back to directory
NEET PHYSICSMOVING CHARGES AND MAGNETISMEasy

Question

The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is:

A

40 \Omega

B

25 \Omega

C

250 \Omega

D

500 \Omega

Step-by-Step Solution

  1. Definitions:
  • Current Sensitivity (IsI_s) is the deflection per unit current: Is=ϕII_s = \frac{\phi}{I} .
  • Voltage Sensitivity (VsV_s) is the deflection per unit voltage: Vs=ϕVV_s = \frac{\phi}{V} .
  1. Relationship: According to Ohm's Law, V=IRV = IR, where RR is the resistance of the galvanometer. Substituting VV into the voltage sensitivity equation: Vs=ϕIR=1R(ϕI)=IsRV_s = \frac{\phi}{IR} = \frac{1}{R} \left( \frac{\phi}{I} \right) = \frac{I_s}{R} Rearranging for Resistance (RR): R=IsVsR = \frac{I_s}{V_s}
  2. Calculation:
  • Given Is=5 div/mA=5×103 div/AI_s = 5 \text{ div/mA} = 5 \times 10^3 \text{ div/A} (converting mA to A).
  • Given Vs=20 div/VV_s = 20 \text{ div/V}.
  • Substitute the values: R=5 div/mA20 div/V=520VmA=0.25 kΩ=250ΩR = \frac{5 \text{ div/mA}}{20 \text{ div/V}} = \frac{5}{20} \frac{\text{V}}{\text{mA}} = 0.25 \text{ k}\Omega = 250 \, \Omega Alternatively, using base units: R=500020=250ΩR = \frac{5000}{20} = 250 \, \Omega.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMcurrentsensitivitymovinggalvanometervoltage

More MOVING CHARGES AND MAGNETISM Questions

View all

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron:

A.will turn towards right of direction of motion
B.will turn towards left of direction of motion
C.speed will decrease
D.speed will increase
EasySolve

A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{-4} \text{ m}^2$ carries a current of $2.0 \text{ A}$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2} \text{ T}$ making an angle of $30^{\circ}$ with the axis of the solenoid. The torque on the solenoid will be

A.$3 \times 10^{-3} \text{ N m}$
B.$1.5 \times 10^{-3} \text{ N m}$
C.$1.5 \times 10^{-2} \text{ N m}$
D.$3 \times 10^{-2} \text{ N m}$
EasySolve

A metallic rod of mass per unit length of 0.5 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction of 0.25 T is acting on it in the vertical direction. What is the current flowing through the rod to keep it stationary?

A.7.14 A
B.5.98 A
C.14.76 A
D.11.32 A
MediumSolve

A voltmeter has a resistance of $G$ ohms and range $V$ volts. The value of resistance used in series to convert it into a voltmeter of range $nV$ volts is:

A.$nG$
B.$(n-1)G$
C.$\frac{G}{n}$
D.$\frac{G}{n-1}$
MediumSolve

A long solenoid of radius $1 \text{ mm}$ has $100$ turns per mm. If $1 \text{ A}$ current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:

A.$6.28 \times 10^{-4} \text{ T}$
B.$6.28 \times 10^{-2} \text{ T}$
C.$12.56 \times 10^{-2} \text{ T}$
D.$12.56 \times 10^{-4} \text{ T}$
EasySolve

A square loop ABCD carrying a current $i$ is placed near and coplanar with a long straight conductor XY carrying a current $I$. The net force on the loop will be:

A.$\frac{\mu_0 I i}{2\pi}$
B.$\frac{2\mu_0 I i L}{3\pi}$
C.$\frac{\mu_0 I i L}{2\pi}$
D.$\frac{2\mu_0 I i}{3\pi}$
HardSolve

An electron moving in a circular orbit of radius $r$ makes $n$ rotations per second. The magnetic field produced at the centre has magnitude:

A.$\frac{\mu_0 n e}{2\pi r}$
B.Zero
C.$\frac{n^2 e}{r}$
D.$\frac{\mu_0 n e}{2r}$
EasySolve

From Ampere's circuital law, for a long straight wire of circular cross-section carrying a steady current, the variation of the magnetic field inside and outside the region of the wire is:

A.A linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one for the outside region.
B.Uniform and remains constant for both regions.
C.A linearly increasing function of distance upto the boundary of the wire and then a linearly decreasing one for the outside region.
D.A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with 1/r dependence for the outside region.
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →