1. ⇒ (MHT CET 2023 13th May Morning Shift )

In a given meter bridge, the current flowing through $40 Ω$ resistor is

A. $I_{2} + I_{g}$

B. $I_{g}$

C. $I_{2} - I_{g}$

D. $I_{2}$

Correct Option is (D)

As the bridge is balanced, no current will flow through the galvanometer. Hence, the current through the $40 Ω$ resistor will be $I_{2}$ .

2. ⇒ (MHT CET 2023 12th May Evening Shift )

Two resistance $X$ and $Y$ are connected in the two gaps of a meterbridge and the null points is obtained at $20 cm$ from zero end. When the resistance of $20 Ω$ is connected in series with the smaller of the two resistance $X$ and $Y$ , the null point shifts to $40 cm$ from left end. The value of smaller resistance in ohm is

A. 6

B. 9

C. 12

D. 15

Correct Option is (C)

For a meterbridge, $\frac{X}{Y} = \frac{l}{100 - l}$

In the first case, $l = 20 cm$

$\begin{aligned} \frac{X}{Y} & = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4} \\ ∴ 4 X & = Y \end{aligned}$

In the second case, $l^{'} = 40 cm$

$\begin{array}{ll} ∴ & \frac{X^{'}}{Y^{'}} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \\ But, X^{'} = X + 20 and Y^{'} = Y \\ ∴ & \frac{X + 20}{Y} = \frac{2}{3} \\ ∴ & \frac{X + 20}{4 X} = \frac{2}{3} \\ ∴ & 8 X = 3 X + 60 \\ ∴ & X = 12 Ω \end{array}$

3. ⇒ (MHT CET 2023 11th May Evening Shift )

Potential difference between the points P and Q is nearly

MHT CET 2023 11th May Evening Shift Physics - Current Electricity Question 14 English

A. 17 V

B. 14 V

C. 12 V

D. 8 V

Correct Option is (A)

Let total current be denoted as I. The given circuit is a Wheatstone bridge.

$\begin{aligned} \Rightarrow R_{1} = 6 + 3 = 9 Ω \\ \Rightarrow R_{2} = 8 + 4 = 12 Ω \end{aligned}$

According to $KCL$ , the current will get divided into two parts $I_{1}$ and $I_{2}$

$∴ I_{1} = \frac{R_{2} I}{(R_{1} + R_{2})}$

Substituting the values,

$\begin{aligned} I_{1} = \frac{12}{9 + 12} \times 5 \\ I_{1} = 2.85 A \end{aligned}$

Potential difference between $P$ and $Q$ is

$\begin{aligned} V = I_{1} R \\ V = 2.85 \times 6 \\ V = 17 V \end{aligned}$

4. ⇒ (MHT CET 2023 11th May Morning Shift )

Potential difference between the points P and Q is nearly

A. 6V

B. 8V

C. 17V

D. 21V

Correct Option is (C)

As, the resistors across each branch are in series.

$\begin{aligned} R_{1} = 6 + 3 = 9 Ω \\ R_{2} = 8 + 4 = 12 Ω \end{aligned}$

According to $KCL$ , the current (i) will get divided into two parts $I_{1}$ and $I_{2}$

$I_{1} = \frac{R_{2}}{(R_{1} + R_{2}) i} = \frac{12}{9 + 12} \times 5 = 2.86 A$

Potential difference between $P$ and $Q$ is $V = I_{1} R = 2.86 \times 6 = 17.14 V$

5. ⇒ (MHT CET 2023 10th May Evening Shift )

In meter bridge experiment, null point was obtained at a distance ' $l$ ' from left end. The values of resistances in the left and right gap are doubled and then interchanged. The new position of the null point is

A. $(100 - l)$

B. $(100 - 2 l)$

C. $(100 - 3 l)$

D. $(100 - \frac{l}{2})$

Correct Option is (A)

For a Metre bridge,

$\frac{X}{R} = \frac{l_{X}}{l_{R}}$

After doubling and interchanging the left and right gap resistances, the ratio of the resistances do not change.

This means the null point does not change.

$∴$ The new position will remain at $(100 - l)$ .

⇒ Current Electricity