MHT-CET PYQ’s 2024 - CURRENT ELECTRICITY

CURRENT ELECTRICITY

1. The figure shows currents in a part of electric circuit. Then current I is:

[MHT-CET 2024, May 15, Shift 2]

• (1) 3.5 A
• (2) 1.5 A
• (3) 4 A
• (4) 2.5 A

2.  In the following electrical network, the value of I is:

[MHT-CET 2024, May 11, Shift 1]

 

• (1) 1 A
• (2) 2 A
• (3) 3 A
• (4) 4 A

3.  In the following circuit, a power of 50 watt is absorbed in the section AB of the circuit. The value of resistance ‘x’ is:

[MHT-CET 2024, May 3, Shift 1]

 

• (1) 10 Ω
• (2) 8 Ω
• (3) 6 Ω
• (4) 4 Ω

4.  Kirchhoff’s second law is based on the law of conservation of:

[MHT-CET 2024, May 11, Shift 2]

• (1) charge
• (2) energy
• (3) momentum
• (4) inter conversion of mass into energy

5. The potential difference (V_A - V_B ) between the points ‘A’ and ‘B’ in the given part of the circuit is:

[MHT-CET 2024, May 10, Shift 2]

 

• (1) –3 V
• (2) 3 V
• (3) 6 V
• (4) 9 V

6. The potential difference (V_A - V_B ) between the points ‘A’ and ‘B’ in the given part of the circuit:

[MHT-CET 2024, May 16, Shift 1]

 

• (1) –13 V
• (2) 13 V
• (3) –23 V
• (4) 23 V

 

7.  In the given circuit, current flowing through the circuit is:

[MHT-CET 2024, May 10, Shift 1]

 

• (1) 2 A
• (2) 3 A
• (3) 4 A
• (4) 5 A

8. In the following circuit, the current I₃  is:

[MHT-CET 2024, May 2, Shift 2]

 

• (1) 5 A
• (2) 3 A
• (3) –3 A
• (4) –\( \frac{5}{6} \) A

9. In a Wheatstone’s bridge, the resistances in four arms are as shown in the figure. The balancing condition of the bridge is:

[MHT-CET 2024, May 4, Shift 2]

 

• (1) \( \frac{P}{Q} = \frac{R}{S_1 + S_2} \)
• (2) \( \frac{P}{Q} = \frac{R(S_1 S_2)}{S_1 + S_2} \)
• (3) \( \frac{P}{Q} = \frac{R(S_1 + S_2)}{2 S_1 S_2} \)
• (4) \( \frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2} \)

10. The current drawn from the battery in the given network is (Internal resistance of the battery is negligible):

[MHT-CET 2024, May 11, Shift 1]

 

• (1) 1.2 A
• (2) 2.4 A
• (3) 4 A
• (4) 4.8 A

11. Potential difference between the points A and B is nearly:

[MHT-CET 2024, May 9, Shift 2]

 

 

• (1) 10 V
• (2) 14 V
• (3) 18 V
• (4) 20 V

12. The resistances in the left and right gap of a metre bridge are 40 Ω and 60 Ω respectively. When the bridge is balanced, the distance of the null point from the centre of the wire towards left is:

[MHT-CET 2024, May 2, Shift 1]

• (1) 5 cm
• (2) 10 cm
• (3) 15 cm
• (4) 20 cm

13. Resistances in the left gap and right gap of a meter bridge are 10 Ω and 30 Ω respectively. If the resistances in the two gaps are interchanged, the balance point will shift to right by:

[MHT-CET 2024, May 3, Shift 2]

• (1) 30 cm
• (2) 40 cm
• (3) 50 cm
• (4) 60 cm

14.  When the two known resistance ‘R’ and ‘S’ are connected in the left and right gaps of a meter bridge respectively, the null point is found at a distance ‘l₁ ’ from the zero end of a meter bridge wire. An unknown resistance ‘X’ is now connected in parallel with ‘S’ and null point is found at a distance ‘l₂ ’ from zero end of meter bridge wire. The unknown resistance ‘X’ is:

[MHT-CET 2024, May 9, Shift 1]

• (1) \( \frac{S l_1 (100 - l_2)}{100 (l_2 - l_1)} \)
• (2) \( \frac{S l_2 (100 - l_1)}{100 (l_1 - l_2)} \)
• (3) \( \frac{100 (l_2 - l_1)}{S l_1 (100 - l_2)} \)
• (4) \( \frac{100 (l_2 - l_1)}{S l_2 (100 - l_1)} \)

15.  In a meter bridge experiment, the balance point is obtained if the gaps are closed by 2 Ω and 3 Ω. A shunt of x Ω is added to 3 Ω resistor to shift the null point by 22.5 cm. The value of ‘x’ is:

[MHT-CET 2024, May 4, Shift 1]

• (1) 1 Ω
• (2) 2 Ω
• (3) 3 Ω
• (4) 4 Ω

16.  A battery of 6 V is connected to the ends of uniform wire 3 m long and of resistance 100 Ω. The difference of potential between two points 50 cm apart on the wire is:

[MHT-CET 2024, May 11, Shift 2]

• (1) 1 V
• (2) 1.5 V
• (3) 2 V
• (4) 3 V

17.  A potentiometer wire of length 1 m is connected in series with 495 Ω resistance and 2 Vbattery. If 0.2 mV/cm is the potential gradient, then the resistance of the potentiometer wire is:

[MHT-CET 2024, May 4, Shift 1]

• (1) 8 Ω
• (2) 7 Ω
• (3) 6 Ω
• (4) 5 Ω

18. In potentiometer experiment, cells of e.m.f. E₁ and E₂ are connected in series (E₁ > E₂) the balancing length is 80 cm of the wire. If the polarity of E₂ is reversed, the balancing length becomes 20 cm. The ratio \( \frac{E_1}{E_2} \) is:

is:

[MHT-CET 2024, May 10, Shift 1]

• (1) 1 : 2
• (2) 2 : 3
• (3) 3 : 4
• (4) 5 : 3

19.  When cell of e.m.f. ‘E₁ ’ is connected to potentiometer wire, the balancing length is l₁ . Another cell of e.m.f. ‘E₂ ’ (E₁ > E₂ ) is connected so that two cells oppose each other, then the balancing length is l₂ . The ratio E₁ : E₂ is:

[MHT-CET 2024, May 2, Shift 2]

• (1) \( \frac{l_1}{l_1 + l_2} \)
• (2) \( \frac{l_1}{l_1 - l_2} \)
• (3) \( \frac{l_1 + l_2}{l_1} \)
• (4) \( \frac{l_1 + l_2}{l_1 - l_2} \)

20.  In the given circuit diagram, in the steady state the current through the battery and the charge on the capacitor respectively are:

[MHT-CET 2024, May 9, Shift 1]

 

• (1) 2 A and 3 μC
• (2) \( \frac{6}{11} \) A and \( \frac{12}{7} \) μC
• (3) 11 A and 3 μC
• (4) zero ampere and 3 μF

 

We will post the solution to this paper in 24 hours. The download link will be activated at that time. In the meantime, students are encouraged to attempt the questions on their own and use the comments section to discuss and collaborate with each other.

Download the solution here.