JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 11)
In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the variation $${{dR} \over {d\ell }}$$ of its resistance R with length $$\ell $$ is $${{dR} \over {d\ell }}$$ $$ \propto $$ $${1 \over {\sqrt \ell }}$$. Two equal resistances are connected as shown in the figure. The
galvanometer has zero deflection when the jockey is at point P. What is the length AP ?
_12th_January_Morning_Slot_en_11_1.png)
0.3 m
0.25 m
0.35 m
0.2 m
Explanation
For the given wire :
dR = C $${{d\ell } \over {\sqrt \ell }}$$,
where C = constant.
Let resistance of part
AP is R1 and PB is R2
$$ \therefore $$ $${{R'} \over {R'}} = {{{R_1}} \over {{R_2}}}$$ or R1 = R2
By balanced WSB concept.
Now $$\int {dR = c\int {{{d\ell } \over {\sqrt \ell }}} } $$
$$ \therefore $$ R1 = C$$\int\limits_0^\ell {{\ell ^{ - 1/2}}d\ell } $$ = C.2.$${\sqrt \ell }$$
R2 = C$$\int\limits_\ell ^1 {{\ell ^{ - 1/2}}d\ell } $$ = C.(2 $$-$$ 2$${\sqrt \ell }$$)
Putting R1 = R2
C2$${\sqrt \ell }$$ = C(2 $$-$$ 2$${\sqrt \ell }$$)
$$ \therefore $$ 2$${\sqrt \ell }$$ = 1
$${\sqrt \ell }$$ = $${1 \over 2}$$
i.e. $$\ell $$ = $${1 \over 4}$$ m $$ \Rightarrow $$ 0.25 m
dR = C $${{d\ell } \over {\sqrt \ell }}$$,
where C = constant.
Let resistance of part
AP is R1 and PB is R2
$$ \therefore $$ $${{R'} \over {R'}} = {{{R_1}} \over {{R_2}}}$$ or R1 = R2
By balanced WSB concept.
Now $$\int {dR = c\int {{{d\ell } \over {\sqrt \ell }}} } $$
$$ \therefore $$ R1 = C$$\int\limits_0^\ell {{\ell ^{ - 1/2}}d\ell } $$ = C.2.$${\sqrt \ell }$$
R2 = C$$\int\limits_\ell ^1 {{\ell ^{ - 1/2}}d\ell } $$ = C.(2 $$-$$ 2$${\sqrt \ell }$$)
Putting R1 = R2
C2$${\sqrt \ell }$$ = C(2 $$-$$ 2$${\sqrt \ell }$$)
$$ \therefore $$ 2$${\sqrt \ell }$$ = 1
$${\sqrt \ell }$$ = $${1 \over 2}$$
i.e. $$\ell $$ = $${1 \over 4}$$ m $$ \Rightarrow $$ 0.25 m
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