Class 12th

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New answer posted

4 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

4 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

4 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

4 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

I010=I0cos? 2? \frac {I_0} {10} = I_0 \cos^2 \theta cos? ? =110=0.31<12which is 0.707\cos \theta = \frac {1} {\sqrt {10} = 0.31 < \frac {1} {\sqrt {2} \quad \text {which is 0.707}

So,

? >45? and90? ? ? <45? \theta > 45^\circ \quad \text {and} \quad 90^\circ - \theta < 45^\circ

so only one option is correct i.e. 18.4°

Angle rotated should be

=90? ? 71.6? =18.4? = 90^\circ - 71.6^\circ = 18.4^\circ

Answer: (A) 18.4°

New answer posted

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R
Raj Pandey

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Raj Pandey

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution
 

 

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Raj Pandey

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A
alok kumar singh

Contributor-Level 10

 Heat generated in the resistance
H = i²RT
H? = 500 = (1.5)²R (20) ⇒ R = 500/45 = 11.11 Ω
H? = (3)²R (20) = 9 * (500/45) * 20 = 2000 J.

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