Q.1
A car battery has emf 12Volts and internal resistance 5×10⁻²Ω. If it draws 60 amp current, the terminal voltage of the battery will be [ CBSE PMT 2000]
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a) 15V
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b) 3 V
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c) 5 V
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d) 9 V
Explanation
E=V + Ir 12=V + 60×5×10⁻² 12=V + 3 V=9 Volts Answer: (d)
Q.2
A moving coil galvanometer of resistance 100Ω is used as an ammeter using resistance 0.1Ω. The maximum deflection current in the galvanometer is 100µA. Find the minimum current in the circuit so that the ammeter shows maximum deflection [ IIT 2005]
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a) 100.1 mA
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b) 1000.1 mA
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c)10.01 mA
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d)1.01 mA
Explanation
Here Ig=100×10⁻⁶ AG=100 ΩS=0.1ΩFrom formula I=Ig [ (G/S) - 1] I=100×10⁻⁶ [ (100/0.1) + 1]I=100.01 mAAnswer: (a)
Q.3
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10-divisions per milliampere and voltage sensitivity is 2 division per millivolt. In order that division reads 1 Volt, the resistance in ohm needed to be connected in series with the coil will be [ AIEEE 2005]
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a) 105
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b) 103
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c)9995
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d)99995
Explanation
Resistance of galvanometer G=Current sensitivity / Voltage sensitivity G=10/2=5Ω br/>for full deflection current Ig=150 10=15 mAVoltage to be measured=150V so that each division can read 1VForm formula R=(V/Ig) - G R=(150/ (15×10⁻³) - 5=9995Ω Answer:(c)
Q.4
In Bohr's model of hydrogen atom the electron moves around the nucleus in a circular orbit of radius 5×10⁻¹¹ metres. Its time period is 1.5×10⁻¹⁶ second. The current associated with the electron motion is [ MNR 1992]
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a) zero
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b) 1.6×10⁻¹⁹ A
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c) 0.17 A
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d) 1.07×10⁻³ A
Explanation
I=q/T I=1.6×10⁻¹⁹ / 1.5×10⁻¹⁶ I=1.07×10⁻³ A Answer: (d)
Q.5
In a neon gas discharge tube 2.9×1018 Ne+ ions are move to the right through a cross section of tube each second, while 1.2×1018 electrons move to the left in this time. The electronic charge is 1.6×10⁻¹⁹ coulomb. Then the net electric current in the tube is [ MPPMT 1999]
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a)1 amp to the right
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b) 0.66 amp to the right
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c)0.66 amp to the left
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d)zero
Explanation
positive charges are more and moving to right and negative charges moving left direction of current is taken as direction of positive charge or opposite to direction of electron thus current will be in right direction net charge moving right=(2.9×1018 + 1.2×1018)×1.6×10⁻¹⁹ Current=4.1×1018 ×1.6×10⁻¹⁹=0.656 C/ sec=0.66amp Answer: (b)
Q.6
A current density of 2.5Am-2 is found to exist in conductor when an electric field of 5×10⁻⁸ Vm-1 is applied across it. The resistivity of a conductor is .. [ GSEB]
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a) 1×10⁻⁸Ωm
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b) 2× 10⁻⁸ Ωm
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c) 0.5×10⁻⁸Ωm
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d) 12.5×10⁻⁸ Ωm
Explanation
We know that J=σE Here σ is conductivity 1 /σ=E/J resitivity ρ=E/J ρ=5×10⁻⁸ / 2.5=2×10⁻⁸ Ωm Answer: (b)
Q.7
At what temperature would the resistance of copper conductor be double its resistance at 0°C? Given α for copper=3.9×10⁻³ °C-1
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a) 256.4 °C
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b) 512.8°C
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c)100°C
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d)256.4K
Explanation
From formula R=Ro[ 1 + αt]Given R=2 Ro∴ 2 Ro=2 Ro [1 + 3.9×10⁻³ × t]2=1 + 3.9×10⁻³ × t t=1/3.9×10⁻³=265.4 °CAnswer: (a)
Q.8
In a region 1019 α-particles and 1019 protons move to the left while 1019 electrons move to the right per second./ The current is
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a) 3.2 amp towards left
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b) 6.4 amp towards left
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c) 9.6 amp towards left
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d) 6.4 amp towards right
Explanation
Direction of current is the direction of positive charge direction of electron is opposite but charge is also opposite thus charge of electrons should be added to charge of positive particle i=(1019 ×2e + 1019 ×e + 1019 ×e ). left i=4×1.6×10⁻¹⁹×1019e=6.4A, left Answer: (b)
Q.9
The belt of an electrostatic generator is 50 cm wide and travels at 30 m/sec. the belt carries charge into the sphere at a rate corresponding to 10-4 ampere. What is the surface density of charge on the belt?
