A cell is connected in the secondary circuit of a potentiometer and a balance point is obtained at 84 cm When the cell is shunted by a 5 - resistor- the balance point changes to 70 cm- If the 5 - resistor is replaced with 3 - resistor then the balance point will be at

R=R [ l_1 l_2/l_2 ]=R^1 [ l_1 l^1_2/l^1_2 ] 5 [ 84 70/70 ]=3 [ 84 l^1_2/l^1_2 ] l^1_2 × 14/14× 3=84 l^1_2 l^1_2+l^1_2/3=84 4l^1_2=84× 3 l^1_2=63cm ...

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Byju's Answer

Standard XII

Physics

Ammeter

A cell is con...

Question

A cell is connected in the secondary circuit of a potentiometer and a balance point is obtained at 84 cm When the cell is shunted by a $5 Ω$ resistor, the balance point changes to 70 cm. If the $5 Ω$ resistor is replaced with $3 Ω$ resistor then the balance point will be at

53 cm

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63 cm

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42 cm

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112 cm

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Solution

The correct option is B 63 cm
$r = R [\frac{l_{1} - l_{2}}{l_{2}}] = R^{1} [\frac{l_{1} - l_{2}^{1}}{l_{2}^{1}}] 5 [\frac{84 - 70}{70}] = 3 [\frac{84 - l_{2}^{1}}{l_{2}^{1}}] \frac{l_{2}^{1} \times 14}{14 \times 3} = 84 - l_{2}^{1} l_{2}^{1} + \frac{l_{2}^{1}}{3} = 84 4 l_{2}^{1} = 84 \times 3 l_{2}^{1} = 63 c m$

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A cell is connected in the secondary circuit of a potentiometer and a balance point is obtained at 84 cm When the cell is shunted by a 5Ω resistor, the balance point changes to 70 cm. If the 5Ω resistor is replaced with 3Ω resistor then the balance point will be at

The correct option is B 63 cmr=R[l1−l2l2]=R1[l1−l12l12]5[84−7070]=3[84−l12l12]l12×1414×3=84−l12l12+l123=844l12=84×3l12=63cm

A cell is connected in the secondary circuit of a potentiometer and a balance point is obtained at 84 cm When the cell is shunted by a $5 Ω$ resistor, the balance point changes to 70 cm. If the $5 Ω$ resistor is replaced with $3 Ω$ resistor then the balance point will be at