potential gradient in terms of resi...
Question
potential gradient in terms of resistivity of the potentiometer wire. Figure shows a long potentiometer wire AB having a constant potential gradient. The null points
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potential gradient in terms of resistivity of the potentiometer wire. Figure shows a long potentiometer wire ( A B ) having a constant potential gradient. The null points for the two primary cells of emfs ( mathrm{E}_{1} ) and ( mathrm{E}_{2} ) connected in the manner shown are obtained at a distance of ( 11=120 mathrm{cm} ) and ( 12=300 mathrm{cm} ) from the end ( mathrm{A} )
Determine: ( E_{1} / E_{2} ) and (ii) position of null point for the cell ( E_{1} ) only.

JEE/Engineering Exams
Physics
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potential gradient in terms of resistivity of the potentiometer wire. Figure shows a long potentiometer wire AB having a constant potential gradient. The null points for the two primary cells of emfs E_1 and E_2 connected in the manner shown are obtained at a distance of 1_1=120 cm and 1_2 = 300 cm from the end A Determine: E_1/E_2 and (ii) position of null point for the cell E_1 only.
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[
begin{array}{l}
varepsilon_{1}-varepsilon_{2}=k(120)
varepsilon_{1}+varepsilon_{2}=K(300)
end{array}
]
( Rightarrow frac{varepsilon_{1}-varepsilon_{2}}{varepsilon_{1}+varepsilon_{2}}=frac{120}{300} )
( Rightarrow frac{2 varepsilon_{1}}{-2 varepsilon_{2}}=frac{420}{-180} )
( Rightarrow mid frac{epsilon_{1}}{varepsilon_{2}}=frac{7}{3} )
( Rightarrow quad varepsilon_{2}=frac{3 varepsilon_{1}}{7} )
( epsilon_{1}-frac{3 varepsilon_{1}}{7}=k(120) )
( Rightarrow quad frac{4 G}{7}=k(120) )
( Leftrightarrow quad G=k(210) )
( Rightarrow ) Null pout for ( varepsilon_{1} ) cell oul ( y=210 mathrm{cm} )

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