Experimental Physics

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54 of 70

Question 54 of 70

The resistance of the meter bridge AB in given figure is 4Ω4 \Omega. With a cell of emf ε=0.5 V\varepsilon=0.5 \mathrm{~V} and rheostat resistance Rh=2Ω\mathrm{R}_{\mathrm{h}}=2 \Omega the null point is obtained at some point J. When the cell is replaced by another one of emf ε=ε2\varepsilon=\varepsilon_{2} the same null point J\mathrm{J} is found for Rh=6Ω\mathrm{R}_{\mathrm{h}}=6 \Omega. The emf⁡ε2\operatorname{emf} \varepsilon_{2} is:

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Answer: (B) 0.3 V0.3 \mathrm{~V}

Given, Emf of cell, ε=0.5v\varepsilon=0.5 \mathrm{v} Rheostat resistance, Rh=2Ω\mathrm{R}_{\mathrm{h}}=2 \Omega Potential gradient is

dvdL=(62+4)×4L\frac{d v}{d L}=\left(\frac{6}{2+4}\right) \times \frac{4}{L}

Let null point be at ℓcm\ell \mathrm{cm} when cell of emfε=0.5v\mathrm{emf} \varepsilon=0.5 \mathrm{v} is used.

thus ε1=0.5 V=(62+4)×4 L×ℓ\varepsilon_{1}=0.5 \mathrm{~V}=\left(\frac{6}{2+4}\right) \times \frac{4}{\mathrm{~L}} \times \ell...(i)

For resistance Rh=6Ω\mathrm{R}_{\mathrm{h}}=6 \Omega new potential gradient is (64+6)×4L\left(\frac{6}{4+6}\right) \times \frac{4}{L} and at null point

(64+6)(4 L)×ℓ=ε2\left(\frac{6}{4+6}\right)\left(\frac{4}{\mathrm{~L}}\right) \times \ell=\varepsilon_{2}...(ii)

Dividing equation (i) by (ii) we get

0.5ε2=106\frac{0.5}{\varepsilon_{2}}=\frac{10}{6} thus ε2=0.3v\varepsilon_{2}=0.3 \mathrm{v}

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