back to directory
NEET PHYSICSUNITS AND MEASUREMENTSEasy

Question

The determination of the value of acceleration due to gravity (gg) by simple pendulum method employs the formula, g=4π2LT2g = 4\pi^2 \frac{L}{T^2}. The expression for the relative error in the value of gg is:

A

Δgg=ΔLL+2(ΔTT)\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\left(\frac{\Delta T}{T}\right)

B

Δgg=4π2[ΔLL2ΔTT]\frac{\Delta g}{g} = 4\pi^2 \left[\frac{\Delta L}{L} - 2\frac{\Delta T}{T}\right]

C

Δgg=4π2[ΔLL+2ΔTT]\frac{\Delta g}{g} = 4\pi^2 \left[\frac{\Delta L}{L} + 2\frac{\Delta T}{T}\right]

D

Δgg=ΔLL2(ΔTT)\frac{\Delta g}{g} = \frac{\Delta L}{L} - 2\left(\frac{\Delta T}{T}\right)

Step-by-Step Solution

Given formula for acceleration due to gravity is: g=4π2LT2g = 4\pi^2 \frac{L}{T^2} In measuring physical quantities, to find the maximum possible relative error, the relative errors of the individual quantities are added, multiplied by their respective powers. Since 4π24\pi^2 is a constant, its error is zero. The relative error in gg is given by: Δgg=ΔLL+2ΔTT\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from UNITS AND MEASUREMENTS. 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.

PHYSICSUNITS AND MEASUREMENTSdeterminationaccelerationgravitysimplependulum

More UNITS AND MEASUREMENTS Questions

View all

In an experiment, four quantities $a$, $b$, $c$ and $d$ are measured with percentage error $1\%$, $2\%$, $3\%$ and $4\%$ respectively. Quantity $P$ is calculated as follows: $P=\frac{a^3b^2}{cd}$. Percentage error in $P$ is:

A.$10\%$
B.$7\%$
C.$4\%$
D.$14\%$
MediumSolve

Dimensional formula for volume elasticity is

A.M¹L⁻²T⁻²
B.M¹L⁻³T⁻²
C.M¹L²T⁻²
D.M¹L⁻¹T⁻²
EasySolve

The velocity $v$ of a particle at time $t$ is given by $v = at + \frac{b}{t+c}$, where $a$, $b$ and $c$ are constants. The dimensions of $a$, $b$ and $c$ are respectively:

A.$[L T^{-2}], [L]$ and $[T]$
B.$[L^2], [T]$ and $[LT^2]$
C.$[LT^2], [LT]$ and $[L]$
D.$[L], [LT]$ and $[T^2]$
MediumSolve

Which one has the dimensions different from the remaining three?

A.Power
B.Work
C.Torque
D.Energy
EasySolve

Taking into account the significant figures, what is the value of $9.99\text{ m} - 0.0099\text{ m}$?

A.$9.98\text{ m}$
B.$9.980\text{ m}$
C.$9.9\text{ m}$
D.$9.9801\text{ m}$
EasySolve

Of the following quantities, which one has dimensions different from the remaining three?

A.Energy per unit volume
B.Force per unit area
C.Product of voltage and charge per unit volume
D.Angular momentum per unit mass
MediumSolve

If the unit of length and force be increased four times, then the unit of energy is:

A.Increased 4 times
B.Increased 8 times
C.Increased 16 times
D.Decreased 16 times
EasySolve

Which of the following pairs of physical quantities has the same dimensions?

A.Work and power
B.Momentum and energy
C.Force and power
D.Work and energy
EasySolve

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 →