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a)6.7×10⁻⁵ C/m²
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b) 6.7×10⁻⁶ C/m²
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c)6.7×10⁻⁷ C/m²
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d)6.7×10⁻⁸ C/m²
Explanation
Charge transferred in 1 sec=it=1×10⁻⁴ CArea of belt traversed in sec=length traversed in 1 sec × breadth Area of belt traversed in sec=30× 0.5 m2=m2 15 charge density σ=q / A=10-4 / 15=6.7×10⁻⁶ C/m2Answer: (b)
Q.10
The dimensions of a manganin ( ρ=4.4×10⁻⁷ ohm ×m) block are 1cm ×1 cm ×100 cm. The resistance between the opposite rectangular faces is
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a)4.4 ×10⁻⁷ ohm
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b) 4.4 ×10⁻³ ohm
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c)4.4 ×10⁻⁵ ohm
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d)4.4 ×10⁻¹ ohm
Explanation
R=ρl /A. for rectangular face l=1 cm and area=100cm2 substitute values to get R=4.4×10⁻⁷ ohmAnswer: (a)
Q.11
In the circuit shown below, the key is pressed at time t =Which of the following statement(s) is(are) true?
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a) The voltmeter displays - 5 V as soon as the key is pressed, and displays + 5 V after a long time
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b) The voltmeter will display 0 V at time t = ln 2 seconds
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c) The current in the ammeter becomes 1/e of the initial value after 1 second
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d) The current in the ammeter becomes zero after a long time
Explanation
40µF and 20µF capacitor will conduct and offer no resistance. Thus Potential Q will be +5 Volts and at P= 0V. Thus voltmeter will display VP – VQ = 0-5V = -5V. After long time current will not conduct through both capacitor and are fully charged, Voltage at Q = 0 and Voltage at P = 5V Since voltmeter is ideal thus have infinite resistance. No current will flow through ammeter option d is correct Option a correct . Charging of capacitor is given by Voltage across capacitor after time t τ = RC called as RC time constant For 40µF capacitor τ = 40×10⁻⁶×25×103 = 1s For 20µF capacitor τ = 20×10⁻⁶×50×103 = 1s Voltage across 40µF capacitor after time t=ln2 Thus voltage at point Q = 5- 2.5 = 2.5V And voltage across 20µF capacitor after time t=ln2 V2 = 2.5 V VP – 2.5V = 0 thus VP =2.5V As VP = VQ voltmeter reading is zero (Option b correct) Option c Total current I1 + I2 after 1s Initial value of current at t=0 , I1 = 0.2×103 + 0.1×103 = 0.3×103 A Thus current after 1 sec = 1/e of initial current Option c is correct Answer:(a, b, c, d)
Q.12
A beam of 16MeV deuterons ( charge=1.6×10⁻¹⁹ C) from a cyclotron falls on a copper block. The beam is equivalent to current of 15µA. At what rate do the deuterons strike the block?
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a)9.4 × 1013
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b) 9.4 × 1011
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c)9.4 × 109
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d)9.4 × 107
Explanation
charge on alpha particle=ei=ne / t For t=1 n=i /e=15×10⁻⁶/1.6×-19=9.375×1013 particle per secondAnswer: (a)
Q.13
A potentiometer wire has length 10 m and resistance 20 Ω. A 2.5V battery of negligible internal resistanceis connected across the wire with an 80Ω series resistance. The potential gradient on the wire will be:
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a) 2.5 × 10 ×10⁻⁴ V/m
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b) 0.62 × 10 ×10⁻⁴ V/mm
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c)1 × 10⁻⁵ V/mm
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d)5 × 10⁻⁵ V/mm
Explanation
Voltage gradiant=V/l=0.5V / 10m=5 × 10⁻⁵ V/mmAnswer: (d)
Q.14
In a potentiometer experiment, the balancing length is 8m when two cells E1 and E2 are joined in series. When the two cells connected in opposite the balancing length is 4 m. The ratio of the emf of two cells ( E1 / E2=is
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a)1:2
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b) 2:1
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c)1:3
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d)3:1
Explanation
(E1 + E2) ∝ 8and (E1 + E2) ∝ solve above equationsAnswer: (d)
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