Question

Solve the below given question with explaination also.

Answer 1

Solution :

⏭ Given:

✏ Length of pendulum = 100cm

✏ Accuracy in measurement of length = 1mm

✏ Time-period of oscillations = 2s

✏ Accracy in measurement of time = 0.1s

⏭ To Find:

✏ Percentage error in measurement of g

⏭ Formula derivation:

$\mapsto \sf \: T = 2\pi \sqrt{ \frac{L}{g} } \\ \\ \mapsto \sf \: {T}^{2} = 4 {\pi}^{2} ( \frac{L}{g} ) \\ \\ \mapsto \sf \: g \propto \: \frac{L}{ {T}^{2} } \\ \\ \therefore \underline{ \boxed{ \bold{\sf{ \purple{ \frac{\% \triangle{g}}{g} = (\frac{ \triangle{L}}{L} + \frac{2 \triangle{T}}{T}) \times 100 }}}}}$

⏭ Conversation:

✏ 1 mm = 0.1 cm

⏭ Calculation:

$\mapsto \sf \: \%\frac{ \triangle{g}}{g} = (\frac{0.1}{100} + \frac{2 \times 0.1}{200} ) \times 100 \\ \\ \mapsto \sf \: \%\frac{ \triangle{g}}{g} = \frac{10}{100} + 0.1 = 0.1 + 0.1 \\ \\ \mapsto \underline{ \boxed{ \bold{ \sf{ \pink{ \%\frac{ \triangle{g}}{g} = 0.2\%}}}}} \: \orange{ \bigstar}$