JEE Main 2024 (8 Apr, Shift 1) Physics Q48: Elasticity

Question

Young's modulus is determined by the equation given by

Y=49000\frac{m}{l}\frac{\mathrm{dyne}}{\mathrm{cm}^2}

where

M

is the mass and

l

is the extension of wire used in the experiment. Now error in Young modules(

Y

) is estimated by taking data from

M

-

l

plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis respectively. If the value of

M

and

l

are 500 g and 2 cm respectively then percentage error of

Y

is :

Official answer

Option 3

NTA final key.

Accepted Answer

Option 3: $2\ \%$

Answer

$0.02\ \%$

Answer

$0.2\ \%$

Answer

$2\ \%$

Answer

$0.5\ \%$

JEE Main 8 April 2024, Shift 1, Physics Q48