1
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+1
-0

The length of a potentiometer wire is ' $L$ '. A cell of e.m.f. ' $E$ ' is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{\mathrm{L}}{3}$, then using the same cell null point is obtained at

A
$\frac{\mathrm{L}}{4}$
B
$\frac{\mathrm{L}}{3}$
C
$\frac{\mathrm{L}}{2}$
D
$\frac{3 L}{4}$
2
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+1
-0

Two cells of e.m.f.s $E_1$ and $E_2\left(E_1>E_2\right)$ are connected as shown in figure.

MHT CET 2025 23rd April Evening Shift Physics - Current Electricity Question 5 EnglishWhen the potentiometer is connected between A and B , the balancing length of the potentiometer wire is 3.60 m . On connecting the potentiometer between A and C , the balancing length is 0.90 m . The ratio $E_1 / E_2$ is

A
$5: 4$
B
$4: 3$
C
$3: 4$
D
$4: 5$
3
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+1
-0

The equivalent resistance of the following circuit when no current flows in the resistance of $5 \Omega$ is nearly

MHT CET 2025 23rd April Morning Shift Physics - Current Electricity Question 10 English
A
$13 \Omega$
B
$17 \Omega$
C
$19 \Omega$
D
$21 \Omega$
4
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+1
-0
In the following circuit, the current through $6 \Omega$ resistor isMHT CET 2025 23rd April Morning Shift Physics - Current Electricity Question 9 English
A
$\frac{1}{5} \mathrm{~A}$
B
$\frac{2}{5} \mathrm{~A}$
C
$\frac{1}{4} \mathrm{~A}$
D
$\frac{3}{4} \mathrm{~A}$
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