Bharti Bhawan Class 10 Physics Solution in Hindi

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Bharti Bhawan Class 10 Physics MCQs in Hindi for Students

A 10 V battery of negligible internal resistance is charged by a 200 V D.C. supply. If the resistance in the charging circuit is 38 22, what is the value of the charging current?

In a potentiometer of 10 wires, the balance point is obtained on the 7th wire. To shift the balance point on the 9th wire, we should :

In the electric circuit terminal potential difference is 1.5 volts and the potential difference across internal resistance is 0.5 volts. The emf of the cell in the circuit will be

The potential gradient along the potentiometer wire is 6 mV/cm. The balancing length for a cell of emf 1.5 volts will be :

The wire of the potentiometer has a resistance of 4Ω and a length of 1 m. It is connected to a cell of emf 2 V and internal resistance 1Ω. The current flowing through the potentiometer wire is :

A voltmeter has a resistance of 50 Ω is connected across a cell of emf 2 V and internal resistance of 10 Ω. The reading of the voltmeter is :

The e.m.f of a cell is 2 V and when it is connected to a wire of 10 Ω resistance the potential difference between the terminal is 1 V. The internal resistance of the cell is :

A potentiometer wire of 1000cm in length, is adjusted to give a null point at 500cm with a standard cell of e.m.f 1.018 V. The e.m.f of a cell that gives a null point at 600cm is :

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cells are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance 5 Ω the balance point is found at 40 cm of the wire for the same cell. The internal resistance of the cell is :

A cell sends a current of 1 A through a circuit of resistance 5 Ω. The emf of the cell of internal resistance 1 Ω will be :

Bharti Bhawan Class 10 Physics MCQs in Hindi for Students

A 10 V battery of negligible internal resistance is charged by a 200 V D.C. supply. If the resistance in the charging circuit is 38 22, what is the value of the charging current?

(A) 10 A

(B) 2.5 A

(D) 2 A

Option c - 5 A

In a potentiometer of 10 wires, the balance point is obtained on the 7th wire. To shift the balance point on the 9th wire, we should :

(A) decrease resistance in the main circuit

(B) decrease resistance in the cell circuit

(D) increase resistance in the cell circuit

Option c - increase resistance in the main circuit

In the electric circuit terminal potential difference is 1.5 volts and the potential difference across internal resistance is 0.5 volts. The emf of the cell in the circuit will be

(A) 1.5 volt

(B) 1 volt

(D) 2 volt

Option d - 2 volt

The potential gradient along the potentiometer wire is 6 mV/cm. The balancing length for a cell of emf 1.5 volts will be :

(A) 150 cm

(B) 200 cm

(D) 300 cm

Option c - 250 cm

The wire of the potentiometer has a resistance of 4Ω and a length of 1 m. It is connected to a cell of emf 2 V and internal resistance 1Ω. The current flowing through the potentiometer wire is :

(A) 0.1 A

(B) 0.2 A

(D) 0.8 A

Option c - 0.4 A

A voltmeter has a resistance of 50 Ω is connected across a cell of emf 2 V and internal resistance of 10 Ω. The reading of the voltmeter is :

(A) 0.1 V

(B) 1.414 V

(D) 1.66 V

Option d - 1.66 V

The e.m.f of a cell is 2 V and when it is connected to a wire of 10 Ω resistance the potential difference between the terminal is 1 V. The internal resistance of the cell is :

(Α) 5Ω

(Β) 10 Ω

(D) 20 Ω

Option b - 10 Ω

A potentiometer wire of 1000cm in length, is adjusted to give a null point at 500cm with a standard cell of e.m.f 1.018 V. The e.m.f of a cell that gives a null point at 600cm is :

(A) 1.22 V

(B) 2.22 V

(D) 4. 22 V

Option a - 1.22 V

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cells are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance 5 Ω the balance point is found at 40 cm of the wire for the same cell. The internal resistance of the cell is :

(A) 5.2 Ω

(Β) 4 Ω

(D) 1Ω

Option c - 1.5Ω

A cell sends a current of 1 A through a circuit of resistance 5 Ω. The emf of the cell of internal resistance 1 Ω will be :

(A) 5 volt

(B) 1 volt

(D) 4 volt

Option c - 6 volt

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Bharti Bhawan Class 10 Physics MCQs in Hindi for Students

A 10 V battery of negligible internal resistance is charged by a 200 V D.C. supply. If the resistance in the charging circuit is 38 22, what is the value of the charging current?

In a potentiometer of 10 wires, the balance point is obtained on the 7th wire. To shift the balance point on the 9th wire, we should :

In the electric circuit terminal potential difference is 1.5 volts and the potential difference across internal resistance is 0.5 volts. The emf of the cell in the circuit will be

The potential gradient along the potentiometer wire is 6 mV/cm. The balancing length for a cell of emf 1.5 volts will be :

The wire of the potentiometer has a resistance of 4Ω and a length of 1 m. It is connected to a cell of emf 2 V and internal resistance 1Ω. The current flowing through the potentiometer wire is :

A voltmeter has a resistance of 50 Ω is connected across a cell of emf 2 V and internal resistance of 10 Ω. The reading of the voltmeter is :

The e.m.f of a cell is 2 V and when it is connected to a wire of 10 Ω resistance the potential difference between the terminal is 1 V. The internal resistance of the cell is :

A potentiometer wire of 1000cm in length, is adjusted to give a null point at 500cm with a standard cell of e.m.f 1.018 V. The e.m.f of a cell that gives a null point at 600cm is :

A cell sends a current of 1 A through a circuit of resistance 5 Ω. The emf of the cell of internal resistance 1 Ω will be :

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