
Assertion : If the current of a lamp decreases by 20%, the percentage decrease in the illumination of the lamp is 40%.
Reason: Illumination of the lamp is directly proportional to square of the current through a lamp.
A) Both (A) and (R) are true and (R) is the correct explanation of A.
B) Both (A) and (R) are true but (R) is not the correct explanation of A.
C) (A) is true but (R) is false
D) (A) is false but (R) is true
Answer
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Hint: Yes, the illumination of lamp is directly proportional to square of the current
through the lamp and the formula \[E = {I^{^2}}RT\] can be used to find the solution.
Complete step by step answer: The illumination from a lamp is measured as a form of energy coming from the lamp.
Using the formula for energy \[E = {I^{^2}}RT\], where I is the current flowing, R is the resistance, T is the time for which current flows.
We can find out the amount of illumination.
Let the initial and final current flowing be $I_1$ and $I_2$ respectively.
Initial Energy, ${E_1} = I_1^2RT$
Final Current,
$
{I_2} = {I_1} - 20\% {I_1} \\
\Rightarrow {I_2} = 80\% {I_1} = 0.8{I_1} \\
$
So, final energy, $E_2$
$
{E_2} = I_2^2RT = {\left( {0.8{I_1}} \right)^2}RT \\
\Rightarrow {E_2} = 0.64I_1^2RT \\
\Rightarrow {E_2} = 0.64{E_1} \\
$
So, the percentage decrease in illumination is given by $\dfrac{{{E_1} - {E_2}}}{{{E_1}}} \times 100 = \dfrac{{{E_1} - 0.64{E_1}}}{{{E_1}}} \times 100 = 36\% $
Therefore, the correct option is D), as the reason is correct but the assertion is false.
Note:Because illumination is directly proportional to the square of the current and not to twice the current, we can directly say that the assertion is false.
through the lamp and the formula \[E = {I^{^2}}RT\] can be used to find the solution.
Complete step by step answer: The illumination from a lamp is measured as a form of energy coming from the lamp.
Using the formula for energy \[E = {I^{^2}}RT\], where I is the current flowing, R is the resistance, T is the time for which current flows.
We can find out the amount of illumination.
Let the initial and final current flowing be $I_1$ and $I_2$ respectively.
Initial Energy, ${E_1} = I_1^2RT$
Final Current,
$
{I_2} = {I_1} - 20\% {I_1} \\
\Rightarrow {I_2} = 80\% {I_1} = 0.8{I_1} \\
$
So, final energy, $E_2$
$
{E_2} = I_2^2RT = {\left( {0.8{I_1}} \right)^2}RT \\
\Rightarrow {E_2} = 0.64I_1^2RT \\
\Rightarrow {E_2} = 0.64{E_1} \\
$
So, the percentage decrease in illumination is given by $\dfrac{{{E_1} - {E_2}}}{{{E_1}}} \times 100 = \dfrac{{{E_1} - 0.64{E_1}}}{{{E_1}}} \times 100 = 36\% $
Therefore, the correct option is D), as the reason is correct but the assertion is false.
Note:Because illumination is directly proportional to the square of the current and not to twice the current, we can directly say that the assertion is false.
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