JEE Main 20149 AprilOnline
- Exam
- JEE Main
- Year
- 2014
- Date
- 9 April
- Mode
- Online
- Subject
- Physics
- Chapter
- Mathematics in Physics
JEE Main 2014 · 9 April · Online · Physics · Mathematics in Physics
An experiment is performed to obtain the value of acceleration due...
An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :
Options
- A4.4%
- B2.27%
- C1.7%
- D2.7%
Correct answer
D. 2.7%
Step-by-step solution
\text{T} = 2 π \sqrt{\dfrac{\text{L}}{\text{g}}}
∴ {\text{T}}^{2} = 4 {π}^{2} \dfrac{\text{L}}{\text{g}}
∴ \text{g} = 4 {π}^{2} \dfrac{\text{L}}{{\text{T}}^{2}}
% Error in g = \left(\dfrac{Δ \text{L}}{{\text{L}}_{0}} × 1 0 0\right) + 2 \left(\dfrac{Δ \text{T}}{{\text{T}}_{0}} × 1 0 0\right)
= 2.7%
∴ {\text{T}}^{2} = 4 {π}^{2} \dfrac{\text{L}}{\text{g}}
∴ \text{g} = 4 {π}^{2} \dfrac{\text{L}}{{\text{T}}^{2}}
% Error in g = \left(\dfrac{Δ \text{L}}{{\text{L}}_{0}} × 1 0 0\right) + 2 \left(\dfrac{Δ \text{T}}{{\text{T}}_{0}} × 1 0 0\right)
= 2.7%