1
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
 

Two identical capacitors A and B are connected in series to a battery of E.M.F., 'E'. Capacitor B contains a slab of dielectric constant $\mathrm{K} . \mathrm{Q}_{\mathrm{A}}$ and $\mathrm{Q}_{\mathrm{B}}$ are the charges stored in A and B . When the dielectric slab is removed, the corresponding charges are $\mathrm{Q}_{\mathrm{A}}^{\prime}$ and $\mathrm{Q}_{\mathrm{B}}^{\prime}$. Then

A
$\mathrm{\frac{Q_A^{\prime}}{Q_A}=\frac{K}{2}}$
B
$\frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2}$
C
$\frac{\mathrm{Q}_{\mathrm{A}}^{\prime}}{\mathrm{Q}_{\mathrm{A}}}=\frac{\mathrm{K}+1}{\mathrm{~K}}$
D
$\frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}}$
2
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference 6 V . Another parallel combination of $\mathrm{n}_2$ capacitors, each of value $\mathrm{C}_2$ is charged by a source of potential difference 2 V . Total energy of both the combinations is same. The value of $\mathrm{C}_2$ in terms of $\mathrm{C}_1$ is

A
$\frac{3 C_1}{n_1 n_2}$
B
$\frac{9 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
C
$\frac{3 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
D
$\frac{9 C_1}{n_1 n_2}$
3
MHT CET 2024 15th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In the circuit shown in the following figure, the potential difference cross $3 \mu \mathrm{~F}$ capacitor is

MHT CET 2024 15th May Evening Shift Physics - Capacitor Question 19 English

A
4 V
B
6 V
C
10 V
D
16 V
4
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Three condensers of capacities ' $\mathrm{C}_1$ ', ' $\mathrm{C}_2$ ', ' $\mathrm{C}_3$ ' are connected in series with a source of e.m.f. ' $V$ '. The potentials across the three condensers are in the ratio

A
$1: 1: 1$
B
$\mathrm{C}_1: \mathrm{C}_2: \mathrm{C}_3$
C
$\mathrm{C}_1^2: \mathrm{C}_2^2: \mathrm{C}_3^2$
D
$\frac{1}{\mathrm{C}_1}: \frac{1}{\mathrm{C}_2}: \frac{1}{\mathrm{C}_3}$
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