JEE Main 2021 (24 Feb, Shift 2) Physics Q1: The period of…

Question

The period of oscillation of a simple pendulum is

T = 2\pi\sqrt{\frac{L}{g}}

. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :

(1) $1.33\%$
(2) $1.03\%$
(3) $1.13\%$ Official answer
(4) $1.30\%$

Official answer

Option 3

NTA final key.

Accepted Answer

Option 3: $1.13\%$

Answer

$1.33\%$

Answer

$1.03\%$

Answer

$1.13\%$

Answer

$1.30\%$

JEE Main 24 February 2021, Shift 2, Physics Q1