Two bulbs operating on standard voltage 110 V have resistances in the ratio 9:16 . The ratio of brightness of light from them is

Question

Two bulbs operating on standard voltage 110 V have resistances in the ratio 9:16 . The ratio of brightness of light from them is (A)  9:16  (B)  16:9  (C)  3:4  (D)  4:3

Tardigrade Admin · Accepted Answer

Power =(V2/R) As V is same for both bulbs, so, Power propto (1/R) so Intensity propto (1/R) ⇒ So, (I1/I2)=(R2/R1) =(16/9)

Q. Two bulbs operating on standard voltage $110 V$ have resistances in the ratio $9 : 16$ . The ratio of brightness of light from them is

$9 : 16$

$16 : 9$

$3 : 4$

$4 : 3$

Power $= \frac{V ^{2}}{R}$
As $V$ is same for both bulbs,
so, Power $\propto \frac{1}{R}$
so Intensity $\propto \frac{1}{R}$
$\Rightarrow$ So, $\frac{I _{1}}{I _{2}} = \frac{R _{2}}{R _{1}}$
$= \frac{16}{9}$

Q. Two bulbs operating on standard voltage 110V have resistances in the ratio 9:16 . The ratio of brightness of light from them is

9:16

16:9

3:4

4:3