Question

Q 2/180 The length and breadth of a rectangle are
measured to be 3.00 cm and 2.00 cm with the help of
Vernier calliper having least count 0.01 cm. The area of
the rectangle with the estimated error will be
(6.00 + 0.05) cm
(6.00 + 0.02) cm
(6.00 + 0.03) cm
(6.00 + 0.01) cm​

Answer 1

Answer

(6.00+0.01)cm

Explanation:

In general

since we know that

Area of rectangle =length x breadth

therefore

A(R)=3x2=6

Therefore estimated error is can be written as 6+0.01 CM.

I hope you got it...

Answer 2

Answer:

(6.00 + 0.05) cm Option 1 is right

Explanation:

Tip

• We utilise percent errors to demonstrate how large our errors are when we measure anything in an analysis process. When we have fewer percent errors, we are coming closer to the acceptable or original value.

Given

Length of rectangle=3.00 cm

Breadth of rectangle=2.00 cm

Find

The area of

the rectangle with the estimated error

Solution

Area of rectangle=l×b=3×2=6

Error=($(\frac{.01}{3} +\frac{.01}{2} )*6$

      =$(\frac{.03+.02}{6} )=.05$

Final Answer

Estimated error will be (6.00 + 0.05) cm

#SPJ3