Mathematics in Physics

JEE Main level, single correct, medium

114 of 121

Question 114 of 121

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 :

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Answer: (D) 2.7%

T=2πLg\text{T}=2\pi \sqrt{\dfrac{\text{L}}{\text{g}}}

∴T2=4π2Lg∴{\text{T}}^{2}=4{\pi }^{2}\dfrac{\text{L}}{\text{g}}

∴g=4π2LT2∴\text{g}=4{\pi }^{2}\dfrac{\text{L}}{{\text{T}}^{2}}

% Error in g =(ΔLL0×100)+2(ΔTT0×100)=\left(\dfrac{\Delta \text{L}}{{\text{L}}_{0}}\times 100\right)+2\left(\dfrac{\Delta \text{T}}{{\text{T}}_{0}}\times 100\right)

=(1×10−30.20×100)+2{190×100}=\left(\dfrac{1\times 1{0}^{-3}}{\text{0.20}}\times 100\right)+2\left\{\dfrac{1}{90}\times 100\right\}

=0.5+2(109)=\text{0.5}+2\left(\dfrac{10}{9}\right)

=0.5+209=\text{0.5}+\dfrac{20}{9}

=0.5+2.22=\text{0.5}+\text{2.22}

= 2.7%

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