A student determined Young's Modulus of elasticity using the formula Y = Mg L ^ 3 4 bd ^ 3 . The value of g is taken to be 9 . 8m s ^ - 2 without any significant error, his observations are as following. Physical

Correct answer: D. 0 . 0155 Y = Mg L ^ 3 4 bd ^ 3 (∆ Y)/(Y) = (∆ M)/(M) + 3(∆ l)/(1) + (∆ b)/(b) + 3(∆ d)/(d) + (∆ )/( ) ⇒ (∆ Y)/(Y) = 0 . 0005 + 3 × 0 . 001 + 0 . 0025 + 3 × 0 . 0025 + 0 . 002 = 0 . 0155

JEE Main 20211 SeptemberEvening Shift

Exam: JEE Main
Year: 2021
Date: 1 September
Shift: Evening Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2021 · 1 September · Evening Shift · Physics · Mathematics in Physics

A student determined Young's Modulus of elasticity using the...

A student determined Young's Modulus of elasticity using the formula $Y = \dfrac{Mg{L}^{3}}{4{bd}^{3}\delta} .$ The value of $g$ is taken to be $9 . 8m{s}^{- 2}$ without any significant error, his observations are as following.

Physical Quantity	Least count of the Equipment used for measurement	Observed Value
Mass $( M )$	$1g$	$2kg$
Length of bar $( L )$	$1mm$	$1m$
Breadth of bar $( b )$	$0 . 1mm$	$4cm$
Thickness of bar $( d )$	$0 . 01mm$	$0 . 4cm$
Depression $( \delta )$	$0 . 01mm$	$5mm$

Then the fractional error in the measurement of $Y$ is :

Options

A $0 . 155$
B $0 . 0083$
C $0 . 083$
D $0 . 0155$

Correct answer

D. $0 . 0155$

Step-by-step solution

$Y = \dfrac{Mg{L}^{3}}{4{bd}^{3}\delta}$

$\dfrac{∆ Y}{Y} = \dfrac{∆ M}{M} + 3\dfrac{∆ l}{1} + \dfrac{∆ b}{b} + 3\dfrac{∆ d}{d} + \dfrac{∆ \delta}{\delta} ⇒ \dfrac{∆ Y}{Y} = 0 . 0005 + 3 \times 0 . 001 + 0 . 0025 + 3 \times 0 . 0025 + 0 . 002 = 0 . 0155$

Practice Mathematics in Physics on Quantrex Academy →

More from Mathematics in Physics

The density ρ of a uniform cylinder is determined by measuring its mass m , length l and… JEE Main 2026 Four persons measure the length of a rod as 20.00 ~cm, 19.75 ~cm, 17.01 ~cm and 18.25 cm.… JEE Main 2026 In an experiment the values of two spring constants were measured as k_ 1 =(10 ± 0.2) N /… JEE Main 2026 Length, breadth and thickness of a strip having a uniform cross section are measured to… JEE Advanced 2025 A quantity Q is formulated as X^ -2 Y^ +(3)/(2) Z^ -(2)/(5) · X, Y and Z are independent… JEE Main 2025 A physical quantity C is related to four other quantities p, q, r and s as follows C=… JEE Main 2025 A physical quantity Q is related to four observables a, b, c, d as follows : Q= ab ^4 ~cd… JEE Main 2025 For an experimental expression y=(32.3 × 1125)/(27.4) , where all the digits are… JEE Main 2025 Full Mathematics in Physics list All JEE Main PYQs