JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 2)
In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10 $$\Omega $$ resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10) $$\Omega $$ such that the null point shifts back to its initial position is :
_11th_January_Evening_Slot_en_2_1.png)
40 $$\Omega $$
30 $$\Omega $$
20 $$\Omega $$
60 $$\Omega $$
Explanation
$${{{R_1}} \over {{R_2}}} = {2 \over 3}\,\,$$ . . .(i)
$${{{R_1} + 10} \over {{R_2}}} = 1 \Rightarrow {R_1} + 10 = {R_2}$$ . . .(ii)
$${{2{R_2}} \over 3} + 10 = {R_2}$$
$$10 = {{{R_2}} \over 3} \Rightarrow {R_2} = 30\Omega $$
& $${R_1} = 20\Omega $$
$${{{{30 \times R} \over {30 + R}}} \over {30}} = {2 \over 3}$$
$$R = 60\,\Omega $$
$${{{R_1} + 10} \over {{R_2}}} = 1 \Rightarrow {R_1} + 10 = {R_2}$$ . . .(ii)
$${{2{R_2}} \over 3} + 10 = {R_2}$$
$$10 = {{{R_2}} \over 3} \Rightarrow {R_2} = 30\Omega $$
& $${R_1} = 20\Omega $$
$${{{{30 \times R} \over {30 + R}}} \over {30}} = {2 \over 3}$$
$$R = 60\,\Omega $$
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