In an experiment to find the emf of a cell us | vedclass

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(B) पोटेंशियोमीटर के सिद्धांत के अनुसार, $E \propto l$, अर्थात $E = kl$, जहाँ $k$ पोटेंशियोमीटर तार का विभव प्रवणता (potential gradient) है।प्रथम स्थि

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) पोटेंशियोमीटर के सिद्धांत के अनुसार, $E \propto l$, अर्थात $E = kl$, जहाँ $k$ पोटेंशियोमीटर तार का विभव प्रवणता (potential gradient) है।प्रथम स्थिति में: $E_1 = 1.5 	ext{ V}$ और $l_1 = 60 	ext{ cm}$। अतः, $1.5 = k 	imes 60$ (समीकरण $1$)।द्वितीय स्थिति में: नल पॉइंट की लंबाई $40 	ext{ cm}$ बढ़ जाती है, इसलिए $l_2 = 60 + 40 = 100 	ext{ cm}$।नए सेल का emf $E_2 = E$ है। अतः, $E = k 	imes 100$ (समीकरण $2$)।समीकरण $2$ को समीकरण $1$ से विभाजित करने पर: $E / 1.5 = 100 / 60 = 10 / 6 = 5 / 3$।अतः, $E = 1.5 	imes (5 / 3) = 0.5 	imes 5 = 2.5 	ext{ V}$।हमें दिया गया है कि $E = x/10 	ext{ V}$, इसलिए $x/10 = 2.5$, जिसका अर्थ है कि $x = 25$।

Similar Questions

In a potentiometer arrangement,a cell of emf $1.5 \ V$ gives a balance point at $150 \ cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $210 \ cm$,what is the emf of the second cell (in $V$)?

$A$ potentiometer is used to measure the potential difference between $A$ and $B$. The null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. The ratio $\frac{E_2}{E_1}$ is $\left(E_1 > E_2\right)$.

$A$ cell can be balanced against $110 \, cm$ and $100 \, cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \, \Omega$. Its internal resistance is ............... $\Omega$.

$A$ potentiometer is used for the comparison of $e.m.f.$ of two cells $E_1$ and $E_2$. For cell $E_1$,the null point is obtained at $20 \ cm$ and for $E_2$,the null point is obtained at $30 \ cm$. The ratio of their $e.m.f.$s will be:

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