JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 15)
A simple pendulum is being used to determine
th value of gravitational acceleration g at a
certain place. Th length of the pendulum is
25.0 cm and a stop watch with 1s resolution
measures the time taken for 40 oscillations to
be 50 s. The accuracy in g is :
4.40%
3.40%
2.40%
5.40%
Explanation
T = $$2\pi \sqrt {{l \over g}} $$
$$ \Rightarrow $$ $$g = {{4{\pi ^2}l} \over {{T^2}}}$$
$$ \Rightarrow $$ $${{\Delta g} \over g} = {{\Delta l} \over l} + 2{{\Delta T} \over T}$$
= $${{0.1} \over {25}} + 2{1 \over {50}}$$ = $${{11} \over {250}}$$
$$ \therefore $$ % accuracy = $${{11} \over {250}}$$ $$ \times $$ 100% = 4.40 %
$$ \Rightarrow $$ $$g = {{4{\pi ^2}l} \over {{T^2}}}$$
$$ \Rightarrow $$ $${{\Delta g} \over g} = {{\Delta l} \over l} + 2{{\Delta T} \over T}$$
= $${{0.1} \over {25}} + 2{1 \over {50}}$$ = $${{11} \over {250}}$$
$$ \therefore $$ % accuracy = $${{11} \over {250}}$$ $$ \times $$ 100% = 4.40 %
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