- Home
- Standard 11
- Physics
1.Units, Dimensions and Measurement
medium
The length and width of a rectangular room are measured to be $3.95 \pm 0.05 \,m$ and $3.05 \pm 0.05 \,m$, respectively. The area of the floor is .................... $m^2$
A$12.05 \pm 0.01$
B$12.05 \pm 0.005$
C$12.05 \pm 0.34$
D$12.05 \pm 0.40$
(KVPY-2016)
Solution
(c)
Area, $A=l \times b$
$=3.95 \times 3.05=12.05\, m ^2$
Now, $A=l \times b$
$\Rightarrow \frac{d A}{A}=\frac{d l}{l}+\frac{d b}{b}$
$\Rightarrow \Delta A=\left(\frac{\Delta l}{l}+\frac{\Delta b}{b}\right) \times A$
$=\left(\frac{0.05}{3.95}+\frac{0.05}{3.05}\right) \times 12.05$
$\approx 0.34$
So, area of floor is $A=12.05 \pm 0.34 \,m ^2$.
Area, $A=l \times b$
$=3.95 \times 3.05=12.05\, m ^2$
Now, $A=l \times b$
$\Rightarrow \frac{d A}{A}=\frac{d l}{l}+\frac{d b}{b}$
$\Rightarrow \Delta A=\left(\frac{\Delta l}{l}+\frac{\Delta b}{b}\right) \times A$
$=\left(\frac{0.05}{3.95}+\frac{0.05}{3.05}\right) \times 12.05$
$\approx 0.34$
So, area of floor is $A=12.05 \pm 0.34 \,m ^2$.
Standard 11
Physics
