HSC Class 11 Physics Solutions. We covered all the HSC Class 11 Physics Solutions in this post for free so that you can practice well for the exam.
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HSC Class 11 Physics MCQs for Students
A battery of emf 1.4 V and internal resistance of 2 ohms is connected to a resistor of 100 ohms, through an ammeter. The resistance of the ammeter is 4/3 ohm. A voltmeter has also been connected to find the potential difference across the resistor. The ammeter reads 0.02 A. The resistance of the voltmeter is :
(Α) 400 Ω
(Β) 300 Ω
(C) 200 Ω
(D) 100 Ω
Option c – 200 Ω
A thick wire is stretched so that its length becomes three times. Assuming that there is no change in density, what is the ratio of the change in the resistance of the wire to the initial resistance of the wire?
(A) 2 : 1
(B) 4 : 1
(C) 3 : 1
(D) 1 : 4
Option c – 3 : 1
Kirchhoff’s second law is based on the law of conservation of
(A) Charge
(B) Energy
(C) Momentum
(D) Current
Option b – Energy
According to Kirchhoff’s law, the algebraic sum of the product of current and resistance as well as emf in a closed loop is :
(A) Zero
(B) Greater than zero
(C) Less than zero
(D) Depends upon emf in a closed loop
Option a – Zero
Kirchhoff’s junction law is equivalent to
(A) conservation of energy
(B) conservation of charge
(C) conservation of electric potential
(D) conservation of electric flux
Option b – conservation of charge
Two resistance wires joining in parallel have a resultant resistance of (6/5)Ω. One of the wires breaks. The effective resistance is 222. The resistance of the broken wire was :
(A) 6/5Ω
(Β) 3Ω
(C) 3/5Ω
(D) zero
Option b – 3Ω
A cell supplies a current of 0.9 A through a 2Ω resistor and a current of 0.3 A through a 7Ω resistor. The internal resistance of the cell is :
(A) 0.5Ω
(B) 1Ω
(C) 2Ω
(D) 4Ω
Option a – 0.5Ω
Ohm law is the relation between :
(A) I = VR
(B) I = V/R
(C) I = V-R
(D) I = V+R
Option b – I = V/R
A galvanometer of resistance 98 Ω is shunted by resistance 2 Ω. The fraction of current through the galvanometer is :
(A) 1/50
(B) 1/49
(C) 1/2
(D) 1/98
Option a – 1/50
In parallel arrangement if (R₁>R₂). The power dissipated in resistance R₁ will be :
(A) more than R₂
(B) less than R₂
(C) equal to R₂
(D) depending on the internal resistance of the cell
Option b – less than R₂
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