Question

The area of square is $9sqcms$ and the error in its is $0.02sq.cm$ The percentage error in the measurement of the length of the diagonal of the square is

• A. $2/9$
• B. $1/9$
• C. $4/9$
• D. $1/3$

Answer 1

The correct option is B $1/9$

The area of the square is, $A_1=9$ sq.m.

Error in the measurement of the area is $0.02$ sq.m.

Thus, the new area of the square is, $9+0.02=9.02$ sq.m.

Now, original area of square is given by,

$A_1=a^2$

$\therefore a^2=9$

$\therefore a=3$

Area of square with error is given by,

$A_2=a_2^2$

$\therefore 9.02=a_2^2$

$\therefore a_2=3.00333$

Length of diagonal in first case-

$d_1=\sqrt{2}a$

$\therefore d_1=3\sqrt{2}$

Length of diagonal in second case-

$d_2=3.0033\sqrt{2}$

Percentage error in the length of diagonal is,

$e=\frac{3.0033\sqrt{2}-3\sqrt{2}}{3\sqrt{2}}\times 100$

$\therefore e=\frac{0.0033\sqrt{2}}{3\sqrt{2}}\times 100$

$\therefore e=0.0011\times 100$

$\therefore e=0.11$

$\therefore e=\frac{1}{9}$