Question

ANSWER FAST....
The length and breadth of a field are measured as 1500±0.5cm and
1000±0.1cm respectively. Calculate the area of the field within the
error limits.

Answer 1

Answer:

=2×(L×B+B×T+T×L)

∴A=2×(4.234×1.005+1.005×0.0201+0.0201×4.234)

∴A=2×(4.2552+0.0202+0.0851)

∴A=8.721m

2

∴A=8.721m

2

to correct significant digits

V=L×B×T

∴V=4.234×1.005×0.0201

∴V=0.0855m

3

to correct significant digits

Explanation:

Mark as branily

Answer 2

★ Given :-

• Length = 1500 ± 0.5 cm

• Breath = 1000 ± 0.1 cm

★ To Find ⇒

• Area of Field with Error Limits.

★ Solution ⇒

$\sf\bold\green{For\:Value\:⇒}$

Area of Field (A) = L × b

⇒ 1500 × 1000

⇒ 1500000 cm²

$\sf\bold\blue{Now,}$

$\sf\bold\green{For\:Error\:⇒}$

$\sf\pink{\dfrac{∆A}{A} = \dfrac{∆L}{L} + \dfrac{∆b}{b}}$

⇒$\sf{\dfrac{∆A}{1500000} = \dfrac{0.5}{1500} + \dfrac{0.1}{1000}}$

⇒$\sf{\dfrac{∆A}{1500000} = \dfrac{5}{15000} + \dfrac{1}{10000}}$

⇒$\sf{\dfrac{∆A}{1500000} = \dfrac{10}{30000} + \dfrac{3}{30000}}$

⇒$\sf{\dfrac{∆A}{1500000} = \dfrac{10 + 3}{30000}}$

$\sf{∆A = \dfrac{1500000 × 13}{30000}}$

$\sf{∆A = 50 × 13}$

$\sf\bold\pink{\fbox{∆A = 650\:cm}}$

$\sf\bold\green{Therefore,}$

Area of Field with error = 1500000 ± 650 cm