Question

pls tell me the correct answer step by step ​

Answer 1

Answer:

$\boxed{\sf Maximum \ percentage \ error \ in \ measuring \ g = 12\%}$

Given:

$\sf g = 4 \pi^2 (\frac{l}{T^2})$

Percentage error in measuring l = 2%

Percentage error in measuring T = 5%

To find:

Maximum percentage error in measuring g

Explanation:

$\sf Percentage \: error \: in \: measuring \: g = \frac{ \Delta g}{g} \times 100 \%$

$\sf Percentage \: error \: in \: measuring \: l = \frac{ \Delta l}{l} \times 100 \%$

$= 2\%$

$\sf Percentage \: error \: in \: measuring \: T = \frac{ \Delta T}{T} \times 100 \%$

$= 5\%$

So,

$\sf \implies \frac{ \Delta g}{g} \times 100 \% = \frac{ \Delta l}{l} \times 100 \% + 2 \times \frac{ \Delta T}{T} \times 100 \%$

$\sf \implies \frac{ \Delta g}{g} \times 100 \% = 2 \% + 2 \times 5 \%$

$\sf \implies \frac{ \Delta g}{g} \times 100 \% = 2 \% + 10 \%$

$\sf \implies \frac{ \Delta g}{g} \times 100 \% = 12 \%$

Therefore,

Maximum percentage error in measuring g = 12 %

Answer 2

Answer:

12%

Explanation:

Please refer the attached image.