NCERT Exemplar Q & A

Hint

(b) Key concept: Current per unit area (taken normal to the current), $I / A$, is called current density and is denoted by $\vec{J}$.
The SI unit of the current density are $\mathrm{A} / \mathrm{m}^2$. The current density is also directed along $E$ and which is also a vector quantity and the relationship is given by
$\vec{J}=\sigma \vec{E}=\frac{\vec{E}}{\rho}$
where $\sigma=$ conductivity and $\rho=$ resistivity or specific resistance of the substance.
The $\vec{J}$ changes due to the electric field produced by charges accumulated on the surface of wire.

Hint

(a) The equivalent emf of this combination is given by
$\varepsilon_{\text {eq }}=\frac{\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_1}{r_2}}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}=\frac{\varepsilon_1\left(\frac{1}{r_1}+\frac{\varepsilon_2 / \varepsilon_1}{r_2}\right)}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}=\frac{\varepsilon_2\left(\frac{\varepsilon_1 / \varepsilon_2}{r_1}+\frac{1}{r_2}\right)}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}$
$\text { As } \frac{\varepsilon_2}{\varepsilon_1}>1 \Rightarrow \Rightarrow \frac{\left(\frac{1}{r_1}+\frac{\varepsilon_2 / \varepsilon_1}{r_2}\right)}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}>1 \text { or } \varepsilon_{\mathrm{eq}}>\varepsilon_1 \text { also } \frac{\varepsilon_1}{\varepsilon_2}<1 \Rightarrow \frac{\left(\frac{\varepsilon_1 / \varepsilon_2}{r_1}+\frac{1}{r_2}\right)}{\left(\frac{1}{r_1}+\frac{1}{r_2}\right)}<1 \text { or } \varepsilon_{\mathrm{eq}}<\varepsilon_2$
$\begin{aligned} &\text { Hence }\\ &\varepsilon_1<\varepsilon_{\mathrm{eq}}<\varepsilon_2 . \end{aligned}$

Hint

(c) The percentage error in $R$ can be minimised by adjusting the balance point near the middle of the bridge, i.e., when $l_1$ is close to 50 cm . This requires a suitable choice of $S$.
Since,
$\frac{R}{S}=\frac{R l_1}{R\left(100-l_1\right)}=\frac{l_1}{100-l_4}$
Since here, $R: S:: 2.9: 97.1$ imply that the $S$ is nearly 33 times to that of $R$. In orded to make this ratio $1: 1$, it is necessary to reduce the value of $S$ nearly $\frac{1}{33}$ times i.e., nearly $3 \Omega$.

Hint

(a) The resistance of wire is given by
$R=\rho \frac{l}{A}$
For greater value of $R, l$ must be higher and $A$ should be lower and it is possible only when the battery is connected across $1 \mathrm{~cm} \times\left(\frac{1}{2}\right) \mathrm{cm}$ (area of cross-section $A$ ).

Hint

(a) The relationship between current and drift speed is given by
$I=n e A v_d$
Here, $I$ is the current and $v_d$ is the drift velocity.
So, $I \propto v_d$
Thus, only drift velocity determines the current in a conductor.

Hint

(b, d)
Kirchhoff’s junction rule is also known as Kirchhoff’s current law which states that the algebraic sum of the currents flowing towards any point in an electric network is zero. i.e., charges are conserved in an electric network.
So, Kirchhoff’s junction rule is the reflection of conservation of charge.

Hint

(a, d)
Here, the potential drop is taking place across $A B$ and $r$. Since the equivalent resistance of parallel combination of $R$ and $R^{\prime}$ is always less than $R$, therefore $I \geq \frac{V}{r+R}$ always.
Note In parallel combination of resistances, the equivalent resistance is smaller than smallest resistance present in combination.

Hint

(a, b) Resistivity is the intrinsic property of the substance.
For a metallic conductor, resistivity is given by
$\rho=\frac{m}{n e^2 \tau}$
where $n$ is the number of charge carriers per unit volume (number density) which can change with temperature $T$ and $\tau$ is relaxation time (time interval between two successive collisions) which decreases with the increase of temperature $\left(T \propto \frac{1}{\tau}\right)$.

Hint

(b, c)
Given, for first student, $R_2=10 \Omega, R_1=5 \Omega, R_3=5 \Omega$
For second student, $R_1=500 \Omega, R_3=5 \Omega$
Now, according to Wheatstone bridge rule,
$\frac{R_2}{R}=\frac{R_1}{R_3} \Rightarrow R=R_3 \times \frac{R_2}{R_1} \dots(1)$
Now putting all the values in Eq. (i), we get $R=10 \Omega$ for both students. Thus, we can analyse that the Wheatstone bridge is most sensitive and accurate if resistances are of same value.
Thus, the errors of measurement of the two students depend on the accuracy and sensitivity of the bridge, which inturn depends on the accuracy with which $R_2$ and $R_1$ can be measured.
When $R_2$ and $R_1$ are larger, the currents through the arms of bridge is very weak. This can make the determination of null point accurately more difficult.

Hint

(a, c) Key concept: Meter bridge: In case of meter bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point $B$, bridge is balanced. If in balanced position of bridge $A B=l, B C=(100-l)$ so that $Q / P=(100-l) / l$. Also $P / Q=R / S=>S=\frac{100-l}{l} \times R$

When there is no deflection in galvanometer there is no current across the galvanometer, then points B and D are at same potential. That point at which galvanometer shows no deflection is called null point, then potential at B and neutral point D are same. When the jockey contacts a point on the meter wire to the right of D , the potential drop across AD is more than potential drop across AB, which brings the potential of point D less than that of B, hence current flows from B to D in the galvanometer wire.