JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 1)
In a meter bridge experiment S is a standard
resistance. R is a resistance wire. It is found
that balancing length is $$l$$ = 25 cm. If R is
replaced by a wire of half length and half
diameter that of R of same material, then the
balancing distance $$l'$$ (in cm) will now
be________.
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Answer
40
Explanation
$${S \over R} = {{75} \over {25}}$$ = 3
R = $${{\rho L} \over A}$$ = $${{4\rho L} \over {\pi {d^2}}}$$
R' = $${{4\rho \left( {{L \over 2}} \right)} \over {\pi {{\left( {{d \over 2}} \right)}^2}}}$$ = 2R
Then $${S \over {R'}} = {{100 - l} \over l}$$
$$ \Rightarrow $$ $${{100 - l} \over l}$$ = $${S \over {2R}}$$ = $${3 \over 2}$$
$$ \Rightarrow $$ $$l$$ = 40 cm
R = $${{\rho L} \over A}$$ = $${{4\rho L} \over {\pi {d^2}}}$$
R' = $${{4\rho \left( {{L \over 2}} \right)} \over {\pi {{\left( {{d \over 2}} \right)}^2}}}$$ = 2R
Then $${S \over {R'}} = {{100 - l} \over l}$$
$$ \Rightarrow $$ $${{100 - l} \over l}$$ = $${S \over {2R}}$$ = $${3 \over 2}$$
$$ \Rightarrow $$ $$l$$ = 40 cm
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