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एक आयताकार कमरे की लम्बाई और चौड़ाई क्रमश: $3.95 \pm 0.05 \,m$ एवं $3.05 \pm 0.05 \,m$ मापी गयी है. कमरे के फर्श का क्षेत्रफल ..................... $m^2$ होगा
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
